Q114 / 160 · JIPMER 2010 · Physics
A capacitor of capacitance $5 \mu \mathrm{F}$ 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$80 \mu \mathrm{C}$
- B$40 \mu \mathrm{C}$
- C$20 \mu \mathrm{C}$
- Correct answerD$10 \mu \mathrm{C}$
Solution
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,
$i=\frac{V}{R}=\frac{2.5}{2.5}=1 \mathrm{~A}$
Potential drop across two parallel branches
$\begin{aligned}
V=E-i r & =2.5-1 \times 0.5 \\
& =2.5-0.5 \\
& =2.0 \mathrm{~V}
\end{aligned}$
So, charge on the capacitor plates
$\begin{aligned}
q=C V & =5 \times 2 \\
& =10 \mu \mathrm{C}
\end{aligned}$
Hence, the correct answer is option: 'D'.