Solution:
$$I = \frac{V}{R} = \frac{100}{20} = 5 \text{ A}$$
$$U = \frac{1}{2} L I^2 = \frac{1}{2} \times 5 \times 5^2 = 62.5 \text{ J}$$
Energy stored = 62.5 J
A coil resistance 20 Ω and inductance 5 H is connected with a 100 V battery. Energy stored in the coil will be :
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