question_answer

In the circuit shown switch S is connected to position 2 for a long time and then joined to position 1. The total heat produced in resistance \[{{R}_{1}}\] is -

A) \[\frac{L{{E}^{2}}}{2R_{2}^{2}}\]               

B) \[\frac{L{{E}^{2}}}{2R_{1}^{2}}\]

C) \[\frac{L{{E}^{2}}}{2{{R}_{1}}{{R}_{2}}}\]             

D) \[\frac{L{{E}^{2}}{{({{R}_{1}}+{{R}_{2}})}^{2}}}{2R_{1}^{2}R_{2}^{2}}\]

Correct Answer: B

Solution :

[A] When the key is at position (2) for a long time; the energy stored in the inductor is: \[{{U}_{B}}=\frac{1}{2}L{{i}_{{{O}^{2}}}}=\frac{1}{2}.L.{{\left( \frac{E}{{{R}_{2}}} \right)}^{2}}=\frac{L.E{{.}^{2}}}{2{{R}_{2}}^{2}}\] This whole energy will be dissipated in the form of heat when the inductor is connected to \[{{R}_{1}}\] and no source is connected.