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In figure switch S is closed for a long time. At t = 0, if it is opened, then

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heat produced in resistor R₁ after opening the switch is $\frac{1}{2} \frac{R_{2} L ε^{2}}{(R_{1} + R_{2}) R^{2}}$

total heat produced in resistor R₁ after opening the switch is $\frac{1}{2} \frac{L ε}{R^{2}} (\frac{R_{1}}{R_{1} + R_{2}})$

no heat will be produced in R₁

total heat produced in resistor R after opening the switch is $\frac{1}{2} \frac{L ε}{R^{2}}$

answer is C.

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Detailed Solution

Just before opening the switch, the current in the inductor is $\frac{E}{R}$

Energy stored in it $= \frac{1}{2} L {(\frac{ε}{R})}^{2}$

This energy will dissipate in resistors R₁ and R₂ in the ratio $\frac{1}{R_{1}} a n d \frac{1}{R_{2}}$

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