Question

An $L-R$ circuit is having $L=2.5\text{H}$, $R=1\Omega$ and $E=4\text{V}$ is switched on at $t=0.$ Find the power dissipated at $t=2.5\text{s}.$

• A. $16\text{W}$
• B. $25.50\text{W}$
• C. $8.75\text{W}$
• D. $6.25\text{W}$

Answer 1

The correct option is D $6.25\text{W}$

Given:
$L=2.5\text{H};R=1\Omega ;E=4\text{V};t=2.5\text{s}$

The time constant of the given circuit is,

$\tau =\frac{L}{R}=\frac{2.5}{1}=2.5\text{s}$

Using the relation for current growth,

$i=i_0(1-e^{-t/\tau })$

$\therefore$ The current at $t=2.5\text{s}$ is,

$i=\frac{E}{R}(1-e^{-t/\tau })$

$i=\frac{4}{1}(1-e^{-2.5/2.5})=4(1-e^{-1})$

$i=4\times 0.63\approx 2.5\text{A}$

The power dissipated in $t=2.5\text{s}$ is,

$P=i^{2}R=(2.5{)}^2\times 1=6.25\text{W}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{D}\right)$ is the correct answer.