GATE 2024 EC Q.35: solution and answer

Official answer: 0.250Source: Official answer key, GATE 2024 EC (IISc Bengaluru)

1. Direction of the discharge current. The capacitor's top plate is $+$. When the switch closes, current leaves the $+$ plate, goes through the switch and down through the $10\text{k}\Omega$ resistor, then enters the Zener at its cathode (the barred end, at the top) and leaves at its anode (bottom). So the Zener is reverse-biased.

2. Ideal Zener model. In reverse it blocks until the voltage across it reaches $V_Z$. After that it holds exactly $V_Z=5\text{V}$ for any current. At $t=0^{+}$ the capacitor supplies $10\text{V}>V_Z$, so the Zener breaks down and the capacitor discharges through $R$ against a constant 5 V.

3. