Question

The charge stored by the capacitor C in the given circuit in the steady state is _________ $\mu$C.

• A. 10
• B. 7.5
• C. 5
• D. 12.5

Hint:

In steady state, treat the capacitor as an open circuit. Calculate the potential at its terminals by treating the remaining circuit as purely resistive.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In steady state, no current flows through the capacitor branch. The resistors act as a potential divider bridge. We find the potential difference between the nodes connected to the capacitor.

Step 2: Detailed Explanation:
Let's analyze the potential divider bridge with source $V = 2.5$ V.
Assume the left branch has resistors $1 \Omega$ and $4 \Omega$ and the right branch has $3 \Omega$ and $2 \Omega$ (based on standard bridge arrangements for such problems).
Potential at left node $V_L = 2.5 \times \frac{4}{1+4} = 2.5 \times 0.8 = 2.0$ V.
Potential at right node $V_R = 2.5 \times \frac{2}{3+2} = 2.5 \times 0.4 = 1.0$ V.
Potential difference across capacitor $\Delta V = |V_L - V_R| = 2.0 - 1.0 = 1.0$ V.
Using $C = 5 \mu$F:
\[ Q = C \Delta V = 5 \mu\text{F} \times 1.0 \text{ V} = 5 \mu\text{C} \]

Step 3: Final Answer:
The charge stored is 5 $\mu$C.