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Q. A coil having a resistance of 20 ohm and inductance 20 henry is connected to 120 V battery. Find the value of energy stored in the magnetic field when the steady state is reached

Electromagnetic Induction

Solution:

Here, R = 20$\Omega$, L = 20 H and E = 120 V
$\therefore $ $I_0 = \frac{E}{R} = \frac{120}{20}$ = 6A
Energy stored = $\frac{1}{2} LI_0\,{}^2 = \frac{1}{2} \times 20 \times 36$ = 360 J