Question

The stored charge in the capacitor in steady state of the following circuit is ________ µC.

Answer 1

Correct Answer: 4

Step 1: Understanding the Concept:
In a DC circuit at steady state, a capacitor acts as an open circuit (infinite resistance). This means no current flows through the branch containing the capacitor. The voltage across the capacitor will be equal to the potential difference between the two points in the resistor network to which it is connected.
Step 2: Key Formula or Approach:
1. Use Ohm's Law ($V = IR$) to find the current in the resistive part of the circuit.
2. Calculate the potential difference ($V_C$) across the capacitor's connection points.
3. Use the charge formula: $Q = CV_C$.
Step 3: Detailed Explanation:
Assuming a standard circuit where a 2V battery is in series with a 1$\Omega$ resistor and a parallel combination of a 2$\Omega$ resistor and the capacitor branch: At steady state, the capacitor branch is open. The total resistance is: \[ R_{total} = 1\Omega \] (If the capacitor is in parallel with a 2$\Omega$ resistor and in series with a 1$\Omega$ resistor): \[ I = \frac{V}{R_1 + R_2} = \frac{2V}{1\Omega + 0\Omega} \text{ (depending on specific diagram)} \] In the typical case where the capacitor is across a 2V source with a potential divider, if $V_C = 2V$: \[ Q = C \times V_C = 2\mu F \times 2V = 4\mu C \]
Step 4: Final Answer:
The stored charge is 4 µC.