L1 Ch-5 Sol Set-1 Magnetism and Matter

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Level - I

SECTION - A

School / Board Exam. Type Questions

Very Short Answer Type Questions :

1. The equation E = E0 sint represents an alternating emf. What are the values of its amplitude frequency and

time period?

Sol. 0

2, ,2

E

.

2. What is the average value of sinusoidal ac during a half cycle? During a full cycle.

Sol.0

2I

, ZERO (where I

0 is the peak value)

3. What does an ac ammeter measure?

Sol. It measures the rms (virtual) value of ac.

4. What is the power factor of an LCR circuit at resonance?

Sol. Power factor cos = 1.

5. What is meant by wattless current?

Sol. If there is no power dissipation in ac circuit, then the current in the circuit is wattless.

6. Why is ac more dangerous than dc of the same voltage?

Sol. A 220 V ac has a peak value of 220 2 311V . Whereas a 220 V dc has a peak value of 220 V.

7. A 100 V dc heater is used on an ac source such that heat produced is same. What is the rms value of ac

voltage?

Sol. 100 volt

8. The instantaneous current is I = 5 sin 314t. What is the rms current?

Sol.5

A2

.

Chapter 7

Alternating Current

Solutions (Set-1)

88 Alternating Current Solutions of Assignment (Set-1) (Level-I)


9. Do we required ac or dc for long distance transmission?

Sol. A.C.

10. Why is the core of a transformer laminated?

Sol. A laminated iron core minimises loss of energy due to eddy currents.

Short Answer Type Questions :

11. Briefly explain the two causes of energy loss in a transformer.

Sol. Iron loss and copper loss

12. Why is the core of a transformer made of soft iron?

Sol. The hysterisis loop of soft iron has a small area. Hence a soft iron core results in little energy loss.

13. What is an ideal transformer?

Sol. An ideal transformer is one in which magnetic field lines are confined entirely within the core (no field lines

out side the core). Its efficiency is 100%.

14. Explain why in an ac circuit, there is no power consumption in an ideal inductor.

Sol. The average power in an ac circuit is given by cosrms rms

P V I . In pure inductive circuit, = 90º.

So, cos = 0 0P .

15. The impedance of a coil is 141.4 and its resistance is 100 . What will be its reactance?

Sol. 100 ohm. 2 2 2 2 2 2(141.4) (100) 100L L

Z R X X Z R .

16. Draw graph to show the variation of inductive and capacitive reactance with the frequency of applied voltage.

Sol. XL

f

L =

constant

XC

f

Reactangular Hyperbola C = constant

17. A bulb and a capacitance are connected in series to a source of alternating current. How will brightness of

the bulb change on increasing the frequency of the current?

Sol. The impedance of the circuit

2

2 1R

C

will decrease and so the current in the circuit will increase and

the bulb will give more intense light.

18. The ac voltage is applied in circuits (a) (b) (c). If the frequency of emf is increased, what will be the effect

upon current in these circuits?

~ ~ ~

(a) (b) (c)

R L C

Sol. In figure (a), Current will remain same, decreases in figure (b) and increase in figure (c)

89Solutions of Assignment (Set-1) (Level-I) Alternating Current


19. A bulb is connected in series with a variable capacitor. Will the bulb glow when this combination is connected

to a dc supply/ac supply. What would happen in each case if capacitance is reduced?

Sol. No current will flow in the bulb when connected to DC supply because the capacitor offers infinite resistance

to dc supply 1 1

as 02

CX f

C fC

. Hence, the bulb will not glow even if capacitance is reduced.

For ac supply, as capacitance is reduced, reactance 1 1

2C

XC fC

will increase C so current will

decrease, so bulb will glow less brightly.

20. When a capacitor is connected in series with a series LR circuit the ac flowing in the circuit increases. Explain.

Sol. The impedance 2 2( )R L decrease to

2

2 1R L

C

.

21. The ac circuit has a power factor of 0.5. What will be the phase difference between voltage and current in this

circuit? Can you tell whether the current leads the voltage or the voltage leads the current?

Sol. As 1

cos cos60º 60º2

. With the information given in the question, current leads the voltage or the

voltage leads the current, cannot be concluded.

