Question

Capacitor $C_1$ of capacitance $1\mu F$ and capacitor $C_2$ of capacitance $2\mu F$ are separately charged fully by a common battery. The two capacitors are then separately allowed to discharge through equal resistors at time $t=0$. Then,

• A. The current in each of the two discharging circuits is zero at $t=0$.
• B. The current in the two discharging circuits at $t=0$ are equal but not zero.
• C. The current in the two discharging circuits at $t=0$ are unequal.
• D. Capacitor $C_1$ loses 50% of its initial charge sooner than $C_2$ loses 50% of its initial charge.

Answer 1

Answer: (B), (D)

The correct options are
B The current in the two discharging circuits at $t=0$ are equal but not zero.
D Capacitor $C_1$ loses 50% of its initial charge sooner than $C_2$ loses 50% of its initial charge.

Charging circuit


The discharging current in the circuit is,

For a discharging circuit of capacitor $C$
$I=I_0{e}^{-t/RC}$
Hence, $I_0=$ initial current $=\frac{V}{R}$
$I=\frac{V}{R}{e}^{-t/RC}$
When, $t=0,I=I_0=\frac{V}{R}$
Hence option $a,c$ are incorrect, $b$ is correct.

$d-$
Since, potential difference between the plates is same initially. Therefore, I is same in both the cases $t=0$ and is equal but not zero.
$I=\frac{q_0}{RC}=\frac{V}{R}$
Also, for discharging circuit $q=q_0{e}^{-t/RC}$
When $q=\frac{q_0}{2}$ then $\frac{q_0}{2}=q_0{e}^{-t/RC}$
$e^{t/RC}=2\Rightarrow \therefore t=RC{\log }_{e}2$

$t_1$ be time taken for charge to be half on capacitor $C_1=1\mu F$
$t_2$ be time taken for charge to be half on capacitor $C_2=2\mu F$
Now ratio of times for capacitors is
$\frac{t_1}{t_2}=\frac{C_1}{C_2}=0.5$
$\therefore t_1=0.5t_2$
So, the time taken for the first capacitor for discharging $50\%$ of initial charge, will be less.