physics-2iit (spark) worksheet-10 q + soln (ashwani)

2IIT1516P(P1)WS10(Spark) 1

PHYSICS Class II IIT-JEE Achiever 2015 - 16 (Spark batch) Solutions to Worksheet – 10 Topics Alternative current

Date 06-09-2015

Subjective questions 1. A 10F capacitor, an inductor and a resistor of 100 are connected to an alternating source of emf

200 2 sin100t as shown in the figure. What is the r.m.s current in the circuit and the value of the inductance if the current and the source voltage attain their maxima simultaneously? What is the average power consumed in this case?

100

10F L

200 2 sin100t

Solution The current and the voltage attain their maxima and minima simultaneously. Hence they are in phase and the circuit is in resonance. Z = R = 100

I = V 200 2 2 2AZ 100 , I r.m.s. = 2A

The power factor = R 1Z

Average power consumed = VI cos = 200 2 1 = 400 watt and at resonance = 1LC

100 = 6

1L 10 10

L = 4 6

1 10H10 10 10

2. A 200Km long telegraph wire has capacity of 0.014F /km. If it carries an alternating current of 50 KHz, what should be the value of an inductance required to be connected in series so that impedance is minimum? Solution

Irms = 9036

= 52

= 2.5 Amp

VL = 2 2120 36 156 84 = 114.47 Volt

cos = 30120

= 0.3

= 72.5

tan72.5 = 2 60 L36

L = 0.303 H


C R

C

~

)R)L((I 22

I

IL

IR

IXC

IXL

I IR

3. An alternating current of 1.5mA and angular frequency = 300 rad/s flows through 10K resistor and a 0.50 F capacitor in series. Find the r.m.s. voltage across the capacitor and impedance of the circuit? Solution i = 1.5 mA, = 300 rad/sec

XC = 5

6

1 1 10C 300 0.50 10 15

= 6.6 103

Z = 22 2 2CR X (10000) 6600 = 11.98 k 12 k

VC = iXC = 1.5 103 6.6 103 = 9.9 10 V

4. In the circuit shown there is a box and capacitance C connected to alternating power source of angular

frequency of 2 rad/s. Box has power factor 12

and circuit has overall power factor 1. Find the

impedance of the box. Solution

tan = LR

1 = LR (cos = 1/ 2 ) tan = 1)

L = R and overall circuit has power factor 1.

so, L = 1C

Z = 2 2R ( L) = L 2

Z = 1 12C C 2

5. In the given circuit, power factor between A and B is 0.5 and over all power factor of the circuit is 1. Find the value of R and L.

A B

V0sin10t C = 2mF

L R

Solution Phasor diagram of the given circuit will be In first case Power factor between A and B is 0.5

cos = L 0.5R

… (1)

In second case Overall power factor is one so XL = XC … (2)


2 = 1LC

or L = 1 5H100 0.002

put the value of L in (i)

R = L 10 5 1000.5 0.5

6. A 10F capacitor is connected with 1-henry inductance in series with a 50Hz source of alternating current. Calculate the impedance of the combination. Solution The impedance of the L-C circuit is

Z = 6

1 1L ~ 1 2 50 ~C 10 10 2 50

Z = 3 2 310 100 ~ 10100 ~

Or Z = 4.47

7. 200V A.C. is applied at the ends of an LCR circuit. The circuit consists of an inductive reactance XL = 50, capacitive reactance XC = 50 and ohmic resistance R = 10. Calculate the impedance of the circuit and also potential differences across L and R. What will be the potential difference across L-C? Solution The impedance of L-C-R circuit is

Z = 2 2 22L CX ~ X R 50 50 10 10

Now the r.m.s. current flowing in the circuit is

i.e. E 200 20ampZ 10

The potential difference across the inductance is VL = iXL = 2050 = 1000V, VR = iR = 2010 = 200V The potential difference across L-C is (50 50)20 = 0

8. To a circuit of 1 resistance and 0.01 henry inductance is connected a 200Volt line of frequency 50 cycles/ second. Calculate the reactance, the impedance and the current in the circuit and also the lag in phase between alternating voltage and current. Solution The reactance of the circuit = XL = L = 2fL XL = 0.01250 = = 3.14

