Alternating current

143

C

03 apter h ontents

Alternating Current (AC)

01.

Alternating current and Voltage

145

02.

Different Type of AC Circuits

151

03.

Inductance, Capacitance and Resistance in Series

160

04.

Power in AC Circuits

165

05.

LC Oscillation

170

06.

Exercise-I (Conceptual Questions)

173

07.

Exercise-II (Previous Years Questions)

182

08.

Exercise-III (Analytical Questions)

185

NEET SYLLABUS Alternating currents, peak and rms value of alternating current/ voltage; reactance and impedance; LC oscillations (qualitative treatment only), LCR series circuit, resonance; power in AC circuits, wattles current.

144

NICOLA TESLA (1836 –1943) Yugoslov scientist, inventor and genius. He conceived the idea of the rotating magnetic field, which is the basis of practically all alternating current machinery, and which helped usher in the age of electric power. He also invented among other things the induction motor, the polyphase system of ac power, and the high frequency induction coil (the Tesla coil) used in radio and television sets and other electronic equipment. The SI unit of magnetic field is named in his honour.

GEORGE WESTINGHOUSE (1846 – 1914) A leading proponent of the use of alternating current over direct current. Thus, he came into conflict with Thomas Alva Edison, an advocate of direct current. Westinghouse was convinced that the technology of alternating current was the key to the electrical future. He founded the famous Company named after him and enlisted the services of Nicola Tesla and other inventors in the development of alternating current motors and apparatus for the transmission of high tension current, pioneering in large scale lighting.

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ALTERNATING CURRENT 1.

145

ALTERNATING CURRENT AND VOLTAGE Voltage or current is said to be alternating if it changes continously in magnitude and perodically in direction. It can be represented by a sine curve or cosine curve I = I0 sin w t

I = I0 cos w t

or

where I = Instantaneous value of current at time t, w = Angular frequency w = I0

I0 = Amplitude or peak value

2p = 2pf T

T = time period

I

I0 T 2

3T 4

I T 2

T

T 4

f = frequency

t

T

3T 4

T 4

t

–I 0

–I 0 I as a sine function of t

I as a cosine function of t

1.1 Amplitude of AC The maximum value of current in either direction is called peak value or the amplitude of current. It is represented by I0. Peak to peak value = 2I0

1.2 Periodic Time The time taken by alternating current to complete one cycle of variation is called periodic time or time period of the current.

1.3 Frequency The number of cycle completed by an alternating current in one second is called the frequency of the current. UNIT : (cycle/s) or (Hz) In India : f = 50 Hz , supply voltage = 220 volt

In USA : f = 60 Hz ,supply voltage = 110 volt

1.4 Condition required for current/ voltage to be Alternating • Amplitude is constant.

• Alternate half cycle is positive and half negative.

• The alternating current continuously varies in magnitude and periodically reverses its direction. sinusoidal AC

I

triangular AC

I +

+ t

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–

E

I

square wave AC

saw tooth wave

I

t

I0

–

t

I0

I0

t

t

t mixture of AC and DC

Not AC (direction not change)

Not AC (not periodic)

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146 Pre-Medical : Physics 1.5 Average Value or Mean Value

The mean value of A.C over any half cycle (either positive or negative) is that value of DC which would send same amount of charge through a circuit as is sent by the AC through same circuit in the same time. T /2

average value of current for half cycle I =

ò

Idt

ò

dt

0 T/2

0

Average value of I = I0 sin wt over the positive half cycle :

Iav =

T 2 0 0

ò

I sin wt dt T 2 0

ò

dt

=

< sin q > = < sin 2q >=0 < cos q >= < cos 2q >= 0 < sin q cos q > = 0 2 2 < sin q > = < cos q >= 1 2

T 2 I0 2I é- cos wt ûù02 = 0 ë wT p

• For symmetric AC, average value over full cycle = 0, Average value of sinusoidal AC Full cycle

(+ve) half cycle

(–ve) half cycle

0

2I0 p

–2I0 p

As the average value of AC over a complete cycle is zero, it is always defined over a half cycle which must be either positive or negative

1.6 Maximum Value Þ

• I = a sinq

IMax. = a

• I = a sinq + b cos q ÞIMax. =

a 2 + b2

• I = a + b sinq Þ

IMax. = a + b ( if a and b > 0)

• I = a sin2 q

IMax. = a (a > 0)

Þ

1.7 Root Mean square (rms) Value It is value of DC which would produce same heat in given resistance in given time as is done by the alternating current when passed through the same resistance for the same time. T

Irms =

ò I dt ò dt 2

0

rms value = virtual value = apparent value

T

0

rms value of Irms =

I = I0 sin wt :

ò

T

0

(I 0 sin wt)2 dt

ò

T

0

I20 T 2 sin wt dt = I 0 T ò0

T

I 1 T é1 - cos 2wt ù 1 é t sin2wt ù dt == I 0 = 0 ò ê ú ê ú 0 T ë 2 T ë 2 2 ´ 2w û 0 2 û

If nothing is mentioned then values printed in a.c circuit on electrical appliances, any given or unknown values, reading of AC meters are assumed to be RMS.

Current

Average

Peak

I1 = I0 sin wt

0

I0

0

I0 2

I2 = I0sin wt cos wt =

I0 sin2wt 2

I3 = I0sinwt + I0coswt •

For above varieties of current rms =

Peak value 2

0

RMS

I0

2 I0

2 I0 2 2 I0

Angular fequency w 2w w

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•

dt

=

E

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147

1.8 Measurement of A.C. Alternating current and voltages are measured by a.c. ammeter and a.c. voltmeter respectively. Working of these instruments is based on heating effect of current, hence they are also called hot wire instruments. Terms

D.C. meter

A.C. meter

Name

moving coil instrument

hot wire instrument

Based on

magnetic effect of current

heating effect of current

Reads

average value

r.m.s. value

If used in

A.C. circuit then they reads zero

A.C. or D.C. then meter works

Q average value of A.C. = zero

properly as it measures rms value

deflection µ current

deflection µ heat

f µ I (linear)

f µ I2rms (non linear)

Scale

Uniform Seperation

Non uniform sepration

f = Number

I - 1 2 3 4 5

I - 1 2 3

of divisions

f - 1 2 3 4 5

f - 1 4 9 16 25

Deflection

4

5

1.9 Phase and phase difference (a) Phase I = I0 sin (wt + f) Initial phase = f

(it does not change with time)

Instantaneous phase = wt + f

(it changes with time)

• Phase decides both value and sign.

l

UNIT: radian

(b) Phase difference Voltage V = V0 sin ( wt + f1) Current I = I0 sin (wt +f2) • Phase difference of I w.r.t. V

f = f2 – f1

• Phase difference of V w.r.t. I

f = f1 – f2

1.10 Lagging and leading Concept

I=I0 sin (wt - f)

(a) V leads I or I lags V ® It means, V reach maximum before I Let if

V = V0 sin wt

then I = I0 sin (wt – f)

and if

V = V0 sin (wt+f )

then I = I0 sin wt

wt

V,I

V=V0sinwt I=I0 sin (wt + f)

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(b) V lags I or I leads V ® It means V reach maximum after I

E

Let if

V = V0 sin wt

then I = I0 sin (wt + f)

and if

V = V0 sin (wt – f )

then I = I0 sin wt

wt

V,I

V=V0sinwt

1.11 Phasor and Phasor diagram A diagram representing alternating current and voltage (of same frequency) as vectors (phasor) with the phase angle between them is called phasor diagram. Let

V = V0 sin wt

and

I = I0 sin (wt +f)

In figure (a) two arrows represents phasors. The length of phasors represents the maximum value of quantity. The projection of a phasor on y-axis represents the instantaneous value of quantity. In figure (b) two arrows represents phasor. Their length represents maximum value.

Y V I

Y I0

I0

V0

f wt fig (a)

X

f fig (b)

V0

X

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148

1.12 Advantages of AC •

A.C. is cheaper than D.C

•

It can be easily converted into D.C. (by rectifier)

•

It can be controlled easily (choke coil)

•

It can be transmitted over long distance at low power loss.

•

It can be stepped up or stepped down with the help of transformer.

GOLDEN KEY POINTS •

AC can't be used in (a) Charging of battery or capacitor (as its average value = 0) (b) Electrolysis and electroplating (Due to large inertia, ions can not follow frequency of A.C) Minimum, at that instant when they are near their peak values

•

The rate of change of A.C.

•

For alternating current I0 > Irms > Iav.

•

Average value over half cycle is zero if one quarter is positive

Maximum, at that instant when they change their direction. + –

and the other quarter is negative. •

Average value of symmetrical AC for a cycle is zero thats why average potential difference on any element in A.C circuit is zero.

•

The instrument based on heating effect of current are works on both A.C and D.C supply and also provides same heating for same value of A.C (rms) and D.C. that's why a bulb bright equally in D.C. and A.C. of same value.

•

If the frequency of AC is f then it becomes zero, 2f times in one second and the direction of current changes 2f times in one second. Also it become maximum 2f times in one second.

•

Some Important wave forms and their RMS and Average Value Wave–form

Sinusoidal

0

Half wave rectifier

Full wave rectifier

+

RMS Value

I0 p

0

p

0

p

2p

–

2

= 0.707 I0

Average or mean Value 2I0 = 0.637 I0 (Half) p

2p

I0 = 0.5 I0 2

I0 = 0.318 I0 (Full) p

2p

I0 = 0.707 I 0 2

2I0 =0.637 I 0 p

(Half and Full) +

Square or Rectangular

Saw Tooth wave

–

0

p 2p

I0

I0 (Half)

I0

I0 (Half) 2

3

•

D.C meter in AC circuit reads zero because < AC > = 0 ( for complete cycle)

•

AC meter works in both AC and DC

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Nature of wave form

E

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149

Illustrations Illustration 1. If I = 2

t ampere then calculate average and rms values over t = 2 to 4 s

Solution 4

I =

ò2

t.dt

2

4

ò dt

3

4 (t 2 )24 2 é = = 8 -2 2ù û 3 (t)24 3ë

and Irms =

ò

4

2

(2 t )2 dt

ò

4

2

dt

=

ò

4

2

4t dt 2

4

é t2 ù = 2ê ú = 2 3 A ë 2 û2

2

Illustration 2. If E = 20 sin (100p t) volt then calculate value of E at t =

1 s 600

Solution 1 ù é épù 1 1 s E = 20 Sin ê100p ´ = 20 sin ê ú = 20 × = 10V ú 600 û 600 2 ë ë6 û

At t =

Illustration 3. A periodic voltage wave form has been shown in figure. Determine. (a) Frequency of the wave form. (b) Average value.

Solution (a) (b)

After 100 ms wave is repeated so time period is T = 100 ms. Þ

f =

1 = 10 Hz T

(1 / 2) ´ 100 ´ 10 = 5 volt (100)

Average value = Area/time period =

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Illustration 4. If a direct current of value a ampere is superimposed on an alternating current I = b sinwt flowing through a wire, what is the effective value of the resulting current in the circuit ?

E

a

I

DC t

+I

b

AC t

=?

Solution As current at any instant in the circuit will be I = IDC + IAC = a + b sinwt T

T

1 (a + b sin wt) 2 dt = T ò0

T

1 (a2 + 2ab sin wt + b2 sin2 wt)dt T ò0

\ Ieff =

1 2 I dt = T ò0

but as

T 1 1 1 sin2 wtdt = sin wtdt = 0 and ò ò T0 2 T0

T

\ Ieff =

a2 +

1 2 b 2

150

ALLEN

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Illustration 5. pù é The Equation of current in AC circuit is I = 4sin ê100p t + ú A. Calculate. 3û ë

(i) RMS Value

(ii) Peak Value

(iii) Frequency

(iv) Initial phase

(v) Current at t = 0

Solution

I0

4

(i)

Irms =

(ii)

Peak value I0 = 4A

(iii)

Q

(iv)

Initial phase =

(v)

pù é At t = 0, I = 4sin ê100p ´ 0 + ú = 4 × 3û ë

2

=

2

= 2 2A

w = 100 p rad/s

\

frequency f =

100p = 50 Hz 2p

p 3

3 = A 2 2 3

Illustration 6. If I = I0 sin wt,

pù é E = E0 cos êwt + ú . Calculate phase difference between E and I 3 ë û

Solution I = I0 sin wt and

pù ép E = E0sin ê + wt + ú 2 3 ë û

\

phase difference =

p p 5p + = 2 3 6

Illustration 7. If E = 500 sin (100 pt) volt then calculate time taken to reach from zero to maximum. Solution 2p 1 T 1 = s, time taken to reach from zero to maximum = = s 100p 50 4 200

Illustration 8.

