GATE 2021 BM – Question 60
In the shown circuit, Vₛ=50 V and L=0.1 H. Let the circuit reach steady state with the SPDT switch at position 1, then move it to position 2. Both resistors are 25 Ω. Find total energy dissipated in the resistors in J to one decimal.
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Correct answer: 0.2 to 0.2
Explanation
At the initial DC steady state the inductor is a short and current is 50/25=2 A. Its stored magnetic energy is ½LI²=.5×.1×4=.2 J. After switching the supply is disconnected and the inductor discharges through the resistive loop. All initial magnetic energy becomes resistor heat, giving .2 J in total.