Question

Find charge on capacitor at steady state.

• A. \(18\,\mu C\)
• B. \(16\,\mu C\)
• C. \(10\,\mu C\)
• D. \(8\,\mu C\)

Answer 1

The Correct Option is A

Concept: At steady state in a DC circuit:
• Capacitor behaves as an open circuit.
• No current flows through the capacitor branch.
• First solve the resistive network to find the potential difference across the capacitor.
• Then use \[ q = C\Delta V \]

Step 1:Open the capacitor branch. Since steady state is reached, \[ i_C = 0 \] Thus the remaining circuit contains only resistors.
Step 2:Find equivalent resistance of the network. After simplification, \[ R_{eq} = 10\Omega \]
Step 3:Find total current from the battery. \[ I = \frac{V}{R} \] \[ I = \frac{12}{10} \] \[ I = 1.2\,A \]
Step 4:Determine branch currents.
Current divides equally through the two \(20\Omega\) resistors: \[ I_1 = I_2 = \frac{I}{2} = 0.6\,A \] Further current division gives \[ I_3 = I_4 = \frac{I_2}{2} = 0.3\,A \]
Step 5:Find potential difference across the capacitor.
Voltage across the \(10\Omega\) resistor: \[ \Delta V = IR \] \[ \Delta V = 0.3 \times 10 \] \[ \Delta V = 3\,V \]
Step 6:Calculate charge on the capacitor. \[ q = C\Delta V \] \[ q = 6\mu F \times 3 \] \[ q = 18\,\mu C \] \[ \boxed{q = 18\,\mu C} \]