Question

An ideal coil of $10\text{H}$ is connected in series with a resistance of $5\Omega$ and a battery of $5\text{V}.2\text{sec}$ after the connection is made, the current flowing (in ampere) in the circuit is

• A. $(1-e)$
• B. $e$
• C. $e^{-1}$
• D. $(1-e^{-1})$

Answer 1

The correct option is D $(1-e^{-1})$

Given:
$R=5\Omega ,L=10\text{H},E=5\text{V},(t=2\text{sec}$



Rise of current in $\text{L-R}$ circuit is given by , $I=I_0(1-e^{-t/\tau })$

Where , $I_0=\frac{E}{R}=\frac{5}{5}=1\text{A}$

Now, time constant , $\tau =\frac{L}{R}=\frac{10}{5}=2\text{sec}$

After $2\text{sec}$

Rise of current $I^{\prime }=1\times [1-e^{-\frac{2}{2}}]$

$I^{\prime }=(1-e^{-1})\text{A}$

Hence, option (d) is the correct answer.