If Phase Difference between E and I is

p and f = 50 Hz then calculate time difference. 4

Solution Phase difference time difference = 2p T

Q

2p º T\

Þ

Time difference =

T p T 1 × = = = 2.5ms 2p 4 8 50 ´ 8

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Qw = 100 p ÞT =

E

ALLEN

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151

BEGINNER'S BOX-1 1.

Explain why A.C. is more dangerous than D.C. ?

2.

Show that average heat produced during a cycle of AC is same as produced by DC with I = Irms.

3.

An ordinary moving coil ammeter used for d.c., cannot be used to measure a.c. even if its frequency is low why ?

4.

Find the time required for a 50Hz alternating current to change its value from zero to rms value.

5.

The current and voltage in a circuit is given by i = 3.5 sin (628t + 30°) A , V = 28 sin (628t–30°) volt. Find (a)

time period of current

(b)

phase difference between voltage and current.

6.

The figure given below shows the variation of an alternating emf with time. What is the average value of the emf for the shaded part of the graph?

2.

DIFFERENT TYPES OF AC CIRCUITS In order to study the behaviour of A.C. circuits we classify them into two categories : (a) Simple circuits containing only one basic element i.e. resistor (R) or inductor (L) or capacitor (C) only. (b) Complicated circuit containing any two of the three circuit elements R, L and C or all of the three elements.

2.1

AC circuit containing pure resistance

R

Let at any instant t, the current in the circuit = I.

V = IR

Potential difference across the resistance = I R

s

with the help of kirchoff’s circuital law E – I R = 0 ÞE0 sin wt = I R Þ I=

E = E 0sinwt

E0 E sin wt = I0 sin wt ( I0 = 0 = peak or maximum value of current) R R

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Alternating current developed in a pure resistance is also of the sinusoidal

E

nature. In a.c. circuits containing pure resistance, the voltage and current are in the same phase. The vector or phasor diagram which represents the phase relationship between alternating current and alternating e.m.f. are as shown in figure.

E and I

O

E = E0 sintw I=I0 sinwt

2p

In the a.c. circuit having R only, as current and voltage are in the same phase, hence in fig. both phasors E0 and I0 are in the same direction, making an angle wt with OX. Their projections on Y-axis represent the instantaneous values of alternating current and voltage. i.e.

Since

I = I0 sinwt and

E = E0 sinwt.

I0 E E = 0 I0 = 0 , hence R 2 R 2

Þ

I rms

E = rms R

wt

3p/2

p p/2

X

Y E

wt

I

I0

O

wt

P E0

X

152

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2.2 AC circuit containing pure inductance

L

A circuit containing a pure inductance L (having zero ohmic resistance) connected with a source of alternating emf. Let the alternating e.m.f. E = E0 sin wt

s

dI E = E 0sinw t dt Note : Negative sign indicates that induced emf acts in opposite direction to that of applied emf.

When a.c. flows through the circuit, emf induced across inductance = -L

Because there is no other circuit element present in the circuit other than inductance so with the help of dI ö E æ dI p Kirchoff’s circuital law E + ç -L ÷ = 0 Þ E = L so we get I = 0 sin æç wt - ö÷ è è dt ø wL 2ø dt E0 E pö æ ´ 1 = 0 , Hence, I = I 0 sin ç wt - ÷ è wL wL 2ø

In a pure inductive circuit current always lags behind the emf by or alternating emf leads the a. c. by a phase angle of

p 2

p . 2

E = E0 sintw

E and I

Maximum current I0 =

p/2

O

3p/2

p

wt 2p

I = I0 sin (wt -p/2)

E E Expression I0 = 0 resembles the expression = R. I wL This non-resistive opposition to the flow of A.C. in a circuit is called the inductive reactance (XL) of the circiut.

E

Y

P E 0

XL = wL = 2 p f L where f = frequency of A.C.

wt

O

Unit of XL : ohm (wL) = Unit of L × Unit of (w =2pf) = henry × sec–1

X

p/2 – w t

I

I0 Q

volt volt = ´ sec -1 = = ohm ampere / sec ampere Inductive reactance XL µ f Higher the frequency of A.C. higher is the inductive reactance offered by an inductor in an A.C. circuit.

XL

For d.c. circuit, f = 0 \ XL = wL = 2 p f L = 0

f

Hence, inductor offers no opposition to the flow of d.c. where as a resistive path to a.c.

2.3 AC circuit containing pure capacitance

E and I O

s

The two plates of the capacitor become alternately positively and negatively charged and the magnitude of the charge on the plates of the capacitor varies sinusoidally with time. Also the electric field between the plates of the capacitor varies sinusoidally with time.Let at any instant t charge on the capacitor = q

E = E0 sinwt

E = E0sintwt p/2

p

Instantaneous potential difference across the capacitor E = q/C Þ q = C E Þ q = CE0 sin wt

I= I0sinw (t + p /2)

3p /2

2p

wt

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C

A circuit containing an ideal capacitor of capacitance C connected with a source of alternating emf as shown in fig. The alternating e.m.f. in the circuit E = E0 sin wt. When alternating e.m.f. is applied across the capacitor a similarly varying alternating current flows in the circuit.

E

ALLEN

Pre-Medical : Physics

The instantaneous value of current I =

Y

P

E

E0 pö æ sin ç wt + ÷ = I0 sin æ wt + p ö where I0 = wCE0 ç è ø 1/ w C 2 ( ) 2 ÷ø è

Q I0

E0

I 90 °

ÞI =

dq d = ( CE 0 sin wt ) = CE 0 w cos wt dt dt

153

In a pure capacitive circuit, the current always leads the e.m.f. by a phase

wt

X

O

angle of p/2. The alternating emf lags behinds the alternating current by a phase angle of p/2. IMPORTANT POINTS

f

E/I is the resistance R when both E and I are in phase, in present case they differ in phase by

p 1 , hence is not the resistance of the capacitor, the capacitor 2 wC

XC

offer opposition to the flow of A.C. This non-resistive opposition to the flow of A.C. in a pure capacitive circuit is known as capacitive reactance X C . XC =

1 1 = wC 2pfC

Unit of XC : ohm Capacitive reactance XC is inversely proportional to frequency of A.C. XC decreases as the frequency increases. \ XC =

For d.c. circuit f = 0

1 = ¥ but has a very small value for a.c. 2pfC

This shows that capacitor blocks the flow of d.c. but provides an easy path for a.c. INDIVIDUAL COMPONENTS (R or L or C) TERM

R

L

C

L

R

C

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Circuit

E

Supply Voltage

V = V0sin wt

V = V0 sin wt

Current

I = I0 sin wt

I = I0 sin (wt –

Peak Current

I0 =

Impedance (W )

V0 =R I0

V0 = wL = X L I0

V0 1 = = XC I0 wC

R = Resistance

XL=Inductive reactance.

XC=Capacitive reactance.

Z=

V0 Vrms = I0 Irms

V0 R

I0 =

V = V0 sin wt p ) 2

V0 wL

I = I0 sin (wt+

I0 =

p ) 2

V0 =V0wC 1 wC

154

ALLEN

Pre-Medical : Physics Phase difference

+

zero (in same phase)

p (V leads I) 2

-

p (V lags I) 2

V

I

I

Phasor diagram

V

I

R

XL

f XL µ f

Variation of Z with f

XC µ 1 f

f

f

G,SL,SC

V

G=1/R = conductance.

(mho, seiman)

XC

SL = 1/XL

SC = 1/XC

Inductive susceptance

Capacitive susceptance

Behaviour of device

Same in

L passes DC easily

C - blocks DC

in D.C. and A.C

A C and D C

(because XL = 0) while gives a high impedance

(because XC =¥) while provides an easy path

for the A.C. of high

for the A.C. of high

frequency (XLµ f)

frequency êX C µ f ú

VL = IXL

VC = IXC

Ohm's law

VR = IR

é

1ù

ë

û

2.4 Resistance and inductance in series (R-L circuit) A circuit containing a series combination of a resistance R and an inductance L, connected with a source of alternating e.m.f. E as shown in figure. L

R

s

•

phasor diagram For L-R circuit Let in a L-R series circuit, applied alternating emf is E = E0 sinwt. As R and L are joined in series, hence current flowing through both will be same at each instant. Let I be the current in the circuit at any instant and VL and VR the potential differences across L and R respectively at that instant. Then VL = IXL

and

VR = IR

Now, VR is in phase with the current while VL leads the current by

p . 2

Q

VL

Y

R

E

I VR

P

X

So VR and VL are mutually perpendicular (Note : E ¹ VR + VL) The vector OP represents VR (which is in phase with I), while OQ represents VL (which leads I by 90°). The resultant of VR and VL = the magnitude of vector OR E =

VR2 + VL2

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E = E0 sinwt

E

ALLEN Thus

Pre-Medical : Physics E² = VR² + VL² = I² (R² + XL² ) Þ I =

E

XL

2

R + X L2 The phasor diagram shown in fig. also shows that in L-R circuit the applied

ZL

emf E leads the current I or conversely the current I lags behind the e.m.f. E by a phase angle f tan f =

•

VL IX L X L wL = = = IR VR R R

155

f

R

æ wL ö Þ f = tan -1 ç è R ÷ø

Inductive Impedance ZL : In L-R circuit the maximum value of current I0 =

E0 2

2 2

R +w L

Here

R 2 + w 2 L2 represents the effective

opposition offered by L-R circuit to the flow of a.c. through it. It is known as impedance of L-R circuit and is represented by ZL. ZL = R2 + w 2 L2

= R 2 + (2pfL)2

The reciprocal of impedance is called admittance YL =

1 = ZL

1 2

R + w 2 L2 C

2.5 Resistance and capacitor in series (R-C circuit)

R

A circuit containing a series combination of a resistance R and a capacitor C, connected with a source of e.m.f. of peak value E0 as shown in fig. phasor diagram For R-C circuit

s

•

E = E0 sinwt

Current through both the resistance and capacitor will be same at every instant and the instantaneous potential differences across C and R are VC = I XC and VR = I R where XC = capacitive reactance and I = instantaneous current. Now, VR is in phase with I, while VC lags behind I by 90°. The phasor diagram is shown in fig.

