Question

In an electrical circuit drawn below, the amount of charge stored in the capacitor is ______ $\mu C$.

Answer 1

Correct Answer: 60

Circuit Description:

The circuit comprises a 10V battery, three resistors with values \( R_1 = 4 \, \Omega \), \( R_2 = 5 \, \Omega \), \( R_3 = 6 \, \Omega \), and a capacitor \( C = 10 \, \mu F \).

Steady State Analysis:

In steady state, no current flows through the capacitor branch. Consequently, the voltage drop across \( R_2 = 5 \, \Omega \) is zero, meaning \( I_2 = 0 \).

Step 1: Current Calculation:

The current in the remaining circuit is calculated as:

\[ I_1 = I_3 = \frac{10}{4 + 6} = 1 \, \text{A} \]

Step 2: Voltage Distribution:

The voltage drop across \( R_3 \) is related to the capacitor and \( R_2 \) voltages by:

\[ V_{R_3} = V_c + V_{R_2}, \quad V_{R_2} = 0 \]

Step 3: Capacitor Voltage Derivation:

As no current passes through the capacitor, the voltage across \( R_3 \) equals the voltage across the capacitor:

\[ I_3 R_3 = V_c \]

Step 4: Numerical Voltage Result:

Substituting the known values yields:

\[ V_c = 1 \times 6 = 6 \, \text{V} \]

Step 5: Charge Calculation:

The charge stored on the capacitor is determined by:

\[ q_c = C V_c = 10 \times 6 = 60 \, \mu C \]

Conclusion:

The charge accumulated on the capacitor is \( 60 \, \mu C \).