GATE EC 2024 Set 1 — Question 35

Given:
Capacitor $C = 10 \mu F = 10 \times 10^{-6}$ F
Initial capacitor voltage $V_C(0^-) = 10$ V
Zener voltage $V_Z = 5$ V
Resistor $R = 10 \text{ k}\Omega = 10 \times 10^3 \Omega$ $E_{initial} = \frac{1}{2} C V_C(0^-)^2 = \frac{1}{2} (10 \times 10^{-6} \text{ F}) (10 \text{ V})^2$
$E_{initial} = \frac{1}{2} \times 10 \times 10^{-6} \times 100 = 500 \times 10^{-6} \text{ J} = 0.5 \text{ mJ}$ When the switch is closed at $t=0$, the capacitor voltage is $10$ V. Since $V_C(0^-) = 10 \text{ V}$ is greater than the Zener voltage $V_Z = 5 \text{ V}$, the ideal Zener diode will immediately go into breakdown and clamp the voltage across itself to $5$ V. The capacitor will discharge until its voltage reaches the Zener voltage, $V_Z = 5$ V. At this point, the Zener diode will continue to conduct, maintaining $5$ V across itself. $E_{final} = \frac{1}{2} C V_Z^2 = \frac{1}{2} (10 \times 10^{-6} \text{ F}) (5 \text{ V})^2$
$E_{final} = \frac{1}{2} \times 10 \times 10^{-6} \times 25 = 125 \times 10^{-6} \text{ J} = 0.125 \text{ mJ}$ The charge that leaves the capacitor and flows through the Zener diode (and resistor) is the change in charge on the capacitor:
$\Delta Q = Q_{initial} - Q_{final} = C V_C(0^-) - C V_Z$
$\Delta Q = (10 \times 10^{-6} \text{ F}) (10 \text{ V}) - (10 \times 10^{-6} \text{ F}) (5 \text{ V})$
$\Delta Q = 100 \times 10^{-6} \text{ C} - 50 \times 10^{-6} \text{ C} = 50 \times 10^{-6} \text{ C} = 50 \mu C$ For an ideal voltage source (like an ideal Zener in breakdown), the energy dissipated is the product of the voltage across it and the total charge that flows through it.
$E_Z = V_Z \times \Delta Q$
$E_Z = 5 \text{ V} \times (50 \times 10^{-6} \text{ C})$
$E_Z = 250 \times 10^{-6} \text{ J} = 0.250 \text{ mJ}$This energy is dissipated in the Zener diode as the capacitor discharges from 10V to 5V. The remaining energy difference ($E_{initial} - E_{final} - E_Z = 0.5 - 0.125 - 0.250 = 0.125 \text{ mJ}$) is dissipated in the $10 \text{ k}\Omega$ resistor.The total energy dissipated in the ideal Zener diode is $0.250 \text{ mJ}$.