Question

A coil of inductance $2H$ having negligible resistance is connected to a source of supply whose voltage is given by $V=3t$ volt (where $t$ is in second). If the voltage is applied when $t=0s$, then the energy stored in the coil after $4s$ is _______ $J$.

Answer 1

Step1: Given data.

Inductance of the coil, $L=2H$

Supply voltage, $e=3tv$

Time interval $t$ from $0sto4s$

Step2. Finding the current flowing in the coil.

We know that,

$e=L\frac{dI}{dt}$ ….$\left(i\right)$

Where $dI$change in current in the coil with respect to time $t$.

Now,

$\Rightarrow$$e\times dt=L\times dI$ ….$\left(ii\right)$

Integrating equation $\left(ii\right)$ both side we get.

$\Rightarrow$$L{\int }_0^{I}dI={\int }_0^{4}3t.dt$

$\Rightarrow$ $2I={\left[\frac{3t^2}{2}\right]}_0^4$

$\Rightarrow$ $I=\frac{3\times 16}{2\times 2}$

$\Rightarrow$ $I=12A$ ….$\left(iii\right)$

Step3: Finding the stored energy in the coil.

Formula used:

$U=\frac{1}{2}L{I}^2$ $\left[\because U=storedenergy\right]$

Now,

$\Rightarrow$$U=\frac{1}{2}\times 2H\times {12}^2{A}^2$

$\Rightarrow$$U=144J$

Hence, the energy stored in the coil after $4s$ is $144$ $J$.