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and the vector OQ represents VC (which lags behind I by

VC

E

=

V

2

+V

R

C

p ). 2

X

f E

The vector OP represents VR (which is in phase with I)

P

VR

O

2

Q E (applied emf)

S

The vector OS represents the resultant of VR and VC = the applied e.m.f. E. Hence

The term

VR2 + VC2 = E 2 Þ E = VR2 + VC2 Þ E² = I² (R² + XC² ) Þ I =

(R 2 + X 2C ) represents the effective resistance of the R-C

E 2

R + X 2C

O

Z=

circuit and called the capacitive impedance ZC of the circuit. Hence, in C-R circuit

æ 1 ö ZC = R + X = R + ç è wC ÷ø 2

2 C

2

2

P

f

R R2 +X

C

Q

XC

2

S

X

156 •

ALLEN

Pre-Medical : Physics Capacitive Impedance ZC : In R-C circuit the term

R 2 + X 2C effective opposition offered by R-C circuit to the flow of a.c. through it. It is known as impedance of R-C circuit and is represented by ZC.The phasor diagram also shows that in R-C circuit the applied e.m.f. lags behind the current I (or the current I leads the emf E) by a phase angle f given by tan f =

VC X C 1/ wC 1 X 1 1 ö -1 æ = = = tan f = C = Þ f = tan ç wCR , VR R R è wCR ÷ø wCR R

2.6 Combination of components (R-L or R-C or L-C) TERM

R-L

R-C

L

R

R

L-C

L

C

C

Circuit

I is same in R & L

I is same in R & C

VL

VR

Phasor diagram

VL

I

V

I VR

V2 = VR2 + VL2

I is same in L & C

I

VC

VC

V2 = VR2 + VC2

V = VL – VC (VL>VC) V = VC – VL (VC>VL)

V leads I (f = 0 to

p ) 2

V lags I (f = –

p to 0 ) 2

V lags I (f = -

V leads I (f = +

in between V & I

Z=

R2 + ( X C )

2

p ,if XL>XC) 2

Z = X L - Xc

Impedance

Z=

Variation of Z

as f ­, Z ­

as f ­, Z ¯

as f ­, Z first ¯ then ­

Z

Z

Z

R

R

with f

R 2 + X 2L

f

At very low f At very high f

Z ; R (XL ® 0) Z ; XL

f

Z ; XC

Z ; R (XC ® 0)

f

Z ; XC

Z ; XL

GOLDEN KEY POINTS • • •

p ,if XC>XL) 2

Phase diference between capacitive and inductive reactance is p Inductor is called Low pass filter because it allows low frequency signal to pass. Capacitor is called high pass filter because it allows high frequency signal to pass.

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Phase difference

E

ALLEN

Pre-Medical : Physics

157

Illustration 9. What is the inductive reactance of a coil if the current through it is 20 mA and voltage across it is 100 V. Solution Q VL = IXL

\ XL =

VL 100 = = 5 kW I 20 ´ 10-3

Illustration 10. The reactance of capacitor is 20 ohm. What does it mean? What will be its reactance if frequency of AC is doubled? What will be its, reactance when connected in DC circuit? What is its consequence? Solution The reactance of capacitor is 20 ohm. It means that the hinderance offered by it to the flow of AC at a specific frequency is equivalent to a resistance of 20 ohm. The reactance of capacitance X C =

1 1 = wC 2pfC

Therefore by doubling frequency, the reactance is halved i.e., it becomes 10 ohm. In DC circuit f = 0. Therefore reactance of capacitor = ¥ (infinite). Hence the capacitor can not be used to control DC. Illustration 11. A capacitor of 50 pF is connected to an a.c. source of frequency 1kHz. Calculate its reactance. Solution XC =

1 1 107 = = W 3 -12 2p ´ 10 ´ 50 ´ 10 wC p

Illustration 12. In given circuit applied voltage V = 50 2 sin (100pt) volt and

L

ammeter reading is 2A then calculate value of L Solution

A

Q Reading of ammeter = Irms

Vrms = Irms XL XL =

Vrms = Irms

X 25 50 2 = = 25 WÞ L = L = w 100p 2 ´2

V0 2 I rms

V

=

1 H 4p

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Illustration 13.

E

Calculate the impedance of the circuit shown in the figure.

30W

40W

Solution Z=

R 2 + (X c )2 =

(30)2 + (40)2 = 2500 = 50 W

Illustration 14. If XL = 50 W and XC = 40 W. Calculate effective value of current in given circuit. Solution XL=50W

XC=40W

Z = XL – XC = 10 W I0 =

V0 40 = =4A Z 10

Þ

Irms =

4 2

= 2 2 A

V=40sin 100p volt

158

ALLEN

Pre-Medical : Physics

Illustration 15. VR=60

In given circuit calculate, voltage across inductor

VL=?

Solution Q V 2 = VR2 + VL2 VL =

V 2 - VR2 =

\ VL2 = V2 – VR2 (100)2 - (60)2 =

V=100 2 sinw t volt

6400 = 80 V

Illustration 16. In given circuit find out

(i) impedance of circuit

8W

6W

(ii) current in circuit

Solution (i)

Z=

R 2 + X 2C =

V = IZ Þ I =

(ii)

(6)2 + (8)2 = 10 W V=20sinwt volt

V0 20 2 = = 2A , so Irms = = Z 10 2

2A

Illustration 17. When 10V, DC is applied across a coil current through it is 2.5 A, if 10V, 50 Hz A.C. is applied current reduces to 2 A. Calculate reactance of the coil. Solution For 10 V D.C. QV = IR \ Resistance of coil R =

10 = 4W 2.5

For 10 V A.C. : V = I Z Þ Z =

V 20 = = 5W I 10

Q Z = R 2 + X 2L = 5 Þ R2 + X 2L = 25 Þ X 2L = 52 – 42Þ XL= 3 W Illustration 18. When an alternating voltage of 220V is applied across a device X, a current of 0.5 A flows through the circuit and is in phase with the applied voltage. When the same voltage is applied across another device (a)

Name the devices X and Y.

(b)

Calculate the current flowing in the circuit when same voltage is applied across the series combination of X and Y.

Solution (a)

X is resistor and Y is a capacitor

(b)

Since the current in the two devices is the same (0.5A at 220 volt) When R and C are in series across the same voltage then R = XC =

220 = 440 W Þ Irms = 0.5

Vrms 2

R +

2 XC

=

220 2

(440) + (440)

2

=

220 440 2

= 0.35A

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Y, the same current again flows through the circuit but it leads the applied voltage by p/2 radians.

E

ALLEN

Pre-Medical : Physics

159

BEGINNER'S BOX-2 1.

A voltage V = 60 sin pt volt is applied across a 20 W resistor. What will an ac ammeter in series with the resistor read ?

2.

An alternating current source E = 100 sin (1000t) volt is connected through a inductor of 10 mH then write down the equation of current.

3.

An alternating voltage E = 200 2 sin (100t) volt is applied to 2H inductor through an a.c. ammeter. What will be reading of the ammeter ?

4.

A 15.0 mF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current?

5.

A 60 mF capacitor is connected to a 110 V. 60 Hz a.c. supply. Determine the rms value of the current in the circuit.

6.

The given graphs (a) and (b) represent the variation of the opposition offered by the circuit element to the flow of alternating current, with frequency of the applied emf. Identify the circuit element corresponding to each graph.

(a)

(b)

7.

When a series combination of inductance and resistance are connected with a 10V, 50 Hz a.c. source, a current p of 1A flows in the circuit. The voltage leads the current by a phase angle of radian. Calculate the values 3 of resistance and inductive reactance.

8.

The current in the shown circuit is found to be 4 sin 314t -

FG H

IJ K

p A. 4

Find the value of inductance. 9.

A current of 4A flows in a coil when connected to a 12V dc source. If the same coil is connected to a 12V,

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50 rad/s a.c. source, a current of 2.4A flows in the circuit. Determine the inductance of the coil.

E

10.

An alternating voltage E=200 sin (300t) volt is applied across a series combination of R=10W and inductance of 800mH. Calculate the impedance of the circuit.

11.

12.

A coil of reactance 100 W and resistance 100 W is connected to a 240 V, 50 Hz a.c. supply. (a)

What is the maximum current in the coil ?

(b)

What is the time lag between the voltage maximum and the current maximum ?

An inductance has a resistance of 100 W. When a.c. signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by 45°. Calculate the self inductance of the coil.

160 13.

ALLEN

Pre-Medical : Physics Match the following options – Circuit component across an ac source (w = 200 rad/sec) (A)

10W

(B)

Phase difference between current and source voltage

5 0 0µ F

5H

(C)

(p)

p 2

(q)

p 6

(r)

p 4

(s)

p 3

(t)

None of the above

5 00µ F

(D) (E)

3.

4H

3 µF

1kW

5H

INDUCTANCE, CAPACITANCE AND RESISTANCE IN SERIES

3.1 L-C-R series circuit

L

C

R

A circuit containing a series combination of an resistance R, a coil of inductance L and a capacitor of capacitance C, connected with a source of alternating e.m.f. of peak value of E0, as shown in figure.

A.C. source

Phasor Diagram For Series L-C-R circuit

E = E0 sinwt

Let in series LCR circuit applied alternating emf is E = E0 sin wt.As L,C and R are joined in series, therefore, current at any instant through the three elements has the same amplitude and phase. However voltage across each element bears a different phase relationship with the current. Y VL Q

Let at any instant of time t the current in the circuit is I. Let at this time t the potential differences across L, C, and R VL = I XL, VC = I XC and VR = I R Now, VR is in phase with current I but VL leads I by 90° While VC lags behind I by 90°.

VR X P

I

O VC

The vector OP represents VR (which is in phase with I) the vector OQ represent VL (which leads I by 90°) Y VL Q T

VL and VC are opposite to each other.

Thus E = Impedance

VR2 + (VL - VC )2 = I R 2 + (X L - X C )2 Z = R 2 + (X L - X C )2 =

Þ I=

1 ö æ R 2 + ç wL è wC ÷ø

2

R + (X L - X C )

2

leads the current I by a phase angle f

E

(a

lie pp

d

f)

f

O

VR

P

X

VC

E

The phasor diagram also shown that in LCR circuit the applied e.m.f. XL - XC tanf = R

(VL-VC)

If VL > VC (as shown in figure) the their resultant will be (V L – VC) which is represented by OT. Finally, the vector OK represents the resultant of VR and (VL – VC), that is, the resultant of all the three = applied e.m.f.

K em

2

Y XL Q T

K

O XC

Z f

R P

X

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and the vector OS represents VC (which legs behind I by 90°)

(XL-XC)

•

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ALLEN

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161

3.2 Series LCR and parallel LCR combination SERIES L-C-R CIRCUIT 1.

PARALLEL L-C-R CIRCUIT

Circuit diagram R

R

L

C

L C

I same for R, L & C 2.

V same for R, L and C

Phasor diagram VL

IC I

VC

(i)

V

VR

IR

IL

If VL > V C then VL–V C

if IC > IL then I C -I L

(i)

I

V

VR

(ii)

If VC > VL then VC–VL

(iii)

V=

I

VR

(ii)

IR

if IL > IC then IL-IC

V

VR2 + (VL - VC )2

Impedance Z = tanf = (iv)

IR

R 2 + (X L - X C )2

XL - XC V - VC = L R VR

Impedance triangle Z f

X=XL–XC

R

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3.3 Resonance

E

A circuit is said to be resonant when the natural frequency of circuit is equal to frequency of the applied voltage. For resonance both L and C must be present in circuit. There are two types of resonance : (i) Series Resonance

(ii) Parallel Resonance

3.4 Series Resonance (a) At Resonance VL = VC

(iii)

f = 0 (V and I in same phase)

(impedance minimum)

(v)

Imax =

(i) XL = XC (iv) Zmin = R

(ii)

V (current maximum) R

(b) Resonance frequency Q XL = XC

Þ

1 wrL = w C r

2 Þ wr =

1 LC

Þ

wr =

1 LC

Þ fr =

1 2p LC

162

ALLEN

Pre-Medical : Physics (c) Variation of Z with f (i) If f< fr

circuit nature capacitive, f (negative)

then XL < XC

(ii) At f = fr then XL = XC

circuit nature, Resistive, f = zero

(iii) If f > fr then XL > XC

circuit nature is inductive,f (positive)

Z R

(d) Variation of I with f as f increase, Z first decreases then increase

f

fr

I max= V R

I max I max 2 I

as f increase, I first increase then decreases

Df f1

fr

f2

f

•

At resonance, impedance of the series resonant circuit is minimum so it is called 'acceptor circuit' as it most readily accepts that current out of many currents whose frequency is equal to its natural frequency. In radio or TV tuning we receive the desired station by making the frequency of the circuit equal to that of the desired station.

