Question

Calculate the stored charge and potential difference between the plates in steady state for both the capacitors as shown in the circuit:

Hint:

For capacitors in parallel, the total capacitance is the sum of the individual capacitances, and the voltage across each capacitor is the same.

Answer 1

Step 1: Given Values.
The circuit consists of two capacitors in parallel, with the following values: - \( C_1 = 2 \, \mu F \), - \( C_2 = 3 \, \mu F \), - \( R = 1 \, \Omega \), - \( V = 12 \, V \). Since the capacitors are in parallel, the total capacitance \( C_{\text{total}} \) is the sum of the individual capacitances: \[ C_{\text{total}} = C_1 + C_2 = 2 \, \mu F + 3 \, \mu F = 5 \, \mu F \]
Step 2: Finding the Charge on the Capacitors.
The charge stored on each capacitor in steady state is given by: \[ Q = C_{\text{total}} \times V \] \[ Q = 5 \, \mu F \times 12 \, V = 60 \, \mu C \]
Step 3: Finding the Potential Difference Across Each Capacitor.
Since the capacitors are in parallel, the potential difference across both capacitors is the same as the applied voltage: \[ V_{\text{across capacitors}} = 12 \, V \]
Step 4: Conclusion.
Thus, the stored charge on the capacitors is \( 60 \, \mu C \), and the potential difference across each capacitor is \( 12 \, V \).