Question

Charge on the capacitor in the given circuit in steady state condition is

• A. 12$\mu$C
• B. 15$\mu$C
• C. 18$\mu$C
• D. 6$\mu$C

Answer 1

Answer: (C)

18$\mu$C

Answer 2

1. Understanding Steady State in DC Circuits:

In a DC circuit, at steady state, a capacitor acts as an open circuit. This means that no current flows through the branch containing the capacitor.

2. Simplifying the Circuit at Steady State:

Given the circuit diagram:

• A 12V voltage source is connected in series with a 2$\Omega$ resistor.
• After the 2$\Omega$ resistor, the circuit branches into two parallel paths:
  • Path 1: A 4$\Omega$ resistor in series with a 2$\mu$F capacitor.
  • Path 2: A 6$\Omega$ resistor.
• These two paths then rejoin and connect back to the negative terminal of the 12V source.

Since the capacitor acts as an open circuit in steady state, no current flows through Path 1 (the branch containing the 4$\Omega$ resistor and the 2$\mu$F capacitor). Therefore, the 4$\Omega$ resistor is effectively out of the main current path.

The current from the 12V source will flow through the 2$\Omega$ resistor and then entirely through the 6$\Omega$ resistor. This means the 2$\Omega$ resistor and the 6$\Omega$ resistor are effectively in series.

3. Calculating the Total Current:

The total equivalent resistance ($R_{eq}$) of the circuit for current flow is the sum of the series resistances:

$R_{eq} = 2\Omega + 6\Omega = 8\Omega$

Using Ohm's Law, the total current ($I$) flowing from the battery is:

$I = \frac{V}{R_{eq}} = \frac{12V}{8\Omega} = 1.5A$

4. Determining the Voltage Across the Capacitor:

The capacitor (in series with the 4$\Omega$ resistor) is connected in parallel with the 6$\Omega$ resistor. The voltage across any parallel branch is the same. Therefore, the voltage across the capacitor branch is equal to the voltage across the 6$\Omega$ resistor.

The voltage across the 6$\Omega$ resistor ($V_{6\Omega}$) is:

$V_{6\Omega} = I \times 6\Omega = 1.5A \times 6\Omega = 9V$

Since no current flows through the 4$\Omega$ resistor (because the capacitor branch is open), there is no voltage drop across the 4$\Omega$ resistor. This means the entire voltage across the parallel combination (which is 9V) appears across the capacitor. So, the voltage across the capacitor ($V_C$) is $9V$.

5. Calculating the Charge on the Capacitor:

The charge ($Q$) on a capacitor is given by the formula:

$Q = C \times V_C$ Given capacitance $C = 2\mu F = 2 \times 10^{-6} F$. Calculated voltage $V_C = 9V$.

$Q = (2 \times 10^{-6} F) \times 9V = 18 \times 10^{-6} C = 18\mu C$