KKMishra's Tutorials: H C Verma solutions, ELECTRIC CURRENT IN CONDUCTORS, Exercises, Q81-Q84, Chapter-32, Concepts of Physics, Part-II

Electric Current in Conductors

Exercises, Q81 - Q84

81. A capacitor of capacitance 100 µF is connected across a battery of emf 6·0 V through a resistance of 20 kΩ for 4·0 s. The battery is then replaced by a thick wire. What will be the charge on the capacitor 4·0 s after the battery is disconnected?

ANSWER: emf, ℰ =6.0 volt, R =20 kΩ =2x10⁴ Ω, C = 100 µF =1x10⁻⁴ F, t = 4 s.

t/RC =4/(2x10⁴*1x10⁻⁴)

=4/2 =2

Hence the charge stored on the capacitor after 4.0s,

q =ℰC(1-e^-t/RC)

=ℰC(1 -e⁻²)

=6x1x10⁻⁴(1 -0.135) C

=6x10⁻⁴*0.865 C

=5.18x10⁻⁴ C

When the battery is displaced by a thick wire, the capacitor starts discharging. In this case the charge after time t,

q' =q*e^-t/RC

=5.18x10⁻⁴*e⁻²

=5.18x10⁻⁴*0.135 C

=0.70x10⁻⁴ C

=70x10⁻⁶ C

=70 µC.

82. Consider the situation shown in figure (32-E36). The switch is closed at t =0 when the capacitors are uncharged. Find the charge on the capacitor C₁ as a function of time t.
The figure for Q - 82

ANSWER: Equivalent capacitance, C of the two capacitors connected in series is given as,

1/C =1/C₁ +1/C₂

→C =C₁C₂/(C₁+C₂)

emf of the battery = ℇ, Resistance through which the capacitors are being charged =r,

Hence the charge on the equivalent capacitor at time t,

q =ℇC(1 -e^-t/rC)

Since the capacitors are connected in series, the same charge will be there on each capacitor. Hence, the charge on capacitor C₁ =q =ℇC(1 -e^-t/rC)

where C = C₁C₂/(C₁ +C₂).

83. A capacitor of capacitance C is given a charge Q. At t =0, it is connected to an uncharged capacitor of equal capacitance through a resistance R. Find the charge on the second capacitor as a function of time.

ANSWER: Suppose a charge Q is given to the capacitor A. Then Q = CV, where V is the potential difference across the plates.

After time t the charge on the 2nd capacitor B = q (say).

Now q =CV' where V' is the potential difference across the capacitor B.

So, V' =q/C.

If the remaining charge on the capacitor A after time t =q', then q' =CV". V" being potential difference across B after time t. Then, V" = q'/C.
The diagram for Q-83

Now the potential difference between the plates of A and B after time t = V" -V'.

Hence the current in the resistance R, =(V" -V')/R. But also current = dq/dt, thus,

dq/dt = (V" -V')/R

=(q'/C -q/C)/R.

=(q' -q)/RC

=(Q-q-q)/RC

=(Q -2q)/RC.

Hence dt/RC = dq/(Q -2q)

Integrating both sides,.

[t/RC] = -½*[ln(Q -2q)]

Putting the limits of integration from t =0, q =0 to t = t, q =q.

t/RC =-½[ln(Q -2q) -ln Q]

→-2t/RC =ln[(Q -2q)/Q]

→e^-2t/RC =(Q -2q)/Q

→2q = Q -Qe^-2t/RC

→q = ½Q(1 -e^-2t/RC)

84. A capacitor of capacitance C is given a charge Q. At t =0, it is connected to an ideal battery of emf ℇ through a resistance R. Find the charge on the capacitor at time t.

ANSWER: Suppose, after time t, a charge q remains on the capacitor. The charge flowing out of the capacitor

=Q -q.

The potential difference across the capacitor, V =q/C.

The potential difference between battery and capacitor = V -ℇ

=q/C -ℇ

The current at this instant

=(q/C -ℇ)/R.

The current is the charge flowing out of the capacitor, i =d(Q -q)/dt =-dq/dt

Hence (q/C -ℇ)/R =-dq/dt

→-dt/CR =dq/(q -ℇC)

Integrating both sides,

[-t/CR] =[ln(q -ℇC)]

Limits of integration, at t =0, q =Q.

