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. $10\mu C$

Answer 1

Answer: (D)

$10\mu C$

In steady state, there will be no current in the capacitor branch. Net resistance of the circuit
$R=1+1+0.5=2.5\Omega$
Current drawn from the cell
$i=\frac{V}{R}=\frac{2.5}{2.5}=1A$
Potential drop across two parallel branches
$V=E-ir=2.5-1\times 0.5$
$=2.5-0.5=2.0$V
So, charge on the capacitor plates
$q=CV=5\times 2=10\mu C$.