Question

Calculate the amount of charge on capacitor of 4 \(\mu\)F. The internal resistance of battery is 1 \(\Omega\): 

 

• A. Zero
• B. 4 \(\mu\)C
• C. 8 \(\mu\)C
• D. 16 \(\mu\)C

Hint:

The most important concept for DC circuits with capacitors is that in the steady state (after a long time), capacitors act like a break or an open switch in the circuit. No current flows *through* the capacitor's branch.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the charge stored on the 4 \(\mu\)F capacitor in the given DC circuit. The key is to analyze the circuit in the steady-state condition.
Step 2: Key Formula or Approach:
In a DC circuit, after a long time (at steady state), a capacitor acts as an open circuit. This is because once the capacitor is fully charged, no more current can flow through it. The charge on a capacitor is given by \(Q = CV\), where \(C\) is the capacitance and \(V\) is the potential difference across it.
Step 3: Detailed Explanation:
1. Steady-State Analysis: We assume the circuit has been connected for a long time, so it has reached a steady state. In this state, the capacitors are fully charged and block the flow of direct current. Therefore, the branch containing the 4 \(\mu\)F capacitor and the 6 \(\Omega\) resistor acts as an open circuit. Similarly, the branch with the two 2 \(\mu\)F capacitors also acts as an open circuit.
2. Current Flow: Since the branch with the 4 \(\mu\)F capacitor is an open circuit, the current flowing through this branch is zero. Let's call this current \(I_1 = 0\).
3. Voltage across the Resistor: The voltage drop across the 6 \(\Omega\) resistor is given by Ohm's law, \(V_{6\Omega} = I_1 \times R = 0 \times 6\Omega = 0\) V.
4. Voltage across the Capacitor: The 4 \(\mu\)F capacitor and the 6 \(\Omega\) resistor are in series. The potential difference across this series combination is the same. Since the voltage drop across the 6 \(\Omega\) resistor is zero, the two ends of the resistor are at the same potential. These two ends are also connected to the two plates of the 4 \(\mu\)F capacitor. Therefore, the potential difference across the 4 \(\mu\)F capacitor is also zero. \(V_{4\mu F} = 0\) V.
5. Charge Calculation: The charge stored on the capacitor is \(Q = C \times V\).
\[ Q_{4\mu F} = (4 \times 10^{-6} \text{ F}) \times (0 \text{ V}) = 0 \text{ C} \] Step 4: Final Answer:
The amount of charge on the 4 \(\mu\)F capacitor is zero.
(Note: Current will flow in the main loop consisting of the 5V battery, its 1\(\Omega\) internal resistance, and the 4\(\Omega\) resistor. But this does not affect the voltage across the upper branch).