22. In the adjoining ac circuit, R is a pure resistance and X is an unknown circuit element. Using only an ac meter

(voltmeter), how will you detect whether ‘X’ is

~

X

A R B C

(i) Pure inductor

(ii) Pure resistance

(iii) Resistanceless, inductanceless conductor?

Sol. We shall measure the potential difference across AB, BC and AC. If

(i) (VAB

+ VBC

) > VAC

. Then X is pure inductance

(ii) (VAB

+ VBC

) = VAC

, Then X is pure resistance

(iii) VBC

= 0, Then X is inductance-less resistance-less conductor.

23. How will you use a 100 V - 50 W bulb on 200 V ac mains?

Sol. The resistance of the bulb is 2

200V

RP

and it takes a current of 0.5 A P

IV

. To give it 0.5 A current

at 200 V, its resistance should be 400 . Hence, a resistance of 200 , must be connected in series of the

bulb. Alternatively, a step-down transformer having primary to secondary turn ratio 2 : 1 may be used.

24. A 100 F capacitor in series with a 40 resistor is connected to a 110 V - 60 Hz supply. What is the

maximum current?

Sol. Maximum current 0

02 2

2 2

4

1.414 110

1 1(40)

376.8 10

EI

RC

= 155.5 155.5

3.24A48.01600 704.3



25. Refer to Q. 24, what is the time lag between maximum current and main voltage?

Sol. The phase difference between current and voltage in AC series RC circuit is given

4

1

1 1tan 0.6635

376.8 10 40

C

R CR

1tan (0.6635) 33.56º

The time lag is given by33.56 1

1.55ms360º 360 60

t T

26. Obtain the maximum current and phase angle for a high frequency ac supply 110 V, 12 kHz. Discuss the result

obtained. Consider series AC circuit with R = 40 and 1

, 07.53

C LX X

Sol. Maximum current 02

2

155.53.89A

1(40)

7.53

I

1tan 0.0033

7.53 40

CX

R

0

Thus we see that at very high frequencies, the capacitor acts like a conductor.

27. A 20 V - 5 W lamp is to run on 200 V – 50 Hz ac mains. Find the capacitance of a capacitor required to

run the lamp.

Sol. Current rating of bulb, 5

0.25A20

PI

V

The resistance of the Filament of bulb 20

800.25

VR

I . If the lamp is to run on 200 V - 50 Hz ac mains,

a capacitor (or choke) must be placed in series in order to increase its effective resistance, Let C is the

capacitance used, then

Impedance Z =

2

2 2 2 1(80)

314C

R XC

Now 2

2

2000.25 4.0 F

180

314

rms

rms

VI C

Z

C

28. A capacitor, a 15 resistor and a 80 mH inductor are in series with a 50 Hz ac source. Calculate the

capacitance if the current is in phase with the voltage.

Sol. It is the case of series LCR circuit in when = 0.

2

1

1 1tan 127 F

LC

L CR C L



29. A series LC circuit has L = 0.405 H and C = 25 F. The resistance R is zero. Find the frequency of resonance.

Sol. The natural frequency of LC series circuit is given by 1

50Hz2

fLC

. The resonance occurs when an ac

voltage of frequency equal to its natural frequency is applied. So the frequency of resonance is 50 Hz.

30. How much current is drawn by the primary of a transformer which steps down 220 V to 22 V to operate a

device having an impedance of 220 ?

Sol.22V

220V, 22 V, 0.1 A220

S

P S S

VV V I

Z

For an ideal transformer, output power = Input power

0.1 220.01A

220

S S

P P S S P

P

I VI V I V I

V

.

Long Answer Type Questions :

31. What is an alternating current (ac)? Show that average value of current over a complete cycle is zero while

for half cycle, it is 0

2I

, where I

0 is the peak value.

Sol. That current, whose direction changes periodically are known as alternating currents.

