Question

An emf of $15V$ is applied in a circuit of inductance $5H$ and resistance $10\Omega$. The ratio of currents flowing at $t=\infty$ and $t=1s$ will be

• A. $1-e^{-1}$
• B. $e^{-1}$
• C. $\frac{e^2}{e^2-1}$
• D. $\frac{e^{1/2}}{e^{1/2}-1}$

Answer 1

The correct option is C $\frac{e^2}{e^2-1}$

We have
$T=\frac{L}{R}=\frac{5}{10}=\frac{1}{2}s$
so, a time of $1s$ means two time constants
$\therefore$ Current after $1s$ will be given by the following equation
$I=I_{max}(1-e^{-t/T})=I_{max}(1-e^{-2})$
Now at $t=\alpha$, the current attains steady value
$\therefore \frac{I_{max}}{I}=\frac{1}{(1-e^{-2})}=\frac{e^2}{e^2-1}$