The impedance of the circuit = 2 2LX R

Z = 2 23.14 1 10.86 = 3.3

The phase difference (lag) is tan = L/R = /1 =

= tan1() = tan1(3.14) = 7220


9. A 20 volts 5 watt lamp is used on a.c. mains of 200 volts 50 c.p.s. Calculate the value of (i) capacitance, (ii) inductance to be put in series to run the lamp. (iii) how much pure resistance should be included in place of the above device so that lamp can run on its voltage Solution For lamp

i = 520

= 0.25 A

R = 200.25

= 80

Current through the lamp should be 0.25 A (i) when condenser C is placed in series

i = 2

2

200

1RC

= 0.25

putting the value of = 2 50 C = 4.0 F (ii) when inductor is used

I = 2 2

200R ( L)

= 0.25

L = 2.53 H (iii) When resistance is used

I = 200R r

= 0.25

r = 720

10. When a 15 V dc source was applied across a choke coil then a current of 5 Amp flows in it. If the same coil is connected to a 15 V, 50 rad/s ac source a current of 3 Amp flows in the circuit. Determine the inductance of the coil. Also find the power developed in the circuit and its resonance frequency if a 2500 f capacitor is connected in series with the coil. Solution For a coil, z = 2 2 2R L

I = VZ

= 2 2 2

VR L

For dc source, = 0 I = VR

i.e. R = 155

= 3 . . . (i)

When ac is applied

I = VZ

i.e. Z = 153.0

= 5

R2 + 2Lx = 25

2Lx = 25 – 9 = 16 xL = 4

L = 440

= 0.08 Henry.

Now when the capacitor is connected is series.


xC = 1C

= 6

1 850x 2500 x10

Z = 22L CR x x = 23 4 8 = 5

I = 155

= 3A

Pav = vrms Irms cos = 2rmsI R = (3)2 x 3 = 27 W.

Resonance frequency:

= 12 LC

= 6

12 0.08 x 2500 x10

= 10002 x5x 2 2

= 25 2

= 11.25 Hz.

Multiple choice questions with one correct alternative

11. A sinusoidal voltage Vo sin t is applied across a series combination of resistance R and inductor L. The amplitude of the current in the circuit is

(A) o2 2 2

V

R +ω L (B) o

2 2 2

V

R ω L (C) o

2 2 2

V

R +ω Lsint (D) oV

R

Ans (A) Impedance of the circuit = 2 2 2R L Amplitude of voltage = Vo

Amplitude of current = o2 2 2

VR L

12. An inductance of negligible resistance whose reactance is 22 at 200Hz is connected to 200Volt, 50 Hz power line. The value of inductance is (A) 0.0175 Henry (B) 0.175 Henry (C) 1.75 Henry (D) 17.5 Henry Ans (A)

22 = L2200; L = 22 7 0.0175 H2 22 200

Value of inductance is the same whatever be the frequency.

13. An ideal choke takes a current of 8 A when connected to an a.c. source of 100 volt and 50Hz. A pure resistor under the same conditions takes a current of 10A. If two are connected in series to an a.c. supply of 100V and 40Hz, then the current in the series combination of above resistor and inductor is (A) 10A (B) 8A (C) 5 2 amp (D)10 2 amp Ans (C)

XL = 1008

, R = 100 1010

L 100 = 1008

or L = 18

H

Z = 2

21 2 40 10 10 28

I = E 100 10 5 2AZ 10 2 2


14. A resistor R, an inductor L and a capacitor C are connected in series to a source of frequency n. If the resonant frequency is nr then the current lags behind voltage, when (A) n = 0 (B) n < nr (C) n = nr (D) n > nr

Ans Below resonant frequency the current leads the applied e.m.f., at resonance it is in phase with applied e.m.f. and above resonance frequency it lags the applied e.m.f.

15. An ac source of angular frequency is fed across a resistor R and a capacitor C in series. The current registered is I. If now the frequency of source is changed to /3 (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of reactance to resistance at the original frequency will be.