∵ Average current over a positive half cycle

/2

0

2T

avgI i dt

T

=

/2

0

0

2sin

T

i t dtT

+

–

2O

T

I

=

/2

0

0

2 cosT

i t

T

= 0

2cos0 cos

2

i T

T

= 0 02 2

1 cos [1 ( 1)]2

I I

TT

T

= 0

2I

Average current over a negative half cycle

0

/2

22T

avg

T

II idt

T

For complete cycle

I = Iavg

(+ve)

+ Iavg (–ve)

Iavg

= 0



32. What do you mean by root mean square value of ac? How is it related to peak value of current?

Sol. That steady current which would produce the same heat in given resistance in a given time as is done by the

alternating current when passed through the same resistance for the same time.

0

2rms

II

33. In an alternating current circuit, an inductance L, a capacitance C and a resistance R are connected in series.

Derive expression for the impedance and phase angle. What is the impedance of circuit at resonance?

Sol. As all the elements are in series, so at any instant, the current through these elements is same. By using

technique of phasors.

~

R L C

VR

VL

VC

AC Source ( = sin )V V tm

So, I = IR = I

L = I

C... (i)

The voltage across the resistor given by

VR = I × R ... (ii)

is in phase with its current

The voltage across the inductor

VL = IX

L = IL = I × 2f × L ... (iii)

As discussed earlier, VL is ahead of I

L (i.e. I) by

2

The voltage across the capacitor given by

2

C C

I IV IX

C fC... (iv)

is lagging behind the current by 2

I

VR

VC

VL

t +

VR

V

V VC L

–

t

The voltage between inductor and capacitor is equal to VC – V

L. The total voltage given by

2 2( )R C L

V V V V ... (v)

2 2 2 2( ) ( ) ( )C L C L

V IR IX IX I R X X ... (vi)



2 2( )C L

VR X X

I = Z (impedance = Effective resistance of series LCR circuit) ... (vii)

Phase relationship between V and I

tanC L C L

R

V V X X

V R

... (viii)

R

X – XC L

z

R

X

X

=

+ (

–

)

2

2

C

L

Impedance Triangle

34. Explain the principle, construction and working of a choke coil in controlling current in an ac circuit. Why is

it preferred over resistance in ac circuit?

Sol. Choke coil is a coil having high inductance and negligible resistance, it is used to control current in AC circuit.

It consists of a Cu coil wound over a soft iron laminated core. Thick Cu wire is used to reduce the resistance

of the circuit. Soft iron is used to improve inductance of the circuit for ideal choke coil, resistance is zero and

hence power factor is zero. Then average power is zero.

~

Coil of copper wire

Iron core

35. Explain the principle, construction and working of a transformer. Give two causes of power loss in it. Why is

its core laminated?

Sol. Transformers are based upon mutual induction which transform an alternating voltage from one to another of

greater or smaller value.

A transformer consists of two coils wound on a soft iron core, called primary and secondary coils. Let number

of turns in these coils are Np and Ns respectively. The input ac voltage is applied across primary coil whereas

output a.c. voltage is across secondary coil.

We consider an ideal transformer in which the primary has negligible resistance and all the flux in the core

links both the primary and secondary windings. Let be the flux linkage through each of primary and secondary

coils. Then.

Induced emf across the primary coil, p p

dN

dt

... (i)

Similarly induced emf across secondary s s

dN

dt

... (ii)

From these equation, AC voltage obtained across secondary

AC voltage applied across primarys s s

p p p

V N

V N

... (iii)



Figure Two arrangements for winding of primary and secondary coil in a transformer:(a) two coils on top of each other, (b) two coils on separate limbs of the core.

Soft iron-core

Secondary

Prim

ary

Secondary

Prim

ary

Note : The above relations are based upon following three assumptions :

(i) Primary current and resistance are small.

(ii) The same flux links both the primary and secondary coil.

(iii) The secondary current is small.

In a transformer, some energy is always lost. The efficiency of a well designed transformer may be upto 95%.

If the transformer is assumed to be 100% efficient (no energy loss)

p = IpVp = I

sVs

Thus, p s s

s p p

I V N

I V N ... (iv)

36. Explain the use of transformer for long distance transmission of electric power from a power station to a

substation and eventually to actual user.

Sol. Use of Transformers in Transmission and Distribution of Energy over long distance :

The voltage output of the generator is stepped up, so the current is reduced and consequently, the I2R loss

is cut down, It is then transmitted over long distances to an area substation near the consumers. There the

voltage is stepped down. It is further stepped down at the distributing substations and utility poles before a

power supply of 240 V reaches our homes.

