Question

In steady state, a capacitor of capacitance $2 \mu F$ is charged to $4\, \mu C$, as shown in figure. If the internal resistance of the cell is $0.5\, \Omega$, then the emf of the cell is

• A. 4 V
• B. 5 V
• C. 2.5 V
• D. 2 V

Answer 1

Answer: (C)

$V_{\text {cap }}=\frac{Q}{C}=\frac{4 \times 10^{-0}}{2 \times 10^{-6}}$
$=2\, v$
Now $V_{\text {cap }}= V$ (across $2\, \Omega$ resistor)
$\therefore I=\frac{V}{R}=\frac{2}{2}=1\, A$
Now, using the relation
$E =V+I r$
$=2+1 \times 0.5$
$=2.5\, V$