Question

Under steady state condition the potential difference across the capacitor in the circuit is _____ V. 
 

• A. 0.5
• B. 1.5
• C. 0
• D. 2

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In steady state, a capacitor acts as an open circuit (infinite resistance). No current flows through the branch containing the capacitor. We calculate the potential at the nodes around it using Ohm's Law.
Step 2: Detailed Explanation:
In the steady state, the $2\mu F$ branch has zero current.
The total resistance of the main loop is $R = 6\Omega + 2\Omega = 8\Omega$.
Current in the circuit $I = \frac{V}{R} = \frac{2 V}{8 \Omega} = 0.25 A$.
The potential difference across the $2\Omega$ resistor is $V_{2\Omega} = I \times 2 = 0.25 \times 2 = 0.5 V$.
The capacitor is in parallel with the $2\Omega$ resistor's effective nodes? Let's check the loop.
The loop starts at the battery, goes through $6\Omega$, then splits. Current only goes through the $2\Omega$ resistor.
The potential drop across the $6\Omega$ resistor is $0.25 \times 6 = 1.5 V$.
Potential across the remaining branches is $2 V - 1.5 V = 0.5 V$.
Since the $4\Omega$ resistor has no current, the entire $0.5 V$ potential difference appears across the capacitor.
Step 3: Final Answer:
The potential difference across the capacitor is 0.5 V.