(A) 35

(B) 53

(C) 35

(D) 53

Ans (A) According to given problem,

I = V =VZ

/ [R2 + (1/C2)]1/2 … (1)

and 2 2 2

I V=2 [R +(3/Cω) ]

… (2)

Substituting the value of I from equation (1) in (2),

4 2 22 2 2 2

1 9R + =R +C ω C ω

, i.e., 22 2

1 3= RC ω 5

so that 2 1/2X (1/Cω) [(3/5)R ] 3= = =

R R R 5

16. The frequency for which a 5.0F capacitor has a reactance of 1000 is given by

(A) 1000 cycles/secπ

(B) 100 cycles/secπ

(C) 200 cycle /s (D) 5000 cycles /sec

Ans (B)

C1XC

C

1X C

= -6

11000×5×10

f = -3

1 100=2π×5×10 π

cycle/sec

17. In an a.c. circuit V and I are given by V = 50 sin50t volt and I = 100 sin(50t + /3) mA. The power dissipated in the circuit (A) 2.5 kW (B) 1.25 kW (C) 5.0 kW (D) 500 watt Ans (B)

P = -3 -3 -350×100 π 1cos ×10 =2500× ×10 =1250×102 3 2

= 1.25 watt

18. The average power dissipation in pure inductance in ac circuit, is

(A) 2

12Li

(B) 2Li2 (C) 2Li

4 (D) zero


Ans (D) Average power dissipated in pure inductor is zero.

19. Circuit as shown in figure below, choose the correct statement.

a.c. source

R C L

(A) current in resistance R and current in inductor L will be in 90 phase difference. (B) potential drop across R and potential drop across L will be in same phase. (C) current through C and current through L will be in 90 phase difference. (D) current in R and current in L will be in same phase. Ans (A)

20. In a series L, R, C, circuit which is connected to a.c. source. When resonance is obtained then net impedance Z will be

(A) Z = R (B) Z = L IC

(C) Z = L (D) Z = 1C

Ans (A) At resonance v

L = 1C

and Z = R

Hence (A) is correct.

21. An L, C, R series circuit is connected to a.c. source. At resonance, the applied voltage and the current flowing through the circuit will have a phase difference of (A) /4 (B) zero (C) (D) /2 Ans (B) The circuit will behave as penely resistance circuit so phase difference between connector and voltage applied will be zero. Hence (B) is correct.

22. The reciprocal of impedance is called (A) reactance (B) admittance (C) inductance (D) conductance Ans (D)

23. The root-mean-square value of an alternating current of 50Hz frequency is 10 ampere. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be (A) 2 102 sec and 14.14 amp. (B) 1 102 sec and 7.07 amp. (C) 5 103 sec and 7.07 amp. (D) 5 103 sec and 14.14 amp. Ans (D)

24. A coil of resistance 2000 and self-inductance 1.0 Henry has been connected to an a.c. source of frequency 2000/2 Hz. The phase difference between voltage and current is (A) 30 (B) 60 (C) 45 (D) 75 Ans (C)


cos = 2 2 2 2 2 2

R R 2000Z R L 2000 L

= 2f = 2 20002

cos = 2 2

2000 122000 2000

= 45

Hence (C) is correct.

25. In a series resonant circuit, the a.c. voltage across resistance R, inductance L and capacitance C are 5V, 10V and 10V, respectively. The a.c. voltage applied to the circuit will be (A) 20V (B) 10V (C) 5V (D) 25V Ans (C) Limits is in resonance condition Voltage applied will be same as across resistor V = 5 V Hence (C) is correct.

26. In the given figure, which voltmeter will read zero voltage at resonant frequency rad/sec? V2 V3 V1

V4

R C L

E=Eosint (A) V1 (B) V2 (C) V3 (D) V4

Ans (D)

27. A resistance R is connected in series with capacitance C Farad value of impedance of the circuit is 10 and R = 6 so, find the power factor of circuit. (A) 0.4 (B) 0.6 (C) 0.67 (D) 0.9 Ans (B)

cos = 610

= 0.6

Hence (B) is correct.

