Charging and Discharging of Capacitors

Practice Questions:

Question 1: moderate

In the circuit shown in the figure, the switch S is initially open and the capacitor is initially uncharged. I1, I2 and I3 represent the current in the resistance 2Ω, 4Ω and 8Ω respectively.

1. Just after the switch S is closed, I1 = 3A, I2 = 3A and I3 = 0

2. Just after the switch S is closed, I1 = 3A, I2 = 0 and I3 = 0

3. long time after the switch S is closed, I1 = 0.6 A, I2 = 0 and I3 = 0

4. long after the switch S is closed, I1 = I2 = I3 = 0.6 A.

View Answer

Here's the short solution for the circuit:

Just after the switch is closed:
- The capacitors act as open circuits because they are initially uncharged (capacitor voltage cannot change instantaneously).
- This means no current flows through the branches containing the capacitors.
Current Distribution:
- The total resistance in the circuit is only the sum of the resistors in the main loop (2Ω + 8Ω), as the branches with capacitors are effectively open.
- Total resistance = $2\Omega + 8\Omega = 10\Omega$
  
  .
- Current $I_1 = \frac{\text{Voltage}}{\text{Resistance}} = \frac{6V}{2\Omega + 8\Omega} = 0.6A$
  
  .
Branch currents:
- $I_3 = 0$
  
  since the capacitor branch is open.
- $I_2 = 0$
  
  for the same reason as above.

Final Answer:

$ I_1 = 0.6 A, \ I_2 = 0$

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Question 2: moderate

In the figure shown, the capacity of the condenser C is 2μF. The current in 2Ω resistor in steady state:

1. 9 A

2. 0.9 A

3. 1/9 A

4. 1/0.9 A

View Answer

In steady state no current flows through the capacitor so, current through 4 ohm resistor will be zero.

In absence of 4 ohm resistor, total resistance of circuit is (1.2+2.8)= 4 ohm.

Total current given by battery = 6/4=1.5 Ampere.

In parallel combination current divides in reverse ratio of resistors: (3/5)*1.5= 0.9 Ampere

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Question 3: moderate

The electric field between the plates of a parallel-plate capacitor of capacitance $2.0~\mu\text{F}$ drops to one third of its initial value in $4.4~\mu\text{s}$ when the plates are connected by a thin wire. Find the resistance of the wire.

1. $0.5~\Omega$

2. $0.1~\Omega$

3. $2~\Omega$

4. $1~\Omega$

View Answer

The electric field in a discharging capacitor drops as $E = E_0 e^{-t/RC}$. Given $E = E_0/3$, we have $RC = \frac{t}{\ln 3}$. Solving for $R = \frac{4.4 \times 10^{-6}}{2.0 \times 10^{-6} \times 1.1} = 2~\Omega$.

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Question 4: moderate

A $5.0~\mu\text{F}$ capacitor having a charge of $20~\mu\text{C}$ is discharged through a wire of resistance $5.0~\Omega$. Find the heat dissipated in the wire between 25 to 50 $\mu\text{s}$ after the connections are made.

1. $40\left(\frac{1}{e^2} - \frac{1}{e^4}\right)~\mu\text{J}$

2. $40\left(\frac{1}{e} - \frac{1}{e^2}\right)~\mu\text{J}$

3. $40\left(\frac{1}{e^2} - \frac{1}{e^3}\right)~\mu\text{J}$

4. None of these

View Answer

The remaining energy in the capacitor is $U(t) = \frac{q_0^2}{2C}e^{-2t/tau}$, where $tau = RC = 25~\mu\text{s}$. The heat dissipated is $H = U(t_1) - U(t_2) = \frac{q_0^2}{2C}\left(e^{-2} - e^{-4}\right)$ where $frac{q_0^2}{2C} = 40~\mu\text{J}$.

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Question 5: moderate

Let C be the capacitance of a capacitor discharging through a resistor R. Suppose $t_1$ is the time taken for the energy stored in the capacitor to reduce to half its initial value and $t_2$ is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio $t_1/t_2$ will be :

1. 2

2. 1

3. 1/2

4. ⇒1/4

View Answer

Energy is $U \propto q^2 \propto e^{-2t/RC}$, so $e^{-2t_1/RC} = 1/2 ⇒ t_1 = \frac{RC\ln 2}{2}$. Charge is $q \propto e^{-t/RC}$, so $e^{-t_2/RC} = 1/4 t_2 = 2RC\ln 2$. Thus, the ratio $t_1/t_2 = 1/4$.

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Question 6: moderate

A capacitor of capacitance $C$ is connected to a battery of emf $\varepsilon$ at $t = 0$ through a resistance $R$. Find the maximum rate at which energy is stored in the capacitor. When does the rate has this maximum value ?

1. $\frac{\varepsilon^2}{4R}$

2. $\frac{\varepsilon^2}{2R}$

3. $RC$

4. $CR \ln 2$

View Answer

The energy stored rate is $P = \frac{\varepsilon^2}{R} e^{-t/RC}(1-e^{-t/RC})$. This is maximum when $e^{-t/RC} = \frac{1}{2}$. The maximum rate is $P_{max} = \frac{\varepsilon^2}{4R}$. This occurs at $t = RC \ln 2$.

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