Question

A coil having inductance $L$ and resistance $R$, is connected to a battery of emf $E$ and the circuit is closed at time, $t=0$. If $t_1$ and $t_2$ are time instants for $90\%$ and $99\%$ completion of current growth in the circuit, then $\frac{t_1}{t_2}$ will be equal to -

• A. $1:3$
• B. $3:1$
• C. $2:1$
• D. $1:2$

Answer 1

The correct option is D $1:2$


Current growth in an LR circuit is given by,

$I=I_0(1-e^{-\frac{t}{\tau }})$

So,

$0.9I_0=I_0(1-e^{-\frac{t_1}{\tau }})$

$\Rightarrow e^{-\frac{t_1}{\tau }}=0.1$

$\Rightarrow -\frac{t_1}{\tau }=\ln \left(0.1\right)$

$\Rightarrow t_1=\tau \ln 10...\left(i\right)$

Similarly,

$t_2=2\tau \ln 10...\left(ii\right)$

From equations $\left(i\right)$ and $\left(ii\right)$,

$\frac{t_1}{t_2}=\frac{\tau \ln 10}{2\tau \ln 10}=\frac{1}{2}=1:2$

Hence, option $\left(D\right)$ is the correct answer.