•

Half power frequencies The frequencies at which, power become half of its maximum value is called half power frequencies

•

Band width = Df = f2 – f1

•

Quality factor Q : Q-factor of AC circuit basically gives an idea about stored energy & lost energy.

Q = 2p

maximum energy stored per cycle maximum energy loss per cycle

(i) It represents the sharpness of resonance. (ii) It is unit less and dimension less quantity (iii) Q =

(X L )r

R Magnification

=

(X C )r 2pfr L 1 = = R R R

fr fr L = = C Df band width

At resonance

VL or VC = QE (where E = supplied voltage)

So at resonance

Magnification factor = Q-factor

I

R1

R 1 < R2 < R 3 R2

Sharpness Sharpness µ Quality factor µ Magnification factor R decrease Þ Q increases

Þ Sharpness increases

R3 fr

f

•

In A.C. circuit voltage for L or C may be greater than source voltage or current but it happens only when circuit contains L and C both and in R it is never greater than source voltage or current.

•

In parallel A.C.circuit phase difference between IL and IC is p

•

Series resonance circuit gives voltage amplification while parallel resonance circuit gives current amplification.

•

At resonance current does not depend on L and C, it depends only on R and V.

•

At half power frequencies :

•

As R increases, bandwidth increases

•

To obtain resonance in a circuit following parameter can be altered : (i) L

(ii) C

net reactance = net resistance.

(iii) frequency of source.

•

Two series LCR circuit of same resonance frequency f are joined in series then resonance frequency of series combination is also f

•

The series resonance circuit called acceptor whereas parallel resonance circuit called rejector circuit.

•

Unit of

LC is second

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GOLDEN KEY POINTS

E

ALLEN

Pre-Medical : Physics

Illustrations

9W

4W

Illustration 19.

163 6W

Find out the impedance of given circuit. Solution Z = R 2 + (X L - X C )2 = 42 + (9 - 6)2 =

42 + 32 = 25 = 5W

(Q XL > XC \ Inductive)

Illustration 20. Find out reading of A C ammeter and also calculate the potential difference across, resistance and capacitor. Solution

V0 100 10 = = A Z 10 2 2 10 ammeter reads RMS value, so its reading = = 5A 2 2 VR = 5 × 10 = 50 V and VC = 5 × 10 = 50 V Þ

Z = R 2 + (X L - X C )2 = 10 2 W Q so

10W

20W

10W

I0 =

A

E=100sin wt100pt volt

Illustration 21.

In LCR circuit with an AC source R = 300 W , C = 20 mF, L = 1.0 H, Erms = 50V and f = 50/p Hz. Find RMS current in the circuit. Solution Irms =

E rms = Z

Þ

Irms =

E rms 1 ù é R 2 + êwL wC úû ë

2

50

=

2

50 1 é ù 3002 + ê2p ´ ´1ú 50 p 20 ´ 10-6 ´ 2p ´ ê ú p û ë

50 é 103 ù (300)2 + ê100 2 úû ë

2

=

50 100 9 + 16

=

1 = 0.1A 10

Illustration 22. For what frequency the voltage across the resistance R will be maximum. Solution It happens at resonance

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f=

E

1 2p LC

=

1 1 1 ´ 10-6 ´ 2p p p

1 mF p

R

1H p

= 500 Hz

Illustration 23.

R=220W

A capacitor, a resistor and a 40 mH inductor are connected in series to an AC source of frequency 60Hz, calculate the capacitance of the capacitor, if the current is in phase with the voltage. Also find the reading of voltmeter V3 and Ammeter. Solution

1 1 1 1 , C= 2 = = = 176mF 2 2 -3 2 2 wC w L 4 p f L 4p ´ (60) ´ 40 ´ 10 V 110 = = 0.5 A V3 = VR Þ V3 = 110 V and I = R 220

At resonance wL =

V1

V2

300V

300V

A 110V, 60Hz

V3

164

ALLEN

Pre-Medical : Physics

Illustration 24. A coil, a capacitor and an A.C. source of rms voltage 24 V are connected in series, By varying the frequency of the source, a maximum rms current 6 A is observed, If this coil is connected to a bettery of emf 12 V, and internal resistance 4W , then calculate the current through the coil. Solution At resonance current is maximum. I =

V R

Þ Resistance of coil R =

V 24 = =4 W I 6

When coil is connected to battery, suppose I current flow through it then I =

12 E = = 1.5 A 4+4 R+r

BEGINNER'S BOX-3 1.

In given circuit find the reading of ammeter and voltmeter.

2.

For the circuit shown in figure, write down the instantaneous current through each element.

3.

A variable frequency 230V alternating voltage source is connected across a series combination of L = 5.0H C = 80mF and R = 40W. Calculate (a) The angular frequency of the source at resonance.

4.

(b) Amplitude of current at resonance frequency

A coil of resistance R and inductance L is connected in series with a capacitor C and complete combination is connected to a.c. voltage, Circuit resonates when angular frequency of supply is w = w0.

5.

Find the phase difference between voltage and current in series LCR circuit at half power frequencies.

6.

A series LCR circuit with L = 0.12H, C = 480 nF, R = 23W is connected to a 230 V variable frequency supply Find – (a) Source frequency for which current is maximum.

(b) Q–factor of the given circuit.

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Find out relation betwen w0, L and C

E

ALLEN 4.

Pre-Medical : Physics

165

POWER IN AC CIRCUIT

4.1 The average power dissipation in LCR AC circuit Let

V = V0 sinwt

I = I0 sin (wt – f)

and

Instantaneous power P = (V0 sinwt)(I0 sin(wt – f) = V0I0 sinwt (sinwtcosf – sinfcoswt) T

1 (V0 I0 sin2 wt cos f - V0 I 0 sin wt cos wt sin f)dt Average power

= T ò0 T é T ù = V0 I0 ê 1 sin2 wt cos fdt - 1 sin wt cos wt sin fdt ú = V0 I 0 é 1 cos f - 0 ´ sin f ù ò ò ê2 ú T0 ë û ëT 0 û

Þ

=

V0 I0 cos f 2

= Vrms Irm,s cosf

Instantaneous

Average power/actual power/

Virtual power/ apparent

power

dissipated power/power loss

Power/rms Power

P = VI

P = Vrms Irms cos f

P = Vrms Irms

Peak power

P = V0 I0

l

Irms cosf is known as active part of current or wattfull current, workfull current. It is in phase with voltage.

l

Irms sinf is known as inactive part of current, wattless current, workless current. It is in quadrature (90°) with voltage.

4.2 Power factor : Average power P = E rms Irms cos f = r m s power ´ cos f

Power factor (cos f) =

Average power and r m s Power

cosf =

R Z

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4.3 Choke Coil

E

In a direct current circuit, current is reduced with the help of a resistance. Hence there is a loss of electrical energy I2 R per sec in the form of heat in the resistance. But in an AC circuit the current can be reduced by choke coil which involves very small amount of loss of energy. Choke coil is a copper coil wound over a soft iron laminated core. This coil

tube light rod

starter

is put in series with the circuit in which current is to be reduced.

choke coil

Circuit with a choke coil is a series L-R circuit. If resistance of choke coil = r (very small) The current in the circuit I =

E with Z

Z = (R + r)2 + (wL)2 So due to large inductance L of the coil, the

current in the circuit is decreased appreciably. However, due to small resistance of the coil r, Pav = Vrms Irms cos f ® 0

The power loss in the choke Q

cos f =

r = Z

r 2

2

2

r +w L

»

r ®0 wL

166

ALLEN

Pre-Medical : Physics GOLDEN KEY POINTS

l

Pav < Prms.

l

Power factor varies from 0 to 1

l

Pure/Ideal

V

R

V, I same Phase

L

V leads I by

C

V lags I by

Choke coil

V leads I by

Power factor = cosf

Average power

1 (maximum)

Vrms. Irms

0, lagging

0

0, leading

0

0, lagging

0

(f = 0 so cosf = 1) and

Pav = Vrms Irms

p 2 p 2

p 2

l

At resonance power factor is maximum

l

Choke coil is an inductor having high inductance and negligible resistance.

l

Choke coil is used to control current in A.C. circuit at negligible power loss

l

Choke coil used only in A.C. and not in D.C. circuit

l

Choke coil is based on the principle of wattless current.

l

Iron cored choke coil is used generally at low frequency and air cored at high frequency.

l

Resistance of ideal choke coil is zero

Illustrations Illustration 25. A voltage of 10 V and frequency 103 Hz is applied to

1 mF capacitor in series with a resistor of 500W. Find the p

power factor of the circuit and the power dissipated. Solution XC =

1 = 2p f C

Power factor cosf=

1 10-6 2p ´ 103 ´ p

= 500W \Z =

R 2 + X 2C =

(500)2 + (500)2 = 500 2 W

2 (10)2 1 1 R 500 Vrms 1 ´ = = = , Power =Vrms I cosf = cosf = W ms rms Z 500 2 Z 500 2 2 10 2 dissipated

Illustration 26. If V = 100 sin 100 t volt and I = 100 sin (100 t +

p ) mA for an A.C. circuit then find out 3

(a)

phase difference between V and I

(b)

total impedance, reactance, resistance

(c)

power factor and power dissipated

(d)

components contains by circuits

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Q

E

ALLEN

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167

Solution (a)

Phase difference

(b)

Total impedance

p (I leads V) 3 V0 100 = = 1kW Now resistance R = Z cos 60° = 1000 ´ 1 = 500W Z= I0 100 ´ 10-3 2 f=-

reactance X = Z sin60° = 1000 ´ 3 = 500 3W 2 (c)

f = – 60°

Þ

Z

X

Power factor = cosf = cos (–60°) = 0.5 (leading)

Power dissipated P = Vrms Irms cos f = (d)

R 60°

100 2

´

0.1 2

´

1 = 2.5 W 2

Circuit must contains R as f ¹ p and as f is negative so C must be there, (L may exist but XC > XL) 2

Illustration 27. If power factor of a R-L series circuit is

1 when applied voltage is V = 100 sin 100pt volt and resistance 2

of circuit is 200W then calculate the inductance of the circuit. Solution cos f =

R 1 R Þ = Þ Z = 2RÞ R 2 + X 2L = 2R Z 2 Z

Þ wL =

XL =

3 RÞ

3 R

L=

3 ´ 200 2 3 H = 100p p

3R = w

Illustration 28. A circuit consisting of an inductance and a resistacne joined to a 200 volt supply (A.C.). It draws a current of 10 ampere. If the power used in the circuit is 1500 watt. Calculate the wattless current. Solution Apparent power = 200 × 10 = 2000 W

Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_05\03-ALTERNATING CURRENT\01 THEORY.P65

\ Power factor cos f=

E

1500 3 True power = = Apparent power 2000 4

2

Wattless current = Irms sin f = 10

10 7 æ 3ö 1-ç ÷ = A è 4ø 4

Illustration 29. A choke coil and a resistance are connected in series in an a.c circuit and a potential of 130 volt is applied to the circuit. If the potential across the resistance is 50 V. What would be the potential difference across the choke coil. Solution V=

VR2 + VL2

Þ

VL =

V 2 - VR2 =

(130)2 - (50)2 = 120 V

168

ALLEN

Pre-Medical : Physics

Illustration 30. An electric lamp which runs at 80V DC consumes 10 A current. The lamp is connected to 100 V – 50 Hz ac source compute the inductance of the choke required. Solution Resistance of lamp R = V = 80 = 8W I 10 Let Z be the impedance which would maintain a current of 10 A through the Lamp when it is run on V 100 100 Volt a.c. then. Z = = = 10 W but Z = R 2 + (wL)2 I 10 Þ

(wL)2 = Z2 – R2 = (10)2 – (8)2 = 36ÞwL = 6 Þ L =

6 6 = = 0.02H w 2p ´ 50

Illustration 31. Calculate the resistance or inductance required to operate a lamp (60V, 10W) from a source of (100 V, 50 Hz) Solution (a)

Q

V

Lamp

\

+ VR = 100

Now current througth Lamp is = But (b)

R

Maximum voltage across lamp = 60V

VR= IR

Þ

40 =

VR = 40V 10 1 Wattage = = A 60 6 voltage

1 (R) 6

Þ

100V, 50Hz

L

R = 240 W

Now in this case (VLamp)2 + (VL)2 = (V)2 (60)2 + (VL)2 = (100)2 Þ VL = 80 V Also VL = IXL =

1 X 6 L

so

100V, 50Hz

XL = 80 × 6 = 480 W = L (2pf) Þ L = 1.5 H

Cost of capacitor is much more than the cost of inductance of same reactance that's why choke coil is used. Illustration 32. A choke coil of resistance R and inductance L is connected in series with

L, R

C

a capacitor C and complete combination is connected to a.c. voltage, Circuit resonates when angular frequency of supply is w = w0. (a)

Find out relation betwen w0, L and C

(b)

What is phase difference between V and I at resonance, is it changes when resistance of choke coil is zero.