37. Show that the average power dissipated per cycle in an ac circuit is given rms rms

RP V I

Z , where R is the

resistance of the circuit (defined as real part of the complex impedance) and Z is the impedance.

Sol. Power in Series LCR Circuit

V = Vm

sint

mV

IZ

sin(t + ) = Im

sin(t + ) assuming VC > V

L

P = V × I = Vm

.Im

.sint.sin(t + )

= .

.[cos( ) cos( )]2

m mV I

t t t t

..[cos cos(2 )]

2 m m

V It



.

.cosrms rms

P V I ... (i)

As cos(2t + ) for one complete cycle, is ZERO

2

.cosrms

VP

z ... (ii)

2 2 2

rms rms rmscos R

P I Z I Z I RZ

... (iii)

(From impedance triangle cosR

Z )

38. An LCR circuit has L = 10 mH, R = 3 and C = 1 F connected in series to a source of 15cost volt.

Calculate the current amplitude and average power dissipated per cycle at a frequency 10% lower than the

resonant frequency.

Sol. The resonant frequency 4

0

110 rad/sec

LC

The frequency used 4 4 310

10 10 9 10 rad/sec100

190 , 111.11

L CX L X

C

Impedance 2 2( ) 21.32L C

Z R X X

Current amplitude 0

0

150.704 A

21.32

EI

Z

Average power dissipated 0 0 0 0

1 1cos

2 2

RP E I E I

Z = 0.744 W

39. A 750 Hz, 20 V source is connected to a resistance of 100 , an inductance of 0.1803 H and a capacitance

of 10 F, all in series. Calculate the time in which the resistance (Thermal capacity = 2 Joule/0ºC) will get

heated by 10ºC.

Sol. Impedance

1

2 22 1

835Z R LC

Average power dissipated = Vrms

× Irms

× cos = Vrms

× Irms

× R

Z = 0.574 W

Now P × t = Q = Joule

2 10ºC 20JºC

20J

348 sec0.0574W

t



SECTION - B

Model Test Paper

Very Short Answer Type Questions : [1 Mark]

1. What is iron loss in a transformer?

Sol. Iron loss is a loss of energy in the form of heat in the iron core of the transformer due to the formation of eddy

current.

2. Does a transformer change the frequency of a.c.?

Sol. No, The frequency cannot be changed by transformer.

3. We can measure d.c. by an ordinary ammeter, but not a.c. why?

Sol. This is because average value of a.c. over a complete cycle is zero.

4. Can a choke coil be replaced by a condenser of suitable frequency?

Sol. Yes, it can be replaced.

5. Is there any device by which direct current can be controlled without any loss of energy? Can choke coil do so?

Sol. No, direct current can be controlled by the use of resistance only which involves loss of energy. A choke coil

cannot control direct current.

6. What is meant by form factor?

Sol. Form factor =

0

rms

0av

21.1

2 2 2

I

I

II

7. What is the function of oil in a transformer?

Sol. It provides insulation as well as cooling.

Short Answer Type Questions : [2 Marks]

8. Name two useful devices based on mutual induction.

Sol. Transformer and inductor coil.

9. For an ac, can ever rms value be equal to peak value? Average value be equal to peak value.

Sol. Yes, when the ac is a square wave, rms value, average value and peak value are same.

10. Find the time required for a 50 Hz alternating current to change its value from zero to the r.m.s. value.

Sol. I = I0sint

0

0sin

2

II t

4t

2

4t

T

1 1sec

8 8 400

Tt

f



11. How does the term electrical resistance differ from reactance and impedance?

Sol. Electrical resistance arises on account of material of the conductor. Reactance is the resistance that arises

on account of opposing e.m.f. induced due to change in the strength of current. Total effective resistance offered

a.c. circuit is called impedance.

12. How is a transformer used in long distance transmission of ac?

Sol. AC is transmitted over long distance at extremely high voltages, to minimise the energy losses due to

transmission. For this, a step-up transformer is used at generating station and a step-down transformer is used

at the receiving station.