At t =t, q =q. Let us put these limits.

-t/CR =ln(q -ℇC) -ln(Q -ℇC)

→ln{(q -ℇC)/(Q -ℇC) = -t/CR

→(q -ℇC)/(Q -ℇC) =e^-t/CR

→q -ℇC =(Q -ℇC)e^-t/CR

→q =ℇC +Qe^-t/CR -ℇCe^-t/CR

=ℇC(1 -e^-t/CR) +Qe^-t/CR

It is an interesting result. The first term shows that the capacitor is being charged by a battery of emf ℇ through a resistance R as if is uncharged initially, while the other term shows that the initial charge Q given to the capacitor decays as if the plates are connected through a resistance R.

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CHAPTER- 32- Electric Current in Conductors

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 to Q10

EXERCISES - Q11 to Q20

EXERCISES - Q21 to Q30

EXERCISES - Q31 to Q40

EXERCISES - Q41 to Q50

EXERCISES - Q51 to Q60

EXERCISES - Q61 to Q70

EXERCISES - Q71 to Q80

CHAPTER- 31- Capacitors

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 to Q10

EXERCISES - Q11 to Q20

EXERCISES - Q21 to Q30

EXERCISES - Q31 to Q40

EXERCISES - Q41 to Q50

EXERCISES - Q51 to Q60

EXERCISES - Q61 to Q68

CHAPTER- 30- Gauss's Law

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 - Q10

EXERCISES - Q11 - Q20

EXERCISES - Q21 - Q24

CHAPTER- 29- Electric Field and Potential

Questions for Short Answer

OBJECTIVE-I

OBJECTIVE -II

EXERCISES - Q1 - Q10

EXERCISES - Q11 - Q20

EXERCISES - Q21 - Q30

EXERCISES - Q31 - Q40

EXERCISES - Q41 - Q50

EXERCISES - Q51 - Q60

EXERCISES - Q61 - Q70

EXERCISES - Q71 - Q75

CHAPTER- 28- Heat Transfer

Questions for Short Answer

OBJECTIVE -I

OBJECTIVE - II

EXERCISES - Q1 to Q10

EXERCISES - Q11 to Q20

EXERCISES - Q21 to Q30

EXERCISES - Q31 to Q40

EXERCISES - Q41 to Q50

EXERCISES - Q51 to Q55

CHAPTER- 27-Specific Heat Capacities of Gases

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE -II

EXERCISES - Q1 - Q10

EXERCISES - Q11 - Q20

EXERCISES - Q21 - Q30

EXERCISES - Q31 - Q35

CHAPTER- 26-Laws of Thermodynamics

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 to Q10

EXERCISES - Q11 to Q22

CHAPTER- 25-CALORIMETRY

Questions for Short Answer

OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q1 to Q10

EXERCISES - Q-11 to Q-18

CHAPTER- 24-Kinetic Theory of Gases

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES- Q1 to 10

EXERCISES- Q11 to 20

EXERCISES- Q21 to 30

EXERCISES- Q31 to Q40

EXERCISES- Q41 to Q50

EXERCISES- Q51 to Q62

CHAPTER- 23 - Heat and Temperature

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-34

CHAPTER- 22 - Photometry

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES

CHAPTER- 21 - Speed of Light

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES

CHAPTER- 20 - Dispersion and Spectra

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES

CHAPTER- 19 - Optical Instruments

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q 1 to Q 12

EXERCISES - Q 13 to Q 24

CHAPTER- 18 - Geometrical Optics

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q 1 to Q10

EXERCISES- Q11 to Q20

EXERCISES- Q21 to Q30

EXERCISES- Q31 to Q40

EXERCISES- Q41 to Q50

EXERCISES- Q51 to Q60

EXERCISES- Q61 to Q70

EXERCISES- Q71 TO 79

CHAPTER- 17 - Light Waves

Questions for Short Answers

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1-Q10

