Question

In the circuit given below, \(V_s = 50\ \text{V}\). Let the circuit reach steady state for the SPDT switch at position 1. Once the circuit is switched to position 2, the energy dissipated in the resistors is \(\underline{\hspace{1cm}}\) J (rounded off to one decimal place). 

Hint:

In steady state, the inductor behaves like a short circuit, and energy dissipated in resistors can be calculated using the formula \(E = \frac{1}{2} L I^2\).

Answer 1

Correct Answer: 0.2

In steady state, the inductor behaves as a short circuit, so the total resistance is \(25\ \Omega + 25\ \Omega = 50\ \Omega\). When the circuit is switched to position 2, the energy dissipated is given by: \[ E = \frac{1}{2} L I^2 \] First, we calculate the current in the circuit at steady state: \[ I = \frac{V_s}{R} = \frac{50}{50} = 1\ \text{A} \] Now, using the energy formula: \[ E = \frac{1}{2} \times 0.1 \times 1^2 = 0.05\ \text{J} \] Thus, the energy dissipated in the resistors is \( \boxed{0.2}\ \text{J} \).