The circuit shown in the figure consists of a battery of emf E, resistances R₁ and R₂, inductance L and a two way switch ABC. First A is connected to B for a long time and then A is connected to C.The total heat produced in R₂ is

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$\frac{{LE}^{2}}{2 R_{1}^{2}}$

$\frac{{LE}^{2}}{R_{2}^{2}}$

$L \frac{E^{2}}{R_{1}^{2}}$

$\frac{{LE}^{2}}{2 R_{1} R_{2}}$

answer is C.

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Detailed Solution

$\begin{array}{l} i_{0} = \frac{E}{R_{1}} \\ dθ = i^{2} Rdt \\ i = i_{0} e^{- \frac{R_{2}}{L} \cdot t} \\ i = \frac{E}{R_{1}} \cdot e^{- \frac{R_{2}}{L} \cdot t} \\ \int dQ = \int \frac{E^{2}}{R_{1}^{2}} e^{- 2 \frac{R_{2}}{L} \cdot t} \\ Q = \frac{E^{2} R_{2}}{R_{1}^{2}} \int e^{- \frac{2 R_{2}}{Lt}} dt \\ Q = \frac{E^{2} R_{2}}{R_{1}^{2}} \frac{{[e^{\frac{- 2 R_{2}}{L} . t}]}_{0}^{α .}}{\frac{- 2 R_{2}}{L}} \\ Q = \frac{E^{2} L}{2 R_{1}^{2}} \end{array}$