NEET Charging and Discharging of Capacitors MCQs & PYQs

Here's the short solution for the circuit:

1. 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.
2. 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
     $I_1 = \frac{\text{Voltage}}{\text{Resistance}} = \frac{6V}{2\Omega + 8\Omega} = 0.6A$ 

     .

3. Branch currents:
   • 
     $I_3 = 0$ 

     since the capacitor branch is open.

   • 
     $I_2 = 0$ 

     for the same reason as above.

Final Answer:

• \( I_1 = 0.6 A, \ I_2 = 0\)

At t =0 capacitor behaves as closed circuit to the 1000 ohm resistor connected in parallel with capacitor will get short circuited.

current  through the other resistor = 2/1000 = 2mA

At t = infinite capacitor behaves as open circuit so equivalent resistance becomes R=1000+1000 = 2000 Ohm

current  through the  resistor = 2/2000 = 1 mA

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

The charging of a capacitor through a resistor is described by the following equation:

$$Q(t) = Q_{\text{max}} \left( 1 - e^{-t/\tau} \right)$$

Where:

• 
  $Q(t)$ 

  is the charge on the capacitor at time $t$ 

  ,

• 
  $Q_{\text{max}}$ 

  is the maximum (steady-state) charge the capacitor can hold,

• 
  $\tau$ 

  is the time constant, $\tau = R \cdot C$ 

  , where $R$ 

  is the resistance and $C$ 

  is the capacitance,

• 
  $t$ 

  is the time.

Step 1: Given condition (10% of steady charge)

We are told that at time

$t$

, the capacitor has collected 10% of the steady charge, so:

$$Q(t) = 0.1 \cdot Q_{\text{max}}$$

Step 2: Substitute into the charging equation

Substitute

$Q(t) = 0.1 \cdot Q_{\text{max}}$

into the charging formula:

$$0.1 \cdot Q_{\text{max}} = Q_{\text{max}} \left( 1 - e^{-t/\tau} \right)$$

Cancel

$Q_{\text{max}}$

from both sides:

$$0.1 = 1 - e^{-t/\tau}$$

Step 3: Solve for $t$

Rearrange the equation to solve for

$e^{-t/\tau}$

:

$$e^{-t/\tau} = 1 - 0.1 = 0.9$$

Take the natural logarithm of both sides:

$$-\frac{t}{\tau} = \ln(0.9)$$

$$t = -\tau \ln(0.9)$$

Using the fact that

$\ln(0.9) = -\ln(10/9)$

:

$$t = \tau \ln\left(\frac{10}{9}\right)$$

Final Answer:

The time at which the capacitor has collected 10% of the steady charge is

$t = \tau \ln\left(\frac{10}{9}\right)$

.

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 \(R = \frac{4.4 \times 10^{-6}}{2.0 \times 10^{-6} \times 1.1} = 2~\Omega\).

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 \(frac{q_0^2}{2C} = 40~\mu\text{J}\).

For any capacitor geometry, capacitance is proportional to \(k\epsilon_0\) and resistance is proportional to resistivity \(\rho\), such that the shape factors cancel in \(\tau = RC = \rho k \epsilon_0\).

Using \(V = V_0(1 - e^{-t/RC})\), we get \(4 = 12(1 - e^{-t/RC})⇒ e^{-t/RC} = 2/3\). Taking the natural logarithm, \(\frac{t}{RC} = \ln(1.5) = 0.4\), which yields \(C = \frac{10^{-6}}{10 \times 0.4} = 0.25~\mu\text{F}\).

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 \(t_1/t_2 = 1/4\).

The charging voltage is \( V = V_0(1 - e^{-t/RC})\), so \(120 = 200(1 - e^{-t/RC})\) ⇒ \(e^{t/RC} = 2.5\). This gives \(t/RC = \ln 2.5 = 2.303 \log_{10} 2.5 \approx 0.921\). Solving with \(t = 5~text{s}\) and \(C = 2~\mu\text{F}\) gives \(R \approx 2.7 \times 10^6~\Omega\).