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 plate is

• A. $10\mu C$
• B. $5\mu C$
• C. $6\mu C$
• D. $10.2\mu C$

Answer 1

Answer: (A)

We know that capacitor offers infinite resistance for DC source.
Current taken from the cell
$i=\frac{E}{R+r}$
$\therefore i=\frac{2.5}{1+1+0.5}=1A$
Potential drop across capacitor
$V=E-ir$
$=2.5-1\times 0.5=2V$
The current in $2\Omega$ resistor is zero.
Charge on capacitor
$Q=CV$
$=5\times 2=10\mu C$