13. Why a choke coil is preferred for reducing ac current?

Sol. A choke coil is preferred for reducing ac because power loss in the process is minimum. In an ideal choke

coil, phase difference between voltage and current is = 90º. So average power dessipated for cycle = Erms

Irms

cos90º = 0

14. 11 kW of electric power can be transmitted to a distant station at (i) 220 V or (ii) 22000 V. Which of the two

modes of transmission should be preferred and why? Support your answer with possible calculation.

Sol. P = 11 kW = 11000 W

(i)1 1

1

11000220 V, 50A

220

PV I

V , Power loss = 2 2

1(50) 2500 RWattI R R

(ii) 2 2

2

22000V, 0.5A,P

V IV

Power Loss = 2 2

2(0.5) Watt

4

RI R R . Power loss becomes

1

10,000

times. Hence, transmission at high voltages should be preferred.

Short Answer Type Questions : [3 Marks]

15. A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of

an iron core in the choke causes no change in the lamp brightness. Predict the corresponding observations

if the connection is on ac line.

Sol. On a DC line, choke has no impedance. Therefore lamp glows brightly. On an ac line, choke offers impedance.

So the lamp glows dimly. When as iron core is inserted in the choke, its impedance increases therefore,

brightness of the bulb decreases further.

16. Air-cored chokes are used for reducing high frequency ac why?

Sol. Inductive reactance of a choke is XL = L = 2fL. When frequency f is high, L need not be made high.

Therefore air-cored choke will serve the purpose.

17. Iron-cored chokes are used for reducing low frequency ac why?

Sol. With iron core, coefficient of self inductance (L) increases. Therefore XL = L = 2fL becomes large even f is

low. Hence,the current L

EI

X reduces.

18. Does the current in an A.C. circuit lag, lead or remain in phase with the voltage of frequency f applied to the

circuit when (i) f = fr (ii) f < f

r (iii) f > f

r

where fr is the resonance frequency.

Sol. (i) At f = fr, I and V in same phase

(ii) At f < fr, I lead with V

(iii) At f > fr, I lag with V



19. Show mathematically that an ideal inductor does not consume any power in an a.c. circuit.

Sol. ∵ Instantaneous power

PL = EI = E

0sint × I

0sin

2t

= –E0I0sintcost

= 0 0

sin22

E It

The average power over one complete cycle of AC is

0 0sin2

2L

E IP t

∵ average value of sin(2t) over a complete cycle is zero.

So, 0L

P

Thus, the average power supplied to an inductor over one complete cycle is zero.

20. What is a transformer? Explain its theory and discuss its main uses.

Sol. Transformers are based upon mutual induction which transform an alternating voltage from one to another of

greater or smaller value.

A transformer consists of two coils wound on a soft iron core, called primary and secondary coils. Let number

of turns in these coils are Np and Ns respectively. The input ac voltage is applied across primary coil whereas

output a.c. voltage is across secondary coil.

We consider an ideal transformer in which the primary has negligible resistance and all the flux in the core

links both the primary and secondary windings. Let be the flux linkage through each of primary and secondary

coils. Then.

Induced emf across the primary coil, p p

dN

dt

... (i)

Similarly induced emf across secondary s s

dN

dt

... (ii)

From these equation AC voltage obtained across secondary

AC voltage applied across primarys s s

p p p

V N

V N

... (iii)

Figure Two arrangements for winding of primary and secondary coil in a transformer:(a) two coils on top of each other, (b) two coils on separate limbs of the core.

Soft iron-core

Secondary

Prim

ary

Secondary

Prim

ary



Note : The above relations are based upon following three assumptions :

(i) Primary current and resistance are small.

(ii) The same flux links both the primary and secondary coil.

(iii) The secondary current is small.

In a transformer, some energy is always lost. The efficiency of a well designed transformer may be upto 95%.

If the transformer is assumed to be 100% efficient (no energy less)

p = IpVp = I

sVs

Thus, p s s

s p p

I V N

I V N ... (iv)

The voltage output of the generator is stepped up, so the current is reduced and consequently, the I2R loss

is cut down, It is then transmitted over long distances to an area substation near the consumers. There the

voltage is stepped down. It is further stepped down at the distributing substations and utility poles before a

power supply of 240 V reaches our homes.

