Question

A coil of self inductance $10\text{mH}$ and resistance $0.1\Omega$ is connected through a switch to a battery of internal resistance $0.9\Omega$. After the switch is closed, the time taken for the current to attain $80\%$ of the saturation value is
$[\text{take}\ln 5=1.6]$

• A. $0.002\text{s}$
• B. $0.103\text{s}$
• C. $0.324\text{s}$
• D. $0.016\text{s}$

Answer 1

The correct option is D $0.016\text{s}$

Given:
self inductance $=10\text{mH}$
resistance $R_L=0.1\Omega$
Battery internal resistance $r=0.9\Omega$
Net resistance of the circuit
$R=R_L+r=0.1\Omega +0.9\Omega =1\Omega$
Current flow in closed circuit is given as

$I=I_0(1-e^{-\frac{Rt}{L}})$
For the current to become $80\%$ of the saturation value

$0.8I_0=I_0(1-e^{-\frac{1\times t}{0.01}})$
$\Rightarrow 0.8=1-e^{-100t}$
$\Rightarrow e^{-100t}=0.2=\left(\frac{1}{5}\right)$
$\Rightarrow 100t=\ln 5\Rightarrow t=\frac{1}{100}\ln 5=0.016\text{s}$

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