~

V=V0 sinwt(volt)

Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_05\03-ALTERNATING CURRENT\01 THEORY.P65

A capacitor of suitable capacitance replace a choke coil in an AC circuit, the average power consumed in a capacitor is also zero. Hence, like a choke coil, a capacitor can reduce current in AC circuit without power dissipation.

E

ALLEN

Pre-Medical : Physics

169

Solution (a)

At resonance condition XL = XC Þ w0L =

(b)

Q cos f =

R R = =1 R Z

\ f = 0°

1 Þ w0 = w 0C

1 LC

No, It is always zero.

BEGINNER'S BOX-4 1.

What is the power factor of a circuit that draws 5A at 160 V and whose power consumption is 600W?

2.

In a series LCR circuit as shown in fig.

3.

(a)

Find heat developed in 80 seconds

(b)

Find wattless current

For a series LCR circuit I

4.

= 100sin (100pt – p/3)mA

and

V

(a)

Calculate resistance and reactance of circuit.

(b)

Find average power loss.

= 100sin (100pt)volt,

then

The source voltage and current in the circuit are represented by the following equations – E = 110 sin (wt +

p ) volt, 6

I = 5 sin (wt –

p ) ampere 6

Find :–

Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_05\03-ALTERNATING CURRENT\01 THEORY.P65

5.

E

6.

(a)

Impedance of circuit.

(b)

Power factor with nature

In given circuit R = 100W. If voltage leads current by 60° then find – (a)

Current supply by source.

(b)

Average power

An inductor of reactance 4W and a resistor of resistance 3W are connected in series with 100V ac supply, calculate wattless current in circuit.

7.

8.

A 100 W resistor is connected to a 220 V, 50 Hz a.c. supply. (a)

What is the rms value of current in the circuit?

(b)

What is the net power consumed over a full cycle?

A choke coil and a resistance are connected in series in an a.c. circuit and a potential of 130 volt is applied to the circuit. If the potential across the resistance is 50V. What would be the potential difference across the choke coil.

170

5.

ALLEN

Pre-Medical : Physics LC OSCILLATION

The oscillation of energy between capacitor (electric field energy) and inductor (magnetic field energy) is called LC Oscillation.

5.1 Undamped oscillation When the circuit has no resistance, the energy taken once from the source and given to capacitor keeps on oscillating between C and L then the oscillation produced will be of constant amplitude. These are called undamped oscillation. I

t

L

C

After switch is closed

Q d2 Q d2Q 1 Q di +L 2 =0 Þ + Q=0 +L =0 Þ C C dt dt dt2 LC é d2 x ù 2 By comparing with standard equation of free oscillation ê 2 + w x = 0ú ë dt û

1 1 Frequency of oscillation f = LC 2p LC Charge varies sinusoidally with time q = qm cos wt w2 =

current also varies periodically with t

I=

dq p = qm w cos (wt + ) dt 2

If initial charge on capacitor is qm then electrical energy strored in capacitor is UE =

1 q 2m 2 C

At t = 0 switch is closed, capacitor starts to discharge. As the capacitor is fully discharged, the total electrical energy is stored in the inductor in the form of magnetic energy. UB =

1 2 LIm 2

(Umax)EPE = (Umax)MPE

where Im = max. current Þ

1 q 2m 1 2 = LIm 2 C 2

• In damped oscillation amplitude of oscillation decreases exponentially with time. • At t =

T 3T 5T , , ..... energy stored is completely magnetic. 4 4 4

• At t =

T 3T 5T , , ..... energy is shared equally between L and C 8 8 8

• Phase difference between charge and current is

p when charge is maximum, current minimum 2 when charge is minimum,current maximum

Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_05\03-ALTERNATING CURRENT\01 THEORY.P65

GOLDEN KEY POINTS

E

ALLEN

Pre-Medical : Physics

171

Illustration 33. An LC circuit contains a 20mH inductor and a 50mF capacitor with an initial charge of 10mC. The resistance of the circuit is negligible. Let the instant the circuit is closed to be t = 0. (a)

What is the total energy stored initially.

(b)

What is the natural frequency of the circuit.

(c)

At what minimum time is the energy stored is completely magnetic.

(d)

At what minimum time is the total energy shared equally between inductor and the capacitor.

Solution

1 (10 ´ 10-3 )2 = ´ = 1.0J 2 50 ´ 10-6

(a)

1 q2 UE = 2 C

(b)

w=

(c)

Q q = q0 cos wt Energy stored is completely magnetic (i.e. electrical energy is zero, q = 0) at

(d)

1 LC

t=

=

1 20 ´ 10

-3

´ 50 ´ 10 -6

= 103 rad/sec

Þ

f = 159 Hz

T 1 , where T = = 6.3 ms 4 f

Energy is shared equally between L and C when charge on capacitor become so,

q0 2

T at t = , energy is shared equally between L and C 8

BEGINNER'S BOX-5 1.

(1)

Initially key was placed on (1) till the capacitor got fully charged.

(2)

Key is placed on (2) at t=0. The minimum time when the energy in both capacitor and inductor will be same-

L

E C

2.

An inductor of inductance 2.0 mH is connected across a charged capacitor of capacitance 5.0 µF and the resulting L–C circuit is set oscillating at its natural frequency. Let Q denote the instantaneous charge on the capacitor and I the current in the circuit. It is found that the maximum value of Q is 200 µC. [IIT-JEE 2006] (i) Find the maximum value of I. (ii) When Q = 200 µC, what is the value of I?

Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_05\03-ALTERNATING CURRENT\01 THEORY.P65

(iii) When Q = 100 µC, what is the value of |dI/dt|?

E

(iv) When I is equal to one–half its maximum value, what is the value of |Q|?

ALLEN

Pre-Medical : Physics

172

ANSWERS BEGINNER'S BOX-1 1. There are two reasons for it : E

t 0.01 sec -E 0= -311.08

2. For AC, I = I0 sinwt, the instantaneous value of heat

8. 1H

9. 0.08 henry

10. 240.2 W

11. (a) 2.4 A

H = I2R = I02sin2wt × R

the average value of heat

produced during a cycle is :

BEGINNER'S BOX-3

0

ò

H dt T

0

é êQ ë

ò

T

0

dt

=

ò

T

0

1. Reading of ammeter = 2·5A Reading of voltmeter = 25V 2. The three current equations are,

2 0

2

(I sin wt ´ R)dt

ò

T

dt

0

I20 sin2 wt dt =

=

1 2 I0 R 2

1 2 ù I0 T ú 2 û

V=L

so iR =

V0 sinwt , R

dV 1 = iC dt C

V0 coswt and iC = V0wC coswt wL 3. (a) angular frequancy at resonance wr = 50 rad/s

iL = –

(b) amplitude of current at resonance Im = 8.13 A

2

æ I ö Þ H av = ç 0 ÷ R = I 2rms R .....(i) è 2ø

However, in case of DC,

di L and dt

V = iRR,

4. (a) w0 =

HDC = I R...(ii) 2

1 LC

, (b) f = 0° No, It is always zero.

p 4

Q I = Irms so from equation (i) and (ii) HDC = Hav

5. f =

AC produces same heating effects as DC of value

6. (a) 6.63 × 102 Hz (b) Quality factor Q = 21.7

I = Irms. This is also why AC instruments which are based on heating effect of current give rms value.

BEGINNER'S BOX-4

3. The average value of a.c. for a cycle is zero. So a d.c. ammeter will always read zero in a.c. circuit.

1. 0.75

4. 2.5 m s

3. (a) R = 500 ohm, X = 500 3 ohm (b) 2.5 watts

5. (a) 0.01 s, (b) 60°

6. 200 V

4. (a) Impedance Z = 22W (b) Power factor =

BEGINNER'S BOX-2 1. 2.1 A

2. (a) 4000 joule (b) 2.12 A

pI F 2. 10000 sin G1000t - J A H 2K

5. (a)

1

A, (b) 50W 2 6. Wattless current = 16A

3. 1A

7. (a) 2.2 A

4. The capacitive reactance XC = 212 W

8. 120V

Irms = 1.03 A The peak current

I0 = 1.46 A

If the frequency is doubled the capacitive reactances is halved, the consequently, the current is doubled.

1 (lagging) 2

(b) 484 watt

BEGINNER'S BOX-5 p LC 4 2. (i) 2.0 A, (ii) Zero, (iii) 104 A/s, (iv) 1.732 × 10–4 C 1. t =

Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_05\03-ALTERNATING CURRENT\01 THEORY.P65

ò

(b) 2.5 ms

13. (A) – r, (B) – p, (C) – p, (D) – p, (E) – r

produced (per second) in a resistance R is,

H av =

(b) inductor

12. 15.9 mH

-E

T

6. (a) resistor

7. R = 5W and XL = 5 3 W

+E0= 311.08

O

5. 2.49 A

E

ALLEN

Pre-Medical : Physics

Build Up Your Understanding

EXERCISE-I (Conceptual Questions) 6.

PEAK, AVERAGE AND RMS VALUE 1.

What is the r.m.s. value of an alternating current which when passed through a resistor produces heat which is thrice of that produced by a direct current of 2 amperes in the same resistor :(1) 6 amp

(2) 2 amp

(3) 3.46 amp

(4) 0.66 amp AC0001

2.

3.

(2) 5 3 V

(3) 5 V

(3)

7.

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1 2

(2)

(i12 + i22 )1 2

(4)

1 2

(i1 + i2 )2

1 2 (i + i22)1/2 2 1

AC0006 The relation between an A.C. voltage source and time in SI units is :

(1) 120 volt and 100 Hz (2)

120 2

volt and 100 Hz

(3) 60 volt and 200 Hz

The phase difference between current and voltage

(4) 60 volt and 100 Hz

p radian, If the frequency of 4

(1) 0.78 s

(2) 15.7 ms

(3) 2.5 s

(4) 2.5 ms

AC0007 8.

AC0003

E

2

(i1 + i2 )

V = 120 sin (100 pt) cos (100 pt) volt value of peak voltage and frequency will be respectively :–

(4) 1V

AC is 50 Hz, then the phase difference is equivalent to the time difference:-

If an A.C. main supply is given to be 220 V. What would be the average e.m.f. during a positive half cycle :(1) 198 V

(2) 386 V

(3) 256 V

(4) None of these

A current in circuit is given by i = 3 + 4 sin wt. Then the effective value of current is : (1) 5

(2)

7

(3)

17

(4)

10

AC0008 9.

The hot wire ammeter measures :(1) D.C. current

AC0004 5.

1

AC0002

in an AC circuit is

4.

