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 is ____ J.

Explanation:

Relation among voltage(V), current(i) and inductance(L) is given as:

V = L\(\frac{di}{dt}\)  ----(1)

Energy stored in the coil is given as:

E = \(\frac{1}{2}\)LI²   ----(2)

where I = current in the coil after time t s.

From equation (1) we have,

Ldi = Vdt ⇒\(L\int_{0}^{I}di = \int_{0}^{t}Vdt\)

Given,

L = 2 H, V = 3t, t = 4 s.

∴ \(L\int_{0}^{I}di = \int_{0}^{t}3tdt\) ⇒ LI = 3t²/2 ⇒ I = 3t2/2L  ----(3)

Putting the given values in equation (3) we get-

I = 12 A.

Now using equation (2) we get,

E = \(\frac{1}{2}\)LI2 = 144 J.

Hence E = 144 J is the correct answer.