Question

In the circuit shown switch $S$ is connected to position $2$ for a long time and then joined to the position $1$ . The total heat produced in resistance $R_{1}$ is :

• A. $\frac{\text{LE}^{2}}{2 \text{R}_{2}^{2}}$
• B. $\frac{\text{LE}^{2}}{2 \text{R}_{1}^{2}}$
• C. $\frac{\text{LE}^{2}}{2 \text{R}_{1} \text{R}_{2}}$
• D. $\frac{\left(\text{LE}\right)^{2} \left(\left(\text{R}\right)_{1} + \left(\text{R}\right)_{2}\right)^{2}}{2 \text{R}_{1}^{2} \text{R}_{2}^{2}}$

Answer 1

Answer: (A)

When the key is at position (B) for a long time; the energy stored in the inductor is :
$\left(\text{U}\right)_{\text{B}} = \frac{1}{2} \text{Li}_{0}^{2} = \frac{1}{2} \cdot \text{L} \cdot \left(\frac{\text{E}}{\left(\text{R}\right)_{2}}\right)^{2} = \frac{\left(\text{LE}\right)^{2}}{2 \text{R}_{2}^{2}}$

This whole energy will be dissipated in the form of heat when the inductor is connected to $R_{1}$ and no source is connected.