Question

A coil having inductance $L$ and resistance $R$ is connected to a battery of emf $\epsilon $ at $t=0$ . If $t_{1}$ and $t_{2}$ are the times for $90\%$ and $99\%$ completion of the current growth in the circuit, then $\frac{t_{1}}{t_{2}}$ will be

• A. $1:2$
• B. $2:1$
• C. $1:5$
• D. $5:1$

Answer 1

Answer: (A)

$I=I_{0}\left(1 - \left(\text{e}\right)^{- \frac{t}{\tau}}\right)$
$\frac{9}{1 0}I_{0}=I_{0}\left(1 - \left(\text{e}\right)^{- \frac{t_{1}}{\tau}}\right)$
$\Rightarrow \text{e}^{- \frac{t_{1}}{\tau}}=\frac{1}{1 0}$
$\Rightarrow \text{e}^{\frac{t_{1}}{\tau}}=10$
$\Rightarrow t_{1}=\tau ln10$ ............(1)
and $\frac{9 9}{1 0 0}I_{0}=I_{0}\left(1 - \left(\text{e}\right)^{- \frac{t_{2}}{\tau}}\right)$
$\Rightarrow \text{e}^{- \frac{t_{2}}{\tau}}=\frac{1}{1 0 0}$
$\Rightarrow t_{2}=\tau ln100=2\tau ln10$ ..............(2)
From equations (1) and (2), we have
$\Rightarrow \frac{t_{1}}{t_{2}}=\frac{1}{2}$