Question

A coil of resistance $20\Omega$ and inductance $5H$ has been connected to a $100V$ battery and the switch is closed at $t=0$. After how much time will the current acquire half of its steady state value. ($\ln 2=0.693)$

• A. $0.17s$
• B. $0.34s$
• C. $0.09s$
• D. $0.69s$

Answer 1

The correct option is A $0.17s$

If the current in the inductor is constant, there is no emf induced. But if the current changes, an emf is induced across the inductor.



emf induced $ϵ=-L\frac{di}{dt}=iR$

For the entire circuit,
$V-L\frac{di}{dt}-iR=0$
where $V=100V$ is the battery voltage.

Solving the above differential equation, we get
$i=i_0(1-e^{-\frac{t}{\tau }})$
where $i_0=\frac{V}{R}$ is the steady state current
and $\tau =\frac{L}{R}$

The graph of current w.r.t time is


$i=\frac{i_0}{2}=i_0(1-e^{-\frac{t}{\tau }})$
$\Rightarrow e^{-\frac{t}{\tau }}=2$
$\Rightarrow \frac{t}{\tau }=\ln 2$
$\Rightarrow t=\ln 2\times \tau =0.693\times \frac{L}{R}$
$=0.693\times \frac{5}{20}=0.17s$