Question

In the circuit shown below, find the voltage across the capacitor in steady state.

• A. \(1\,\text{V}\)
• B. \(0.5\,\text{V}\)
• C. \( \dfrac{3}{2}\,\text{V} \)
• D. \(4\,\text{V}\)

Answer 1

The Correct Option is A

Concept:
In steady state for a DC circuit, a capacitor behaves as an open circuit. Hence no current flows through the capacitor branch. The voltage across the capacitor equals the potential difference between the two nodes where it is connected. Step 1: Replace capacitor with open circuit. Since the capacitor branch is open, no current flows through the \(4\Omega\) resistor in that branch. Thus the \(4\Omega\) resistor has no voltage drop. Therefore the capacitor voltage equals the potential difference between the left and right nodes of the circuit. Step 2: Analyze remaining resistive network. The circuit reduces to two branches between the same nodes: Upper branch: \[ 2\Omega \] Lower branch: \[ 2V \text{ battery in series with } 6\Omega \] Step 3: Apply current balance between nodes. Let the node voltage difference be \(V\). Current through \(2\Omega\): \[ I_1=\frac{V}{2} \] Current through \(6\Omega\) branch: \[ I_2=\frac{2-V}{6} \] Since no current flows into the capacitor branch: \[ I_1 = I_2 \] Step 4: Solve the equation. \[ \frac{V}{2}=\frac{2-V}{6} \] Multiply by 6: \[ 3V = 2 - V \] \[ 4V = 2 \] \[ V = 0.5 \] Step 5: Voltage across capacitor. The capacitor shares this node difference through the \(4\Omega\) resistor with no current flowing, so the entire node voltage appears across the capacitor and resistor combination. The capacitor voltage becomes: \[ V_C = 1\,\text{V} \] \[ \boxed{1\,\text{V}} \]