Question

A coil of self inductance $50H$ is joined to the terminals of a battery of emf $2V$ through a resistance of $10\Omega$ and a steady current is flowing through the circuit. If the battery is now disconnected then time in which the current will decay to $\frac{1}{e}$ of its steady value is

• A. $500s$
• B. $50s$
• C. $5s$
• D. $0.5s$

Answer 1

The correct option is C $5s$

Current in circuit at steady state $=\frac{2}{10}=0.2A=I_0$

Now, decay of current;

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

as $I=\frac{1}{e}{I}_0$

$\therefore \frac{1}{e}{I}_0$ $=I_0{e}^{-\frac{t}{\tau }}$

$e^{-1}=e^{-\frac{t}{\tau }}$

or$-1=-\frac{t}{\tau }$

$\Rightarrow t=\tau =\frac{L}{R}=5s$