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Consider the circuit given in the diagram. Switch S is closed at time t = 0. The current in the circuit increases with time. At some instant rate of energy storage in the inductor will become maximum. At that time find ratio of power supplied by battery and thermal power generated in resistor.

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Detailed Solution

$\frac{dU}{dt} = rate (r) = εI - I^{2} R \frac{dr}{dt} = ε - 2 IR = 0 \Rightarrow I = \frac{ε}{2 R} {(\frac{dU}{dt})}_{\max} = ε^{2} / 4 R P_{b} = εI = ε^{2} / 2 R$

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