A coil of inductance 2 H 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 $$t = 0$$, then the energy stored in the coil after 4 s in J.

A coil of inductance $$L = 2\,\text{H}$$ with negligible resistance is connected to a voltage source $$V = 3t$$ volts. The voltage across an inductor is $$V = L\frac{dI}{dt}$$.

Substituting: $$3t = 2\frac{dI}{dt}$$, which gives $$\frac{dI}{dt} = \frac{3t}{2}$$.

Integrating from $$t = 0$$ (where $$I = 0$$): $$I = \frac{3t^2}{4}$$.

At $$t = 4\,\text{s}$$: $$I = \frac{3 \times 16}{4} = 12\,\text{A}$$.

The energy stored in the inductor is $$U = \frac{1}{2}LI^2 = \frac{1}{2} \times 2 \times (12)^2 = 144\,\text{J}$$.

Therefore, the energy stored in the coil after 4 seconds is $$144$$ J.