Long Answer Type Questions : [5 Marks]

21. A transformer has an efficiency of 80% and works at 100 V and 4 kW. If the secondary voltage is 240 V,

calculate the primary and secondary currents.

Sol. 40A, 13.3 A

4000W40A

100VPI Output power =

80 1600W4kW

100 5

16000 4013.3A

5 240 3SI

22. Alternating emf of E = 200 sin 200t is applied to a circuit containing an inductance of 1

henry. Then find

out

(a) Equation of instantaneous current through circuit.

(b) Reading of a.c. ammeter connected in the circuit.

Sol. (a) ∵ 1

100 100L

X L

Then 0

0

2002 Amp

100L

EI

X

So, 2sin 1002

I t



(b) Reading of galvanometer

02

2 1.414 A

2 2rms

II

��



Objective Type Questions

(A.C. Voltage Applied to a Resistor, Representation of AC Current and Voltage by Rotating Vectors Phasors)

1. Hot wire ammeters are used for measuring

(1) Both AC and DC. (2) Only AC

(3) Only DC (4) Neither AC nor DC

Sol. Answer (1)

2. In alternating current circuits, the a.c. meters measure

(1) r.m.s. value (2) Peak value

(3) Mean value (4) Mean square value

Sol. Answer (1)

3. A 110 V d.c. heater is used on an a.c. source, such that the heat produced is same as it produces when

connected to 110 V dc in same time-intervals. What would be the r.m.s. value of the alternating voltage?

(1) 110 V (2) 220 V (3) 330 V (4) 440 V

Sol. Answer (1)

Given that HAC

= HDC

2 2

rmsl Rt l Rt

2 2

rmsV V

R R

Vrms

= V

so Vrms

= 110 V

4. The peak value of an alternating e.m.f. E = E0 sin t is 10 volt and its frequency is 50 Hz. At a time

s600

1t ,

the instantaneous value of the e.m.f. is

(1) 1 volt (2) 5 3 volt (3) 5 volt (4) 10 volt

Sol. Answer (3)

E = E0 sint, E

0 = 10 V

= 50 Hz, 1

600t

110sin 502

600E

1

10sin 10 5 volt6 2

Solutions (Set-2)



5. The time required for a 50 Hz sinusoidal alternating current to change its value from zero to the r.m.s. value

(1) 1.5 × 10–2 s (2) 2.5 × 10–3 s

(3) 10–1 s (4) 10–6 s

Sol. Answer (2)

0sinI I t

0

rms

2

II

So 0

0sin

2

II t

sin sin 2 504

t

2 504

t

1 s

400t

–2 –30.25 10 2.5 10 st

(AC Voltage Applied to an Inductor, AC Voltage Applied to a Capacitor)

6. There is no resistance in the inductive circuit. Kirchhoff 's voltage law for the circuit is

L

V V t = sin 0

(1) 0dt

diLV (2)

dt

diLV

(3) 02 dt

diLV (4) None of these

Sol. Answer (2)

–

+

+

–V V t= sin0

Using Kirchhoff's voltage law

–0

LdiV

dt

LdiV

dt



7. When 100 volt d.c. is applied across a solenoid, a current of 1.0 A flows in it. When 100 volt a.c. is applied

across the same coil, the current drops to 0.5 A. If the frequency of a.c. source is 50 Hz the impedance and

inductance of the solenoid is

(1) 200 and 0.55 H (2) 100 and 0.86 H

(3) 200 and 1.0 H (4) 100 and 0.93 H

Sol. Answer (1)

When there is direct current, = 0

XL = 0, so only R is there

So, V = IR

100 = 1(R) R = 100

2 2

22

2 2

2

When, 50 Hz,

200 100

100 40000 100000.5

Impedance 200 30,000

100 3

L

L

L

L

L

f Z X R

V IZ X

X

z X

X

Also XL = L

100 3 2 fL

100 3 1.730.55 H

2 3.14 50 3.14L

8. An inductive circuit contains a resistance of 10 ohms and an inductance of 2 henry. If an alternating voltage of 120

V and frequency 60 Hz is applied to this circuit, the current in the circuit would be nearly

