Question

In the circuit shown in the figure $C_1=2C_2$, switch $S$ is closed at time $t=0$. Let $i_1$ and $i_2$ be the currents flowing through $C_1$ and $C_2$ at any time $t$, then the ratio $i_1/i_2$ (assume capacitors were initially uncharged.)

• A. is constant.
• B. increases with increase in time $t$.
• C. decreases with increase in time $t$.
• D. first increases then decreases.

Answer 1

The correct option is B increases with increase in time $t$.

The current through $C_1$ at any time t is $i_1=i_{10}{e}^{-t/C_{1}R}$

and current through $C_2$ at any time t is $i_2=i_{20}{e}^{-t/C_{2}R}$

$\frac{i_1}{i_2}=\frac{i_{10}}{i_{20}}{e}^{(t/C_{2}R-t/C_{1}R)}=\frac{i_{10}}{i_{20}}{e}^{t/2C_{2}R}$ as $C_1=2C_2$

Thus $i_1/i_2$ will increase with time.