28. In a R, L, C circuit, three elements is connected in series by an a.c. source. If frequency is less than resonating frequency then net impedance of the circuit will be (A) capacitive (B) inductive (C) capacitive or inductive (D) pure resistive Ans (A)

29. Using an A.C. voltmeter, the potential difference in the electrical line in a house is read to be 234 volts. If the line frequency is known to be 50 cycles per second, the equation for the line voltage is (A) V = 165 sin(100t) (B) V = 331 sin(100t) (C) V = 234 sin(100t) (D) V = 440 sin(100t)


Ans (B)

30. In an a.c. circuit, containing an inductance and a capacitor in series, the current is found to be maximum when the value of inductance is 0.5henry and of capacitance is 8F. The angular frequency of the input A.C. Voltage must be equal to (A) 500 (B) 5 104 (C) 4000 (D) 5000 Ans (A)

= 6

1 1 12LC 0.5 8 10

103 = 500 rad/sec


31. An alternating voltage E (in volts) = 2002 sin (100t) is connected to a 1 F capacitor through an a.c. ammeter. The reading of the ammeter shall be (A) 10 mA (B) 20 mA (C) 40 mA (D) 80 mA Ans (B)

irms = 2001 C

= 200 100 1 106 = 20 mA

Hence (B) is correct.

32. In a series R, L, C circuit XL = 10, XC = 4 and R = 6. Find the power factor of the circuit.

(A) 12

(B) 32

(C) 1/2 (D) none of the these

Ans (A)

tan = 66

= 1

cos = 12


33. In LCR circuit the capacitance is changed from C to 4C. For the same resonant frequency, the inductance should be changed from L to (A) 2L (B) L/2 (C) L/4 (D) 4L Ans (C)

L = 1C

. . . (1)

L = 14 C

. . . (2)

L 1L 4 L = L

4

Hence (C) is correct.

34. A resistance (R) = 12; inductance (L) = 2 henry and capacitive reactance C = 5 mF are connected in series to an ac generator (A) at resonance, the circuit impedance is zero. (B) at resonance, the circuit impedance is 12. (C) the resonance frequency of the circuit is 1/2. (D) at resonance, the inductive reactance is less than the capacitive reactance.


Ans (B)

35. In an A.C. circuit, the current is I = 5 sin(100/2) amp and the A.C. potential is V = 200 sin(100t) volt. Then the power consumption is (A) 20 watts (B) 40 watts (C) 1000Watts (D) 0 watts Ans (D)

Multiple choice questions with one or more than one correct alternatives

36. An L–C circuit has capacitance C1 = C and inductance L1 = L. A second circuit has C2 = C/2 and L2 = 2L and third circuit has C3 = 2C and L3 = L/2. All the three capacitors are charged to the same potential V and then made to oscillate. Then (A) angular frequency of oscillation is same for all the three circuits (B) maximum current is greatest in first circuit (C) maximum current is greatest in second circuit (D) maximum current is greatest in third circuit. Ans (A) and (D)

Angular frequency 1LC

Since product of L and C is same for all the three circuits. Therefore is same for all the three circuits. Hence, choice (A) is correct. From conservation of mechanical energy

2 20

1 1Li CV2 2

or 2 20

Ci VL

or 0Ci .LL

or 0CiL

CL

is maximum for the third circuit.

Hence, maximum current is greatest for third circuit. So, choice (D) is correct.

37. An LCR circuit with 100 resistance is connected to an ac source of 200 V and angular frequency 300 rad/s. When only the capacitance is removed the current lags behind the voltage by 60º. When only the inductance is removed, the current leads the voltage by 60º. Then in LCR circuit the current and power dissipated are (A) 2A (B) 1A (C) 200W (D) 400W Ans (A) and (D) When capacitance is removed then circuit becomes L–R circuit with

LtanR

…(i)

When inductance is removed, the circuit becomes C–R circuit with 1tanCR

…(ii)


from (i) and (ii) we get, 1LC

or XL = XC So, in LCR circuit Z = R and circuit is in resonance.

Hence ViZ

V 200 2AR 100

i = 2A Hence, choice (A) is correct and Choice (B) is wrong.

av rms rmsP V i cos 200 2 1 = 400 W.

