GATE 2024 EC – Question 35
As shown in the circuit, the initial voltage across the capacitor is 10 V, with the switch being open. The switch is then closed at $t=0$. The total energy dissipated in the ideal Zener diode ($V_Z=5$ V) after the switch is closed (in mJ, rounded off to three decimal places) is ______.
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Correct answer: 0.24 to 0.26
Explanation
Following the capacitor's positive terminal, the discharge current enters the Zener from its cathode side, so the Zener is in reverse breakdown with a constant 5 V across it. The capacitor discharges until its voltage falls to 5 V (below that the Zener stops conducting). The charge that flows is $Q=C\Delta V=10\ \mu\text{F}\times5\text{ V}=50\ \mu$C. The Zener dissipates $V_ZQ=5\times50\ \mu\text{C}=250\ \mu$J $=0.250$ mJ (the remaining 125 μJ of the 375 μJ lost by the capacitor is dissipated in the 10 kΩ resistor).