Question

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

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

Answer 1

Answer: (D)

According to the question, The circuit as shown below,

The current in the circuit,
$I(t)=I(\infty )-[I(\infty )-I(0)]e^{-t/\tau }\dots$(i)
$\because$ Time constant, $\tau =\frac{L}{R}=\frac{5}{10}=\frac{1}{2}sec$
at $t\to \infty$, inductor behaves as short circuit hence circuit will be

So, $I(\infty )=\frac{15}{10}=1.5A\dots$(ii)
at $t\to 0$, inductor behaves as open circuit
so, no current flow in the circuit, $I(0)=0$
So from Eq. (i),
$\Rightarrow I(t)=1.5-(1.5-0)e^{\frac{-t}{(1/2)}}=1.5\left(1-e^{-2t}\right)$
at $t=1sec,\Rightarrow I(1)=1.5\left(1-e^{-2}\right)\dots$ (iii)
From Eq. (ii) and (iii), we get
$\Rightarrow \frac{I(\infty )}{I(1)}=\frac{1.5}{1.5\left(1-e^{-2}\right)}=\frac{e^2}{e^2-1}$