EXERCISES - Q11-Q20

EXERCISES - Q21-Q30

EXERCISES - Q31-Q41

CHAPTER- 16 - Sound Waves

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES -Q-21 TO Q-30

EXERCISES -Q-31 TO Q-40

EXERCISES -Q-41 TO Q-50

EXERCISES -Q-51 TO Q-60

EXERCISES -Q-61 TO Q-70

EXERCISES - Q-71 TO Q-80

EXERCISES - Q-81 TO Q-89

CHAPTER- 15 - Wave Motion and Waves on a String

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-30

EXERCISES - Q-31 TO Q-40

EXERCISES - Q-41 TO Q-50

EXERCISES - Q-51 TO Q-57

CHAPTER- 14 - Fluid Mechanics

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES- Q-1 TO Q-10

EXERCISES- Q-11 TO Q-20

EXERCISES- Q-21 TO Q-32

CHAPTER- 13 - Fluid Mechanics

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES Q-1 TO Q-10

EXERCISES- Q11 TO Q20

EXERCISES Q-21 TO Q30

EXERCISES Q-31 TO Q35

CHAPTER- 12 - Simple Harmonic Motion

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES- Q1 TO Q10

EXERCISES- Q11 TO Q20

EXERCISES- Q21 TO Q30

EXERCISES- Q31 TO Q40

EXERCISES- Q41 TO Q50

EXERCISES- Q51 TO Q58 (2-Extra Questions)

CHAPTER- 11 - Gravitation

Questions for Short Answers

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q_1 to Q_10

EXERCISES -Q-11 TO Q-20

EXERCISES -Q 21 TO 30

EXERCISES -Q 31 TO 39

CHAPTER- 10 - Rotational Mechanics

Questions for Short Answers

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 TO Q-15

EXERCISES - Q-16 TO Q-30

EXERCISES Q-31 TO Q-45

EXERCISES Q-46 TO Q-60

EXERCISES Q-61 TO Q-75

EXERCISES Q-76 TO Q-86

CHAPTER- 9 - Center of Mass, Linear Momentum, Collision

Questions for Short Answers

Objective - I

Objective - II

EXERCISES Q-1 TO Q-10

EXERCISES Q-11TO Q-20

EXERCISES Q-21 TO Q-30

EXERCISES Q-31 TO Q-42

EXERCISES Q-43 TO Q-54

CHAPTER- 8 - Work and Energy

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Click here for → Exercises (1-10)

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Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

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CHAPTER- 6 - Friction

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For more practice on problems on friction solve these- "New Questions on Friction".

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CHAPTER- 5 - Newton's Laws of Motion

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CHAPTER- 4 - The Forces

The Forces-

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Monday, June 21, 2021

H C Verma solutions, ELECTRIC CURRENT IN CONDUCTORS, Exercises, Q81-Q84, Chapter-32, Concepts of Physics, Part-II

81. A capacitor of capacitance 100 µF is connected across a battery of emf 6·0 V through a resistance of 20 kΩ for 4·0 s. The battery is then replaced by a thick wire. What will be the charge on the capacitor 4·0 s after the battery is disconnected?

ANSWER: emf, ℰ =6.0 volt, R =20 kΩ =2x10⁴ Ω, C = 100 µF =1x10⁻⁴ F, t = 4 s.

t/RC =4/(2x10⁴*1x10⁻⁴)

=4/2 =2

Hence the charge stored on the capacitor after 4.0s,

q =ℰC(1-e-t/RC)

=ℰC(1 -e⁻²)

=6x1x10⁻⁴(1 -0.135) C

=6x10⁻⁴*0.865 C

=5.18x10⁻⁴ C

When the battery is displaced by a thick wire, the capacitor starts discharging. In this case the charge after time t,

q' =q*e-t/RC

=5.18x10⁻⁴*e⁻²

=5.18x10⁻⁴*0.135 C

=0.70x10⁻⁴ C

=70x10⁻⁶ C

=70 µC.

