Question

A capacitor of capacitance $5\mu F$ is connected as shown in the figure. The internal resistance of the cell is $0.5\Omega$. The amount of charge on the capacitor plates is:

• A. $80\mu C$
• B. $40\mu C$
• C. $20\mu C$
• D. $12.5\mu C$

Answer 1

Answer: (D)

$12.5\mu C$

In steady state, the capacitor will be fully charged and the branch containing the capacitor will behave as open circuit. The voltage drop across the capacitor branch will be 2.5V (because it is in parallel across the battery).

Since the current through the capacitor branch will be zero in the steady state (the branch behaves as open circuit in steady state), all the voltage drop will occur across the capacitor. Therefore, charge in the capacitor will be: $Q=(2.5\times 5)\mu F=12.5\mu C$