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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Links to the Chapters
Links to the Chapters
CHAPTER- 32- Electric Current in ConductorsCHAPTER- 31- CapacitorsCHAPTER- 30- Gauss's Law
CHAPTER- 29- Electric Field and Potential
CHAPTER- 28- Heat Transfer
OBJECTIVE -I
CHAPTER- 26-Laws of Thermodynamics
CHAPTER- 25-CALORIMETRY
Questions for Short Answer
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-11 to Q-18
CHAPTER- 24-Kinetic Theory of Gases
CHAPTER- 23 - Heat and Temperature
CHAPTER- 21 - Speed of Light
CHAPTER- 20 - Dispersion and Spectra
CHAPTER- 19 - Optical Instruments
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 32- Electric Current in Conductors
CHAPTER- 31- Capacitors
CHAPTER- 30- Gauss's Law
CHAPTER- 29- Electric Field and Potential
CHAPTER- 28- Heat Transfer
CHAPTER- 26-Laws of Thermodynamics
CHAPTER- 25-CALORIMETRY
Questions for Short Answer
OBJECTIVE-I
OBJECTIVE-II
CHAPTER- 24-Kinetic Theory of Gases
CHAPTER- 23 - Heat and Temperature
CHAPTER- 21 - Speed of Light
CHAPTER- 20 - Dispersion and Spectra
CHAPTER- 19 - Optical Instruments
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
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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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CHAPTER- 3 - Kinematics - Rest and Motion
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CHAPTER- 2 - "Physics and Mathematics"
CHAPTER- 2 - "Physics and Mathematics"
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