38. A current of 4A flows in a coil when connected to 12V dc source. If the same coil is connected to a 12V, 50 rad/s source, a current of 2.4 A flows in the circuit. Then (A) R = 4 (B) R = 3 (C) L = 4H (D) 0.08 H Ans (B) and (D) In case of a coil 2 2 2Z R L for D.C. = 0 So, Z = R

or V ViZ R

or VRi

12R 34

R = 3 When ac is applied

V 12Z 5i 2.4

i.e., Z = 5 2 2 2

LR X 5 2 2 2 2 2LX 5 R 5 3

or XL = 4 XL = L = 4

or 4 4L H 0.0850

H

or L = 0.08 H.


39. In the circuit shown in the figure R = 50, E1 = 25 3 volt and E2 = 25 6 sin (t) volt where = 100 s–1. The switch is closed at t = 0 and remains closed for 14 minutes, then it is opened

R

1S

2

~

(A) The amount of heat produced in the resistor is 63000 J. (B) The amount of heat produced in the resistor is 7000 J. (C) If total amount of heat produced is used to heat 3 kg of water at 20ºC, the final temperature will be 25ºC. (D) The value of direct current that will produce same amount of heat in same time through same resistor will be 1.5 A. Ans (A), (C) and (D)

1 21I(t) (I I )R

25 3(1 2 sin wt)

Heat produced in one cycle 2 /

0

3R dt J2

Heat produced in 14 minutes 1Q 14 60 50 63000 J3

Q = ms = 5ºC

Tf = 5 + 20 = 25ºC and heat produced by direct current I2RT=Q

I 1.5RT

.

40. An alternating voltage (in volts) varies with time t (in seconds) as V = 200 sin (100 t) (A) The peak value of the voltage is 200 V (B) The rms value of the voltage is 220 V (C) The rms value of the voltage is 100 2 V

(D) The frequency of the voltage is 50 Hz Ans (A), (C) and (D) V = 200 sin (100 t) Compare this equation with V = V0 sin t

V0 = 200 V, 0rms

V 200V 100 22 2

50


E = 15 V L = 5mH

R = 1

I

A B

41. A electric heater is connected to 100 V, 60 Hz ac supply. (A) The peak value of the voltage is 100 V (B) The peak value of the current in the circuit is 2 2 A

(C) The rms value of the voltage is 100 V (D) The rms value of the current is 2 A Ans (B), (C) and (D) Vrms = 100 V. Peak value of voltage =100 2V .

Peak value of current 100 2 2 2 A50

rms2 2I 2A

2 .

42. L, C and R respectively represent inductance, capacitance and resistance. Which of the following combinations have the dimensions of frequency? (A) R/L (b) 1/RC (c) R / LC (d) 1/ LC Ans (A), (B) and (C) The dimensions of L and 1/ C are the same as those of resistance, where 2 v

43. The network shown in figure is part of a circuit. The battery has negligible internal resistance. At a certain instant the current I = 5 A and is decreasing at a rate of 10–3 As–1. At that instant, the potential difference (A) across L is 5 mV (B) across L is 5 V (C) between points A and B is 15 V (D) between points A and B is 25 V Ans (B) and (C)

A BdIV IR E L Vdt

or B AdIV V IR E Ldt

dIdt

is negative, 3 3B AV V 5 1 15 5 10 10 15V and voltage across inductor

VL = 5 V.

44. In a series LCR circuit (A) the voltage VL across the inductance leads the current in the circuit by a phase angle of / 2 (B) the voltage VC across the capacitance lags behind the current by a phase angle of / 2 (C) the voltage VR across the resistance is in phase with the current (D) the voltage across the series combination of L, C and R is V = VL + VC + VR

Ans (A), (B) and (C)

The voltage across the combination is given by 1/22 2

R C LV V V V .

45. To convert mechanical energy into electrical energy, one can use (A) DC dynamo (B) AC dynamo (C) motor (D) transformer Ans (A) and (B) Conceptual


Read the passage given below and answer questions 15 to 17 by choosing the correct alternative In the given circuit at t = 0, switch S is closed.