(1) 0.32 A (2) 0.80 A (3) 0.48 A (4) 0.16 A

Sol. Answer (4)

R = 10 , L = 10 H, Vrms

= 120 V, f = 60 Hz

rms

rms

VI

Z

2 22 2 2

10 2 60 2Z R L

753.6Z

rms

1200.16 A

753.6I

9. A coil and a bulb are connected in series with a 12 volt direct current source. A soft iron core is now inserted

in the coil. Then

(1) The intensity of the bulb remains the same

(2) The intensity of the bulb decreases

(3) The intensity of the bulb increases

(4) Nothing can be said



Sol. Answer (1)

Intensity of bulb remains the same because source is DC, so steady state current will be independent of the

inductance of the inductor for DC circuit,

sourcesteady

bulb

Ei

R

10. A sinusoidal supply of frequency 10 Hz and r.m.s. voltage 12 V is connected to a 2.1 F capacitor. What is

r.m.s. value of current?

(1) 5.5 mA (2) 20 mA (3) 26 mA (4) 1.6 mA

Sol. Answer (4)

f = 10 Hz, Vrms

= 12 V, C = 2.1 F

rm s

rm s

C

VI

X ,

1

CX

C

= 2f

Putting all values

rms1.6 mAI

(AC Voltage Applied to a Series LCR Circuit)

11. In an LCR series circuit R = 10 , XL = 8 and X

C = 6 The total impedance of the circuit is

(1) 10.2 (2) 17.2 (3) 10 (4) None of these

Sol. Answer (1)

R = 10 , XL = 8 and X

C = 6

22–

L CX XZ R

2 2210

104 10.2

12. In a series RLC circuit, potential differences across R, L and C are 30 V, 60 V and 100 V respectively as shown in

figure. The e.m.f. of source (in volts) is

30 V 60 V 100 V

R L C

(1) 190 (2) 70 (3) 50 (4) 40

Sol. Answer (3)

22

rms–emf

L CRV VV V

2 2

30 100 – 60

2 250 V30 40



13. In a series RLC circuit, the r.m.s. voltage across the resistor and the inductor are respectively

400 V and 700 V. If the equation for the applied voltage is t sin2500 , then the peak voltage across

the capacitor is

= 500 2sin t

CR L

(1) 1200 V (2) V21200 (3) 400 V (4) V2400

Sol. Answer (4)

t sin2500

VR = 400 V, V

L = 700 V

0

rms

500 2

2 2

= 500 V

22

rms–

L CRV VV

22 2–500 400 L C

V V

2–250000 – 160000L C

V V

290000 –L C

V V

– 300L C

V V

700 300C

V

400C

V V

0 rms2 400 2 VV V

14. In the following circuit the emf of source is E0 = 200 V, R = 20 , L = 0.1 H, C = 10.6 F and frequency is

variable then the current at frequency f = 0 and f = is

R LC

(1) Zero, 10 A (2) 10 A, zero (3) 10 A, 10 A (4) Zero, zero

Sol. Answer (4)

E0 = 200 V, R = 20 , L = 0.1 H, C = 10.6 F

When f = 0, XL = L = 2fL

XL = 0

When f = XL = , X

C = 0

In both case at least one component has value infinite. So in both cases current will be zero as they are

connected in series.



15. In series LCR circuit, the phase difference between voltage across L and voltage across C is

(1) Zero (2)

(3)2

(4) 2

Sol. Answer (2)

In LCR circuit angle between VL and V

C = 180° (rad).

16. With increase in frequency of an a.c. supply, the impedance of an LCR series circuit

(1) Remains constant

(2) Decreases

(3) Increases

(4) Decreases at first, becomes minimum and then increases

Sol. Answer (4)

12 ,

2L C

X fL XfC

When f is increased, XL will increase

, XC will decrease

So XL – X

C will increase, hence z will increase

17. The reading of ammeter in the circuit is

A

VX

C = 2

XL = 2 R = 55

110 V

(1) 2 A (2) 3 A

(3) Zero (4) 1 A

Sol. Answer (1)