The r.m.s. value of current for a variable current i=i1 cos wt + i2 sin wt :– (1)

The peak value of an alternating e.m.f. which is given by E = E0 coswt is 10 volts and its frequency 1 is 50 Hz. At time t = s, the instantaneous 600 e.m.f. is (1) 10 V

Incorrect statement are :

(2) A.C. current

(a) A.C. meters can measure D.C also

(3) None of above

(b) If A.C. meter measures D.C. there scale must be linear and uniform

(4) both (1) & (2)

(c) A.C. and D.C. meters are based on heating effect of current (d) A.C. meter reads rms value of current (1) a,b

(2) b,c

(3) c,d

173

(4) d,a AC0005

AC0009 10.

Frequency of A.C. in India is – (1) 45 Hz

(2) 60 Hz

(3) 50 Hz

(4) None of the above AC0010

174 11.

ALLEN

Pre-Medical : Physics For an alternating current I = I0cos wt, What is the rms value and peak value of current :(1) I0 ,

(3) I0 ,

I0 2 I0 2

(2)

I0 2

(4) 2I0 ,

15.

, I0

The graphs given below depict the dependence of two reactive impedances X1 and X2 on the frequency of the alternating e.m.f. applied individually to them. We can then say that :

I0

X2

X1

2 AC0011

If a step up transformer have turn ratio 5, frequency 50 Hz root mean square value of potential difference on primary 100 volts and the resistance of the secondary winding is 500 W then the peak value of voltage in secondary winding will be (the efficiency of the transformer is hundred percent) (1) 500 2

(2) 10 2

(3) 50 2

(4) 20 2

(1) X1 is an inductor and X2 is a capacitor (2) X1 is a resistor and X2 is a capacitor

16.

AC0012 SIMPLE AC CIRCUIT 13.

A resonant A.C. circuit contains a capacitor of capacitance 10–6 F and an inductor of 10–4 H. The

17.

frequency of electrical oscillations will be :(1) 105 Hz (3)

14.

105 Hz 2p

(2) 10 Hz (4)

10 Hz 2p

AC0013 1 A resistance of 300W and an inductance of p henry are connected in series to a A.C. voltage of 20 volts and 200 Hz frequency. The phase angle between the voltage and current is :æ4ö (1) tan–1 ç ÷ è3ø

18.

19.

æ3ö (2) tan–1 ç ÷ è4ø æ3ö (3) tan–1 ç ÷ è2ø

(3) X1 is a capacitor and X2 is an inductor (4) X1 is an inductor and X2 is a resistor AC0015 A 12 ohm resistor and a 0.21 henry inductor are connected in series to an AC source operating at 20 volts, 50 cycle/second. The phase angle between the current and the source voltage is: (1) 30° (2) 40° (3) 80° (4) 90° AC0016 A 110 V, 60 W lamp is run from a 220 V AC mains using a capacitor in series with the lamp, instead of a resistor then the voltage across the capacitor is about:(1) 110 V (2) 190 V (3) 220 V (4) 311 V AC0017 The resistance that must be connected in series with inductance of 0.2 H in order that the phase difference between current and e.m.f. may be 45° when the frequency is 50 Hz, is:(1) 6.28 ohm. (2) 62.8 ohm. (3) 628 ohm. (4) 31.4 ohm. AC0018 A student connects a long air cored - coil of manganin wire to a 100 V D.C. supply and records a current of 25 amp. When the same coil is connected across 100 V. 50 Hz a.c. the current reduces to 20 A , the reactance of the coil is :(1) 4 W (2) 3 W (3) 5 W

æ2ö (4)tan–1 ç ÷ è3ø

AC0014

(4) None AC0020

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12.

E

ALLEN 20.

Pre-Medical : Physics

The graph shows variation of I with f for a series R-L-C network. Keeping L and C constant. If R decreases :

(3) V0 sin (wt+ p/2),

Im (current) I

(4) V0 sin (wt + p/2),

24.

(freq.)

AC0021 Alternating current is flowing in inductance L and resistance R. The frequency of source is w/2p. Which of the following statement is correct : (1) For low frequency the limiting value of impedance is L. (2) For high frequency the limiting value of impedance is wL. (3) For high frequency the limiting value of impedance is R. (4) For low frequency the limiting value of impedance is wL. AC0022

coil. When the frequency of the voltage is changed to 400 Hz keeping the magnitude of V the same, the current is now :(1) 8 I in phase with V (2) 4 I and leading by 90° from V

25.

(3)

I and lagging by 90° from V 4

(4)

I and lagging by 90° from V 8

AC0025 A capacitor of capacity C is connected in A.C. circuit. The applied emf is V=V0 sinwt, then the current is : (1) I =

V0 sinwt wL

(2) I =

V0 sin(wt + p/2) wL

A bulb and a capacitor are connected in series to a increased, while keeping the voltage of the source

Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_05\03-ALTERNATING CURRENT\02-ALTERNATING CURRENT_EXERCISE.P65

An a.c. source of voltage V and of frequency 50 resistance. A current of r.m.s value I flows in the

source of alternating current. If its frequency is

E

V0 sin (wt – p/2) wL

Hz is connected to an inductor of 2 H and negligible

(a) Maximum current (Im) increases (b) Sharpness of the graph increases (c) Quality factor increases (d) Band width increases (1) a, b, c (2) b, c, d (3) c, d, a (4) All

22.

V0 sinwt wL

AC0024

fr

21.

constant, then

(3) I = V0 wC sinwt

(1) Bulb will give more intense light.

(4) I = V0 wC sin (wt + p/2)

(2) Bulb will give less intense light.

AC0026

(3) Bulb will give light of same intensity as before (4) Bulb will stop radiating light.

26.

The impedence of a circuit, when a resistance R and an inductor of inductance L are connected in

AC0023 23.

175

series in an A.C. circuit of frequency (f) is :-

In an A.C. circuit resistance and inductance are connected in series. The potential and current in

(1)

R + 4pfL2

(2)

R + 4p2 f 2L2

(3)

R2 + 4p2 f 2 L2

(4)

R 2 + 2p2 f 2 L2

inductance is: V0 (1) V0 sin wt, sinwt wL

(2) V0 sin wt,

V0 sin(wt + p/2) wL

AC0027

176 27.

ALLEN

Pre-Medical : Physics A capacitor of capacity C and reactance X if

32.

capacitance and frequency become double then reactance will be :– X (2) 2

(1) 4X

(3)

X 4

æ1ö (2) tan–1 ç ÷ è pø

æ2ö (3) tan–1 ç ÷ èpø

æ4ö (4) tan–1 ç ÷ è pø

AC0033 33.

The coil of choke in a circuit :

If the current through an inductor of inductance L is given by I = I0 sinwt, then the voltage across inductor will be :-

(1) increases the current

(1) I0 wL sin (wt – p/2)

(2) I0 wL sin (wt + p/2)

(3) I0 wL sin (wt – p)

(4) None of these AC0034

(2) controled the current 34.

(3) has high resistance to d.c. circuit (4) does not change the current AC0029

29.

æ 1 ö (1) tan–1 ç ÷ è 2p ø

(4) 2X AC0028

28.

A capacitor of capacitance 100 mF & a resistance of 100W is connected in series with AC supply of 220V, 50Hz. The current leads the voltage by .......

(1)

1 The inductive reactance of an inductive coil with p

henry and 50 Hz :–

(1)

50 ohm p

(3) 100 ohm

p 2

(2)

p 6

(3)

p 4

(4) 0 AC0035

35. (2)

There is a 5 W resistance in an A.C., circuit. Inductance of 0.1 H is connected with it in series. If equation of A.C. e.m.f. is 5 sin 50 t then the phase difference between current and e.m.f. is :-

p ohm 50

A 50 Hz a.c. source of 20 volts is connected across R and C as shown in figure below. The voltage across R is 12 volts. The voltage across C is

(4) 50 ohm AC0030

In the L–R circuit R = 10W and L = 2H. If 120 V, 60 Hz alternating voltage is applied then the flowing current in this circuit will be :(1) 0.32 A

(2) 0.16 A

(3) 0.48 A

(4) 0.80 A

(1) 8 V (2) 16V (3) 10 V

AC0031 31.

An inductance of 0.4 Henry and a resistance of 100 ohm are connected to a A.C. voltage source of 220 V and 50 Hz. Then find out the phase difference between the voltage and current flowing in the circuit : (1) tan–1 (2.25 p) (3)

tan–1

(1.5 p)

(2) tan–1 (0.4 p) (4)

tan–1

(0.5 p) AC0032

36.

(4) Not possible to determine unless values of R and C are given AC0036 200 W resistance and 1H inductance are connected in series with an A.C. circuit. The frequency of the source is

200 Hz. Then phase difference in between 2p

V and I will be :(1) 30°

(2) 60°

(3) 45°

(4) 90° AC0037

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30.

E

ALLEN 37.

Pre-Medical : Physics

Impedance of the following circuit will be :

42.

177

A circuit contains R, L and C connected in series with an A. C. source. The values of the reactances for inductor and capacitor are 200W and 600W respectively and the impedance of the circuit is Z1. What happens to the impedance of the same circuit if the values of the reactances are interchanged:(1) The impedance will remain unchanged (2) The impedance will increase

(1) 150W

(2) 200W

(3) 250W

(4) 340W

(3) The impedance will decrease

AC0038 38.

(4) Information insufficient

In showing figure find VR :

AC0043 43.

When V = 100 sinwt is applied across a series (RL-C) circuit, At resonance the current in resistance (R=100 W) is i = i0 sinwt, then power dissipation in circuit is:(1) 50 W

(2) 100 W

(3) 25 W

(4) Can't be calculated AC0044

(1) 132V

(2) 396V

(3) 185 V

(4)

44.

220 ´ 176V

(1) Current in the circuit is maximum and phase difference between E and I is p/2

AC0039 39.

If alternating current of 60 Hz frequency is flowing through inductance of L=1 mH and drop in DVL is 0.6 V then alternating current :(1)

1 A p

(2)

5 A p

(3)

50 A p

(4)

(2) Current in the circuit is maximum and phase difference between E and I is zero (3) Voltage is maximum and phase difference between E and I is p/2

20 A p

(4) Current is minimum and phase difference between E and I is zero AC0045

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AC0040

E

LCR SERIES CIRCUIT, RESONANCE 40. An inductance of 1mH, a condenser of 10mF and a resistance of 50W are connected in series. The reactance of inductor and condensers are same. The reactance of either of them will be :-

41.

(1) 100 W

(2) 30 W

(3) 3.2 W

(4) 10 W

(1) LC

(2) (LC)–½

-1 / 2

45.

An alternating voltage is connected in series with a resistance r and an inductance L. If the potential drop across the resistance is 200 volt and across the inductance is 150 volt, the applied voltage: (1) 350 volt

(2) 250 volt

(3) 500 volt

(4) 300 volt AC0046

AC0041 L, C and R represent physical quantities inductance, capacitance and resistance respectively. The combination representing dimension of frequency is æLö (3) ç ÷ èCø

At resosnance in a series LCR circuit, which of the following statements is true:-

C (4) L AC0042

46.

For a series R-L-C circuit :(a) Voltage across L and C are differ by p (b) Current through L and R are in same phase (c) Voltage across R and L differ by p/2 (d) Voltage across L and current through C are differ by p/2 (1) a, b, c

(2) b, c, d

(3) c, d, a

(4) All AC0047

178 47.

ALLEN

Pre-Medical : Physics A series R - L - C (R = 10 W, X L = 20 W, Xc = 20 W) circuit is supplied by V = 10 sin wt volt then power dissipation in circuit is :-

53.

(1) Zero

(2) 10 watt

(1) Only resistance

(3) 5 watt

(4) 2.5 watt

(2) Resistance & inductance. (3) Resistance & capacitance

AC0048 48.

In an AC Circuit decrease in impedance with increase in frequency indicates that circuit has/have :-

The self inductance of the motor of an electric fan is 10 H. In order to impart maximum power at 50Hz. It should be connected to a capacitance of : (1) 2 × 10

–6

F

(3) 10–4 F

(4) Resistance, capacitance & inductance. AC0054 54.

