Question

A capacitor of \( 50 \, \mu F \) is connected in a circuit as shown in figure. The charge on the upper plate of the capacitor is \(\dots\dots\dots\) \( \mu C \). 

Hint:

Always solve the DC resistor network first by treating capacitors as "breaks" in the circuit to find the steady-state voltages at the capacitor nodes.

Answer 1

Correct Answer: 100

Step 1: Understanding the Concept:
In a DC circuit, after a long time (steady state), a capacitor behaves as an open circuit (infinite resistance). No steady current flows through the capacitor branch. The voltage across the capacitor will be equal to the potential difference across the branch it is connected in parallel with.
Step 2: Key Formula or Approach:
1. Ohm's Law: \( V = IR \).
2. Capacitance Equation: \( Q = CV \).
Step 3: Detailed Explanation:
1. Steady State Analysis:
Since the capacitor is an open circuit, the current \( I \) flows only through the three \( 2 \, k\Omega \) resistors which are in series.
\[ R_{total} = 2 \, k\Omega + 2 \, k\Omega + 2 \, k\Omega = 6 \, k\Omega \]
\[ I = \frac{V}{R_{total}} = \frac{6 \, V}{6 \, k\Omega} = 1 \, mA \]
2. Voltage across Capacitor:
The capacitor is connected in parallel with the bottom-most \( 2 \, k\Omega \) resistor. Therefore, the voltage across the capacitor \( V_c \) is:
\[ V_c = I \times 2 \, k\Omega = 1 \, mA \times 2 \, k\Omega = 2 \, V \]
3. Charge Calculation:
\[ Q = C \times V_c = 50 \, \mu F \times 2 \, V = 100 \, \mu C \]
Regarding the polarity: The upper terminal of the \( 6V \) source is positive. Following the potential drop, the junction at the top of the bottom resistor is at a higher potential than the negative terminal. Thus, the upper plate of the capacitor accumulates a positive charge of \( +100 \, \mu C \).
Step 4: Final Answer:
The charge on the upper plate is 100 \( \mu C \).