Question

A capacitor of capacitance $5 \mu F$ is connected as shown in the figure. The internal resistance of the cell is $0.5 O$ 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)

$10 \mu C$

Answer 2

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