(2) 3 × 10 F –6

In given LCR circuit, the voltage across the terminals of a resistance & current will be–

(4) 10–6 F AC0049

49.

In a series resonant R-L-C circuit, if L is increased by 25% and C is decreased by 20%, then the resonant frequency will : (1) Increases by 10%

(2) Decreases by 10%

(3) Remain unchanged

(4) Increases by 2.5% AC0050

50.

The value of quality factor is :wL (1) R

w (2) RC

(3)

(1) 400V, 2A (3) 100V, 2A 55.

Phase of current in LCR circuit – (1) Is in the phase of potential

LC

(2) Leading from the phase of potential

(4) L/R

(3) Lagging from the phase of potential

AC0051 51.

(2) 800V, 2A (4) 100V, 4A AC0055

(4) Before resonance frequency, leading from the phase of potential and after resonance frequency, lagging from the phase of potential

If value of R is changed, then :-

56.

(1) Voltage across L remains same

In LCR circuit, the voltage across the terminals of a resistance, inductance & capacitance are 40V, 30V & 60V, then the voltage across the main source will be –

(2) Voltage across C remains same

(1) 130 volt

(3) Voltage across LC combination remains same

(2) 100 volt (3) 70 volt

(4) Voltage across LC combination changes 52.

AC0052 In a series LCR circuit voltage across resister, inductor and capacitor are 1V, 3V and 2V respectively. At the instant t when the source voltage is given by : V=V0 cos w t, the current in the circuit will be : pö æ (1) I = I0 cos ç wt + ÷ 4ø è

pö æ (2) I=I0 cos ç wt - ÷ 4ø è

pö æ (3) I = I0 cos ç wt + ÷ 3ø è

pö æ (4) I=I0 cos ç wt - ÷ 3ø è

AC0053

(4) 50 volt

57.

AC0057 500 For an alternating current of frequency Hz in Lp C-R series circuit with L = 1H, C = 1 mF, R = 100W, impedance is :(1) 100 W (2) 100 p W (3) 100 2 p W (4) 100 p W AC0058

Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_05\03-ALTERNATING CURRENT\02-ALTERNATING CURRENT_EXERCISE.P65

AC0056

E

ALLEN

Pre-Medical : Physics 63.

POWER IN AC CIRCUIT 58.

(1)

(3) I2R

4 2 IpR p

(4)

1 2 Ip R p2

AC0064 In an alternating circuit applied voltage and flowing

64.

V2 Cw

current are E = E0 sinwt and I = I0 sin(wt+p/2)

An AC circuit draws 5A at 160 V and the power consumption is 600 W. Then the power factor is:-

respectively. Then the power consumed in the

(1) 1

(2) 0.75

(1) Zero

(2) E0I0/2

(3) 0.50

(4) Zero

(3) E0I0/ 2

(4) E0I0/4

Which is not correct for average power P at resonance :

circuit will be :

65.

V

negligible :– (2) Inductance and resistance both low.

I

(3) Low resistance and high inductance

2 2

(4) High resistance and low inductance AC0066 AC0061

66.

p ) A. then find the watt less 6

power in watt :(1) 104

(2) 103

(3) 102

(4) 2.5 × 103 AC0067

67.

(1) VR+VL+Vc

If V = 100 sin100t volt, and I = 100 sin(100t +

In an A.C. circuit inductance, capacitance and resistance are connected. If the effective voltage across inductance is VL , across capacitance is Vc and across resistance is VR, then the total effective value of voltage is :

An A.C. supply gives 30V r.m.s. which passes through a 10W resistance. The power dissipated in it is :-

(2) VR+VL–Vc (3)

VR + ( VL - VC )

(4)

VR - ( VL - VC )

2

2

AC0065 In which of the following case power factor will be (1) Inductance and resistance both high

(4) P=I 2rms R

Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_05\03-ALTERNATING CURRENT\02-ALTERNATING CURRENT_EXERCISE.P65

(4)

AC0059

(3) P=VI

E

(2) I2Lw

1 2 Ip R 2

(2) P =

2

2

AC0062 62.

MNLwL - w1C OPQ

(2)

(1) P=Irms Vrms

61.

V2

(1) I2p R cos q

AC0060 60.

For a series LCR circuit the power loss at resonance is :-

A sinusoidal A.C. current flows through a resistor of resistance R. If the peak current is IP, then the power dissipated is :-

(3)

59.

179

(1) 90 2 W

(2) 90W

(3) 45 2 W

(4) 45 W AC0068

68.

An inductor of inductance L and resistor of

In an a.c. circuit V and I are given by

resistance R are joined in series and connected by

V = 100 sin (100 t) volts

a source of frequency w. Power dissipated in the

I = 100 sin (100t + p/3) mA

circuit is :-

The power dissipated in the circuit is (1) 104 watt

(2) 10 watt

(3) 2.5 watt

(4) 5.0 watt

(1)

(R (

2

+ w2 L2 V V

(3) R 2 + w2 L2 AC0063

) )

(

V2 R

(2) R 2 + w2 L2 (4)

)

R 2 + w2 L2 V2 AC0069

69.

ALLEN

Pre-Medical : Physics 74.

For given circuit the power factor is :

If alternating current of rms value 'a' flows through resistance R then power loss in resistance is : (1) Zero (3)

(1) 0

a2R 2

(2) a2R (4) 2a2R

(2) 1/2

AC0075

(3) 1/ 2

75.

(4) None of these AC0070 70.

Which of the following device in alternating circuit provides maximum power :(1) Only capacitor

In a purely capacitive circuit average power dissipated in the circuit is -

(2) Capacitor and resistor (3) Only inductor

(1) Vrms Irms

(4) Only resistor AC0076

(2) Depends on capacitance LC Oscillation

(3) Infinite

76. (4) Zero AC0071 71.

Energy loss in pure capacitance in A.C. circuit is 1 (1) CV2 2

(3)

1 CV2 4

(1) CL (3) (4) Zero

LI2 2

(2) 2LI2

LI2 (3) 4

(4) Zero

The power factor of L-R circuit is :

(1)

wL R

(3) wLR

1 CL

C L

(4)

L C

AC0077

In an oscillating LC circuit the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is stored equally between the electric and magnetic fields is(1) Q/2

(2) Q/ 3

(3) Q/ 2

(4) Q AC0078

AC0073 73.

77.

Power dissipated in pure inductance will be : (1)

(2)

(2) CV

AC0072 72.

Comparing the L–C oscillations with the oscillations of a spring–block system (force constant of spring = k and mass of block = m), the physical quantity mk is similar to :–

R

(2)

( wL )

(4)

wLR

2

+ R2

AC0074

78.

A fully charged capacitor C with intial charge q0 is connected to a coil of self inductance L at t = 0. The time at which the energy is stored equally between the electric and the magnetic fields is :(1) 2p LC

(2)

LC

(3) p LC

(4)

p LC 4

AC0079

Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_05\03-ALTERNATING CURRENT\02-ALTERNATING CURRENT_EXERCISE.P65

180

E

ALLEN 79.

Pre-Medical : Physics

A LC circuit is in th e stat e of resonance. if C = 0.1 mF and L = 0.25 henry. Neglecting ohmic resistance of circuit what is the frequency of oscillations (1) 1007 Hz

(2) 100 Hz

(3) 109 Hz

(4) 500 Hz

80.

Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_05\03-ALTERNATING CURRENT\02-ALTERNATING CURRENT_EXERCISE.P65

AC0080

E

A 60 µF capacitor is charged to 100 volts. This charged capacitor is connected across a 1.5 mH coil, so that LC oscillations occur. The maximum current in the coil is :(1) 1.5 A

(2) 2 A

(3) 15 A

(4) 20 A AC0081

ANSWER KEY

EXERCISE-I (Conceptual Questions) Que. Ans. Que. Ans. Que. Ans. Que. Ans. Que. Ans. Que. Ans.

181

1 3

2 2

3 4

4 3

5 2

6 3

7 4

8 1

9 4

10 3

11 2

12 1

13 3

14 1

15 3

16 3

17 2

18 2

19 2

20 1

21 2

22 1

23 3

24 4

25 4

26 3

27 3

28 2

29 3

30 2

31 2

32 2

33 2

34 3

35 2

36 3

37 3

38 1

39 2

40 4

41 2

42 1

43 1

44 2

45 2

46 4

47 3

48 4

49 3

50 1

51 3

52 2

53 3

54 3

55 4

56 4

57 1

58 2

59 2

60 3

61 3

62 3

63 3

64 1

65 3

66 4

67 2

68 2

69 3

70 4

71 4

72 4

73 2

74 2

75 4

76 4

77 3

78 4

79 1

80 4

182

ALLEN

Pre-Medical : Physics

AIPMT/NEET

EXERCISE-II (Previous Year Questions) AIPMT 2006 1.

A transistor-oscillator using a resonant circuit with

5.

an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of

(1) e2 R

frequency f. If L is doubled and C is changed to 4C, then frequency will be :(1)

(3)

f 4

1 ö æ R 2 + ç Lw ÷ è Cw ø

2

2 é 2 æ 1 ö ù 2 e + w R R L ê çè ÷ ú (2) Cw ø û ë

(2) 8 f

f

(4)

2 2

2 é 1 ö ù æ (3) e2 êR 2 + ç Lw ÷ ú R è Cw ø û ë

f 2

2 é 1 ö ù æ e2 êR 2 + ç Lw ú ÷ è Cw ø û ë (4) R

AC0082 2.

AIPMT 2009 Power dissipated in an LCR series circuit connected to an a.c. source of emf e is :-

A coil of inductive reactance 31W has a resistance

AC0086

of 8 W. It is placed in series with a condenser of

AIPMT Pre. 2010

capacitative reactance 25W. The combination is connected to an a.c. source of 110 volt. The power

6.

factor of the circuit is :(1) 0.56

(2) 0.64

(3) 0.80

(4) 0.33

In the given circuit the reading of voltmeter V 1 and V2 are 300 volts each. The reading of the voltmeter V3 and ammeter A are respectively : L

C

V1

V2

R=100W

AC0083

4.

(1)

E 0 I0

(3)

E 0 I0

2

2

cos f

(2) E0I0

A

220V, 50Hz

7.

(1) 100 V, 2.0 A

(2) 150 V, 2.2 A

(3) 220 V, 2.2 A

(4) 220 V, 2.0 A AC0087

AIPMT Mains 2010 A condenser of capacity C is charged to a potential difference of V1. The plates of the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to V2 is ? (1)

E I (4 ) 0 0 sin f 2

C(V12 - V22 ) L

æ C(V12 - V22 ) ö (3) ç ÷ø è L

AC0085

V3

(2) 1/2

C(V12 + V22 ) L

æ C(V1 - V2 )2 ö (4) ç ÷ø è L

1/ 2

AC0088

Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_05\03-ALTERNATING CURRENT\02-ALTERNATING CURRENT_EXERCISE.P65

3.

AIPMT 2007 What is the value of inductance L for which the current is a maximum in a series LCR circuit with C=10 µF and w = 1000s–1 ? (1) 10 mH (2) 100mH (3) 1 mH (4) cannot be calculated unless R is known AC0084 AIPMT 2008 In an a.c. circuit the e.m.f. (e) and the current (i) at any instant are given respectively by :e = E0 sinwt i = I0 sin (wt –f) The average power in the circuit over one cycle of a.c. is :-

E

ALLEN 8.

9.

10.

Pre-Medical : Physics

AIPMT Pre. 2011 An ac voltage is applied to a resistance R and an inductor L in series. If R and the inductive reactance are both equal to 3W, the phase difference between the applied voltage and the current in the circuit is :(1) p/6 (2) p/4 (3) p/2 (4) Zero AC0089

13.

(1) 11.

3

14.

Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_05\03-ALTERNATING CURRENT\02-ALTERNATING CURRENT_EXERCISE.P65

E

1 2

(2)

1 8

(3)

1 4

(4)

3 4

A coil of self-inductance L is connected in series with a bulb B and an AC source. Brightness of the bulb decreases when :

T/2

T

(2) V0

(2) frequency of the AC source is decreased.

t

(3)

V0 2

(3) number of turns in the coil is reduced. (4) A capacitance of reactance XC = XL is included in the same circuit.

V (4) 0 2

(2) 4.0 A

(3) 8.0 A

(4)

20 13

AC0098 Re-AIPMT 2015 15.

A

In an electrical circuit R, L, C and an a.c. voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and the current in the circuit is p/3. If instead, C is removed from the circuit the phase difference is again p/3. The power factor of the circuit is : (2)

3 2

(3)

1 2

(4)

A series R-C circuit is connected to an alternating voltage source. Consider two situations :(a) When capacitor is air filled. (b) When capacitor is mica filled. Current through resistor is i and voltage across capacitor is V then :-

AIPMT Pre. 2012

(1) 1

1 sin (100 pt + p/3) volt 2

NEET-UG 2013

AC0092

12.

e=

(1) an iron rod is inserted in the coil.

AC0091 A coil has resistance 30 ohm and inductive reactance 20 ohm at 50 Hz frequency. If an ac source , of 200 volt, 100 Hz, is connected across the coil, the current in the coil will be :(1) 2.0 A

1 sin (100 pt) ampere 2

AC0094

V0

V0

i=

(1)

V

O

AIPMT Mains 2012 The instantaneous values of alternating current and voltages in a circuit are given as

The average power in Watts consumed in the circuit is :-

In an ac circuit an alternating voltage e = 200 2 sin 100 t volts is connected to a capacitor of capacity 1µF. The r.m.s. value of the current in the circuit is:(1) 10 mA (2) 100 mA (3) 200 mA (4) 20 mA AC0090 AIPMT Mains 2011 The r.m.s. value of potential difference V shown in the figure is :-

183

16.

(1) Va = Vb

(2) Va < Vb

(3) Va > Vb

(4) ia > ib

AIPMT 2015 A resitance 'R' draws power 'P' when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes 'Z', the power drawn will be:

1 2

AC0093

AC0103

(1) P

R Z

æ Rö (2) P çè ÷ø Z

(3) P

æ Rö (4) P çè ÷ø Z

2

AC0104

184 17.

18.

ALLEN

Pre-Medical : Physics NEET-I 2016 An inductor 20 mH, a capacitor 50 µF and a resistor 40W are connected in series across a source of emf V = 10 sin 340 t. The power loss in A.C. circuit is :(1) 0.51 W (2) 0.67 W (3) 0.76 W (4) 0.89 W AC0107 A small signal voltage V(t) = V0 sin wt is applied across an ideal capacitor C :(1) Current I (t), lags voltage V(t) by 90°. (2) Over a full cycle the capacitor C does not consume any energy from the voltage source. (3) Current I (t) is in phase with voltage V(t). (4) Current I (t) leads voltage V(t) by 180°.

NEET(UG) 2018 21.

An inductor 20 mH, a capacitor 100 µF and a resistor 50 W are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is (1) 0.79 W

(2) 0.43 W

(3) 2.74 W

(4) 1.13 W AC0119

NEET(UG) 2019 (Odisha) 22.

AC0108

The variation of EMF with time for four types of generators are shown in the figures. Which amongst them can be called AC ?

NEET-II 2016 19.

E

E

Which of the following combinations should be selected for better tuning of an L-C-R circuit used

t

(a)

for communication ?

t

(b)

(1) R = 15 W, L = 3.5 H, C = 30 mF (2) R = 25 W, L = 1.5 H, C = 45 mF

E

(3) R = 20 W, L = 1.5 H, C = 35 mF

E

(4) R = 25 W, L = 2.5 H, C = 45 mF

t

(c)

(1) (a) and (d) (3) (a) and (b)

23.

(2) (a), (b), (c) and (d) (4) only (a) AC0166 A circuit when connected to an AC source of 12 V gives a current of 0.2 A. The same circuit when connected to a DC source of 12 V, gives a current of 0.4 A. The circuit is (1) series LR (2) series RC (3) series LC (4) series LCR AC0167

ANSWER KEY

EXERCISE-II (Previous Year Questions) Que. Ans. Que. Ans.

t

(d)

1 3

2 3

3 2

4 1

5 2

6 3

7 3

8 2

16 4

17 1

18 2

19 1

20 1

21 1

22 2

23 1

9 4

10 3

11 2

12 1

13 2

14 1

15 3

Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_05\03-ALTERNATING CURRENT\02-ALTERNATING CURRENT_EXERCISE.P65

20.

AC0109 The potential differences across the resistance, capacitance and inductance are 80 V, 40 V and 100 V respectively in an L-C-R circuit. The power factor of this circuit is :(1) 0.8 (2) 1.0 (3) 0.4 (4) 0.5 AC0110

E

ALLEN

Pre-Medical : Physics

Master Your Understanding

EXERCISE-III (Analytical Questions) 1.

2.

When 100 volts d.c. is applied across a solenoid a current of 1.0 amp. flows in it. When 100 volt a.c. is applied across the same coil, the current drops to 0.5 amp. If the frequency of the a.c. source is 50 Hz the impedance and inductance of the solenoid are :– (1) 200 ohm and 0.55 H (2) 100 ohm and 0.86 H (3) 200 ohm and 1.0 H (4) 100 ohm and 0.93 H AC0123 Time constant of the given circuit is t. If the battery is replaced by an ac source having voltage V = V0 cos wt, power factor of the circuit will be :L

185

combinations are P1, P2, P3 respectively. Then :(1) P1 > P2 > P3

(2) P1 = P2 < P3

(3) P1 = P2 > P3

(4) P1 = P2 = P3 AC0130

6.

In the L-C circuit shown in figure, the current is in direction shown in the figure and charges on the capacitor plates have sign shown in the figure. At this time :-

+Q –Q

i L

R

(1) i as well as Q increasing

(1) wt (3)

(2)

1 + (wt)

2

(2) i as well as Q decreasing

1

(4) is decreasing but Q is increasing

(4) None AC0124

3.

(3) i is increasing but Q is decreasing

1 + (wt)2 7.

An alternating emf of angular frequency w is applied across an inductance. The instantaneous power developed in the circuit has an angular frequency: (1) w/4

(2) w/2

(3) w

(4) 2w

AC0131 The switch in the circuit pictured is in position a for a long time. At t = 0 the switch is moved from a to b. The current through the inductor will reach its first maximum after moving the switch in a time:-

R e

AC0128 Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_05\03-ALTERNATING CURRENT\02-ALTERNATING CURRENT_EXERCISE.P65

4.

E

The self inductance of a choke coil is 10 mH. when

(1) 2 p LC

it is connected with a 10 V D.C. source, then the loss of power is 20 watt. When it is connected with

(3)

10 volt A.C. source loss of power is 10 watt. The frequency of A.C. source will be : (1) 50 Hz

(2) 60 Hz

(3) 80 Hz

(4)100 Hz

C

(2)

1 LC 4

(4) p LC AC0132

8. AC0129

5.

p LC 2

L

ab

An inductance L, a capacitance C and resistance R may be connected to an AC source of angular

The inductance of the oscillatory circuit of a radio station is 10 milli henry and its capacitance is 0.25mF. Taking the effect of the resistance negligible, wavelength of the broadcasted waves will be (velocity of light = 3.0 × 108 m/s, p = 3.14):

frequency w, in three different combinations

(1) 9.42 × 104 m

(2) 18.8 × 104 m

of RC, RL and RLC in series. Assume that

(3) 4.5 × 104 m

(4) none of these

wL =

1 . The power drawn by the three wC

AC0133

186 9.

ALLEN

Pre-Medical : Physics A coil has an inductance of 0.7 henry and is joined in series with a resistance of 220 W. When the alternating emf of 220 V at 50 Hz is applied to it then the phase through which current lags behind the applied emf and the wattless componenet of current in the circuit will be respectively

12.

The diagram shows a capacitor C and a resistor R connected in series to an AC source, V1 and V2 are voltmeters and A is an ammeter. Consider now the following statements : V1

(1) 30°, 1 A

C

(2) 45°, 0.5 A

~

(3) 60°, 1.5 A (4) none of these

A AC0134

10.

V2

R

(I) Readings in A and V2 are always in phase (II) Reading in V1 is ahead with reading in V2 (III) Readings in A and V1 are always in phase Which of these statements are is correct : (1) I only (2) II only (3) I and II only (4) II and III only AC0138

In the circuit shown in the figure, the A.C. source gives a voltage V = 20 cos (2000 t) volt neglecting source resistance, the voltmeter and ammeter readings will be : 6W A 5 mH, 4W

13.

50 µF

A capacitor of capacitance 2 mF is connected in the tank circuit of an oscillator oscillating with a frequency of 1 kHz. If the current flowing in the circuit is 2 mA,

V

the voltage across the capacitor will be:–

(2) 5.6V, 1.4A (3) 0V, 0.47 A

14.

(4) 1.68 V, 0.47 A 11.

AC0136 An inductor and a resistor in series are connected to an A.C. supply of variable frequency. As the frequency of the source is increased, the phase angle between current and the potential difference across source will be : L

15. l ~ l

R

(1) First increase and then decrease (2) First decrease and then increase

(1) 0.16 V

(2) 0.32 V

(3) 79.5 V

(4) 159 V

AC0139 If an alternating current i = im sin wt is flowing through a capacitor then voltage drop DVC across capacitor C will be ? (1) -

im sin wt wC

(2) -

(3) -

pö im æ sin wt + ÷ wC çè 4ø

(4)

(4) Go on increasing AC0137

pö im æ sin wt - ÷ wC çè 4ø

AC0141 If an alternating current i = imsin wt is flowing through an inductor then voltage drop DVL across inductor L will be :(1) imwL sin wt (2) imwL cos wt

æ è

pö 4 ÷ø

æ è

pö 4 ÷ø

(3) imwL sin ç wt +

(3) Go on decreasing

im cos wt wC

(4) imwL cos ç wt -

AC0142

Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_05\03-ALTERNATING CURRENT\02-ALTERNATING CURRENT_EXERCISE.P65

(1) 0V, 1.4A

E

ALLEN

Pre-Medical : Physics

16.

If frequency of alternating source is made zero then which of the following statement is true : (1) Current through capacitor will be zero (2) Current through resistance will be zero (3) Current through inductance will be zero (4) All AC0143

17.

The figure shows a LCR network connected to

18.

300 V a.c. supply. The circuit elements are such

187

A 1.5 µF capacitor is charged of 60 V. The charging battery is then disconnected and a 15 mH coil is connected in series with the capacitor so that LC oscillations occurs. Assuming that the circuit contains no resistance. The maximum current in this coil shall be close to (1) 1.4 A (2) 1.2 A (3) 0.8 A (4) 0.6 A AC0146

that R = XL = XC = 10W. V1, V2 and V3 are three a.c. voltmeters connected as shown in the figure. Which of the following represents the correct set of readings of the voltmeters ? V1 V2 V3

R

C

L

300 V (1) V1 = 100 V, V2 = 100 V, V3 = 100 V (2) V1 = 150 V, V2 = 0 V, V3 = 150 V (3) V1 = 300 V, V2 = 100 V, V3 = 100 V (4) V1 = 300 V, V2 = 300 V, V3 = 300 V

Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_05\03-ALTERNATING CURRENT\02-ALTERNATING CURRENT_EXERCISE.P65

AC0144

E

EXERCISE-III (Analytical Questions) Que. Ans. Que. Ans.

1 1

2 2

3 4

16 1

17 4

18 4

4 3

5 2

6 3

ANSWER KEY 7 3

8 1

9 2

10 2

11 4

12 1

13 1

14 2

15 2

Pre-Medical : Physics

E

Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_05\03-ALTERNATING CURRENT\02-ALTERNATING CURRENT_EXERCISE.P65

188

ALLEN

IMPORTANT NOTES