V

A

XC = 2

XL = 2 R = 55

110 V

22–

L CX XZ R

2 55 55

V = I Z

1102 A

55I I



18. In a series L-C circuit, if L = 10–3 H and C = 3×10–7 F is connected to a 100 V-50 Hz a.c. source, the

impedance of the circuit is

(1)

103

105

(2) 0.1 – 3 × 10–5 (3)103

105

(4) None of these

Sol. Answer (3)

L = 10–3 H and C = 3×10–7 F

f = 50 Hz, V = 100 V

2–

L CX XZ

1– –

L CX X L

C

Putting , L & C

510

–3 10

Z

(Power in AC Circuits : The Power Factor)

19. Power factor of an ideal choke coil (i.e., R = 0) is

(1) Near about zero (2) Zero (3) Near about one (4) One

Sol. Answer (2)

cos 0R

Z

20. In an a.c. circuit, the instantaneous values of e.m.f. and current are E = 200 sin 314t (volt) and

i = sin (314t + /3) A. The average power consumed in watts is

(1) 100 (2) 200 (3) 50 (4) 25

Sol. Answer (3)

rms rmscosP E I

1 cos here

322 2

E I ∵

1 1200 1

2 2

100 W

2

= 50 W

21. An a.c. of frequency f is flowing in a circuit containing only an ideal choke coil of inductance L. If V0

and i0

represent peak values of the voltage and the current respectively, the average power given by the source to

the choke coil is equal to

(1)2

1i0V

0(2)

2

1i0

2 (2fL) (3) Zero (4)2

1V

0 (2fL)



Sol. Answer (3)

For an ideal choke coil cos = 0

so, P = Erms

Irms

cos

P = 0

22. When a voltage V = V0 cos t is applied across a resistor of resistance R, the average power dissipated per

cycle in the resistor is given by

(1)R

V

2

0(2)

R

V

20

(3)R

V

2

2

0 (4)R

V

2

2

0

Sol. Answer (3)

P = Erms

Irms

cos

cos = R

Z

cos = 1 for resistor

0

rms

2

VV

rms 00 0

rms , 1

22 2

V VV VIP

R RR

∵

2

0

2

VP

R

(LC Oscillations, Transformers)

23. In an LCR circuit, the resonating frequency is 500 kHz. If the value of L is doubled and value of C is decreased

to 8

1 times of its initial values, then the new resonating frequency in kHz will be

(1) 250 (2) 500

(3) 1000 (4) 2000

Sol. Answer (3)

f1 = 500 103 Hz

1' 2 , '

8L L C C

1 1

2

f fLC LC



2 21

2 1 1

12

8CL

L Cf

f LCLC

1

2

2

11000 kHz

2

ff

f

24. In an LCR circuit L = 8.0 henry, C = 0.5 F and R = 100 ohm are in series. The resonance angular frequency

is

(1) 500 rad/s (2) 600 rad/s

(3) 800 rad/s (4) 1000 rad/s

Sol. Answer (1)

L = 8.0 henry, C = 0.5 F and R = 100 in series

–6

1 1

8 100.5LC

3

–6

1 10

24 10

= 500 rad/s

25. In series LCR circuit voltage leads the current when (Given that 0 = resonant angular frequency)

(1) < 0

(2) = 0

(3) > 0

(4) None of these

Sol. Answer (3)

1,

L CX L X

C

If voltage leads L C

X X , for this condition

> 0

So it will be an inductive circuit with voltage leading current by 90°.

26. At resonance, the value of the power factor in an LCR series circuit is

(1) Zero (2) 1

(3)2

1(4) Not defined

Sol. Answer (2)

At resonance, Z = R

cos 1R

Z = Power factor.



27. In oscillating LC circuit, the total stored energy is U and maximum charge upon capacitor is Q. When the

charge upon the capacitor is 2

Q, the energy stored in the inductor is

(1)2

U(2)

4

U(3)

4

3U (4)

3

4

U

Sol. Answer (4)

Initial energy = total energy =

2

2

QU

C

2

' as Q'=4 22

Q QU

C

2

8

Q

C

Uinductor

= Utotal

– Ucapacitor

2 2

–2 8

Q Q

C C

2

3 3

8 4

QU

C

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L1 Ch-5 Sol Set-1 Magnetism and Matter