Question

An inductor (L = 20 H), a resistor (R = 100 $\Omega$) and a battery (E = 10 V) are connected in series. After a long time, the circuit is short-circuited and then the battery is disconnected. Find the current in the circuit at 1 ms after short circuiting.

• A. $4.5\times {10}^{5}A$
• B. $3.2\times {10}^{-5}A$
• C. $9.9\times {10}^{-2}A$
• D. $6.7\times {10}^{-4}A$

Answer 1

The correct option is C $9.9\times {10}^{-2}A$

Given data :

$L=20H$

$R=100\Omega$

$E=10V$

Time after short circuiting the circuit, $t=1ms.$

The initial current, $I^{\prime }$ in the circuit is,

$I^{\prime }=\frac{E}{R}$

$I^{\prime }=\frac{10}{100}$

$I^{\prime }=0.1A$ ........(1)

The time constant, $\tau$ is,

$\tau =\frac{L}{R}$

$\tau =\frac{20}{100}$

$\tau =0.2s$ ........(2)

Now, the current at $t=1ms=0.001s$, is:

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

Using equations (1) and (2) here, we get:

$I=\left(0.1\right)\left(e^{-\frac{0.001}{0.2}}\right)$

$I=\left(0.1\right)\left(e^{-0.005}\right)$

$I=\left(0.1\right)\left(0.995\right)$

$I=0.0995A$

$I=9.95\times {10}^{-2}A$