JEE Mains · Physics · STD 12 -6. Electromagnetic induction
An inductor coil stores \(64\, {J}\) of magnetic field energy and dissipates energy at the rate of \(640\, {W}\) when a current of \(8\, {A}\) is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit in seconds
- A \(0.4\)
- B \(0.8\)
- C \(0.125\)
- D \(0.2\)
Answer & Solution
Correct Answer
(D) \(0.2\)
Step-by-step Solution
Detailed explanation
\(\mathrm{U}=\frac{1}{2} \mathrm{Li}^{2}=64 \Rightarrow \mathrm{L}=2\) \(\mathrm{i}^{2} \mathrm{R}=640\) \(\mathrm{R}=\frac{640}{(8)^{2}}=10\) \(\tau=\frac{\mathrm{L}}{\mathrm{R}}=\frac{1}{5}=0.2\)
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