Question

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. \(4\,\text{V}\)

Hint:

In steady state DC circuits: \[ \text{Capacitor} \rightarrow \text{Open circuit} \] \[ \text{Inductor} \rightarrow \text{Short circuit} \] This greatly simplifies circuit analysis.

Answer 1

The Correct Option is B

Concept: In steady state (DC condition), a capacitor behaves as an open circuit. Thus, no current flows through the capacitor branch. Hence the potential difference across the capacitor equals the potential difference across the two nodes of that branch.
Step 1: Analyze the circuit at steady state. Since the capacitor branch carries no current, only the resistive loop with \(2\Omega\) and \(6\Omega\) remains active. Thus the total resistance in the loop is \[ R_{\text{total}} = 2 + 6 = 8\Omega \]
Step 2: Find circuit current. Using Ohm's law: \[ I = \frac{V}{R} \] \[ I = \frac{2}{8} \] \[ I = 0.25\,\text{A} \]
Step 3: Voltage across the \(2\Omega\) resistor. \[ V_{2\Omega} = IR \] \[ V_{2\Omega} = 0.25 \times 2 \] \[ V_{2\Omega} = 0.5\,\text{V} \] This is the same as the voltage across the capacitor. \[ \boxed{V_C = 0.5\,\text{V}} \]