Self Induction: Practice Problem & Solution
A coil resistance 20 Ω and inductance 5 H is connected with a 100 V battery. Energy stored in the coil will be :
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Self Induction. Understanding the underlying formula is key to arriving at the correct answer below:
$$I = \frac{V}{R} = \frac{100}{20} = 5 \text{ A}$$
$$U = \frac{1}{2} L I^2 = \frac{1}{2} \times 5 \times 5^2 = 62.5 \text{ J}$$
Energy stored = 62.5 J
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