5mH 10 10V 20 F

2

S

46. The current through 10 resistor at any instant t; 0 < t < will be

(A) 1000 t

31 e6

(B)

1000 t35 e

6

(C) 1000 t

31 e6

(D) 1000 t

36 e5

Ans (B) From loop (i) applying Kirchoff’s law 12 I – 10 I = 10 … (i)

From loop (ii) dI 'L 10I 10I ' 0dt

L dI 'I I '10 dt

… (ii)

10

2

L

I’

(i-i )’

i1

2

Solving simultaneously (i) and (ii) we have 5t3LI ' 5 5e

…(iii)

and 5t3L25I 5 e

6

…(iv)

1000t35I I ' e

6

47. The energy stored in the inductor at any instant t; 0 < t < will be

(A) 1000 t 231 (5 5e ) mJ

2

(B) 1000 t 23125 (1 e ) mJ

2

(C) 1000 t 2325 (1 e ) mJ

2

(D) 1000 t 235 (1 e ) mJ

2

Ans (B) 12

L1E LI2

1000t23

L125E (1 e )

2

mJ

48. The energy stored in the capacitor and inductor respectively as t will be (A) 1 mJ and 62.5 mJ (B) 62.5 mJ and 1 mJ (C) 2 mJ and 62.5 mJ (D) 1 mJ and 60 mJ. Ans (A)

3 2L

1E (t ) 5 10 (5 0)2

EL = 62.5 mJ


2C

1E (t ) CV2

61 20 10 100 1mJ2

Read the passage given below and answer questions 15 to 17 by choosing the correct alternative The figure represents variation of peak current i0 with applied frequency of the AC source of three different LCR circuits having different resistances. The value of inductance L and capacitance C are same for all the three circuits.

1

2

3

0

i0

49. If R1, R2 and R3 be the resistance of circuit 1, 2 and 3 respectively, then (A) R1 > R2 > R3 (B) R1 < R2 < R3 (C) R1 > R2 = R3 (D) R1 = R2 = R3

Ans (B) 0

0 22

i1R LC

For maximum current 0

01i L 0

R C .

50. If R1 = 1 , R2 = 5 , R3 = 10 and L = 900

mH, C = 40F, then the value of 0 is

(A) 250 Hz (B) 125 Hz (C) 2506

Hz (D) 2503

Hz

Ans (D)

01

2 LC

= 250 Hz3

51. In the previous question, the frequency with which energy oscillates between Electric Field Energy and Magnetic Field Energy, is

(A) 14 LC

(B) 1LC

(C) 12 LC

(D) the energy in the electric field does not oscillate


Ans (B)

= 12 LC

Choose the appropriate entry/entries from column II to match each of the entries of the column I. It is possible that an option(s) in column II may be valid more than once, for a given entry in column I.

52. Match the Following Column I Column II

(i) For square wave having peak value v0 (P) v0 > vrms > vav

(ii) For sinusoidal wave having peak value v0 (Q) In a pure inductance. (iii) Current leads the voltage by /2 (R) vav = vrms = v0

(iv) Wattless current (S) In a pure capacitance Ans (i) – (R); (ii) – (P); (iii) – (S); (iv) – (Q), (S) For sinusoidal wave ;

0 0rms av

v 2vv ;v2

Wattless current in a pure inductance or pure capacitance When angle between voltage and current will be 90°.

53. Match the following Column I Column II

(i) In L–R series circuit if switch is closed at t = 0 (u sin of DC source)

(P) Current at t = 0 is non–zero

(ii) In L–C series combination switch is closed at t = 0 (if initially the capacitor is fully charged

(Q) Nothing can be said about the current

(iii) If voltage V=V0 sin t is applied to pure inductor at t = 0

(R) Current in the circuit is zero at t = 0

(iv) If voltage V = V0 sin t is applied at t = 0 to L–C–R series circuit

(S) Magnetic field energy in inductor is zero at t = 0

Ans: (i) – (R), (S); (ii) – (R), (S); (iii) – (P); (iv) – (Q)

* * *

physics-2iit (spark) worksheet-10 q + soln (ashwani)

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