82. Consider the situation shown in figure (32-E36). The switch is closed at t =0 when the capacitors are uncharged. Find the charge on the capacitor C₁ as a function of time t. The figure for Q - 82

ANSWER: Equivalent capacitance, C of the two capacitors connected in series is given as,

1/C =1/C₁ +1/C₂

→C =C₁C₂/(C₁+C₂)

emf of the battery = ℇ, Resistance through which the capacitors are being charged =r,

Hence the charge on the equivalent capacitor at time t,

q =ℇC(1 -e-t/rC)

Since the capacitors are connected in series, the same charge will be there on each capacitor. Hence, the charge on capacitor C₁ =q =ℇC(1 -e-t/rC)

where C = C₁C₂/(C₁ +C₂).

83. A capacitor of capacitance C is given a charge Q. At t =0, it is connected to an uncharged capacitor of equal capacitance through a resistance R. Find the charge on the second capacitor as a function of time.

ANSWER: Suppose a charge Q is given to the capacitor A. Then Q = CV, where V is the potential difference across the plates.

After time t the charge on the 2nd capacitor B = q (say).

Now q =CV' where V' is the potential difference across the capacitor B.

So, V' =q/C.

If the remaining charge on the capacitor A after time t =q', then q' =CV". V" being potential difference across B after time t. Then, V" = q'/C. The diagram for Q-83

Now the potential difference between the plates of A and B after time t = V" -V'.

Hence the current in the resistance R, =(V" -V')/R. But also current = dq/dt, thus,

dq/dt = (V" -V')/R

=(q'/C -q/C)/R.

=(q' -q)/RC

=(Q-q-q)/RC

=(Q -2q)/RC.

Hence dt/RC = dq/(Q -2q)

Integrating both sides,.

[t/RC] = -½*[ln(Q -2q)]

Putting the limits of integration from t =0, q =0 to t = t, q =q.

t/RC =-½[ln(Q -2q) -ln Q]

→-2t/RC =ln[(Q -2q)/Q]

→e-2t/RC =(Q -2q)/Q

→2q = Q -Qe-2t/RC

→q = ½Q(1 -e-2t/RC)

84. A capacitor of capacitance C is given a charge Q. At t =0, it is connected to an ideal battery of emf ℇ through a resistance R. Find the charge on the capacitor at time t.

ANSWER: Suppose, after time t, a charge q remains on the capacitor. The charge flowing out of the capacitor

=Q -q.

The potential difference across the capacitor, V =q/C.

The potential difference between battery and capacitor = V -ℇ

=q/C -ℇ

The current at this instant

=(q/C -ℇ)/R.

The current is the charge flowing out of the capacitor, i =d(Q -q)/dt =-dq/dt

Hence (q/C -ℇ)/R =-dq/dt

→-dt/CR =dq/(q -ℇC)

Integrating both sides,

[-t/CR] =[ln(q -ℇC)]

Limits of integration, at t =0, q =Q.

At t =t, q =q. Let us put these limits.

-t/CR =ln(q -ℇC) -ln(Q -ℇC)

→ln{(q -ℇC)/(Q -ℇC) = -t/CR

→(q -ℇC)/(Q -ℇC) =e-t/CR

→q -ℇC =(Q -ℇC)e-t/CR

→q =ℇC +Qe-t/CR -ℇCe-t/CR

=ℇC(1 -e-t/CR) +Qe-t/CR

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CHAPTER- 8 - Work and Energy

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q =ℰC(1-e^-t/RC)

q' =q*e^-t/RC

82. Consider the situation shown in figure (32-E36). The switch is closed at t =0 when the capacitors are uncharged. Find the charge on the capacitor C₁ as a function of time t.
The figure for Q - 82

q =ℇC(1 -e^-t/rC)

Since the capacitors are connected in series, the same charge will be there on each capacitor. Hence, the charge on capacitor C₁ =q =ℇC(1 -e^-t/rC)

If the remaining charge on the capacitor A after time t =q', then q' =CV". V" being potential difference across B after time t. Then, V" = q'/C.
The diagram for Q-83

→e^-2t/RC =(Q -2q)/Q

→2q = Q -Qe^-2t/RC

→q = ½Q(1 -e^-2t/RC)

→(q -ℇC)/(Q -ℇC) =e^-t/CR

→q -ℇC =(Q -ℇC)e^-t/CR

→q =ℇC +Qe^-t/CR -ℇCe^-t/CR

=ℇC(1 -e^-t/CR) +Qe^-t/CR

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CHAPTER- 6 - Friction

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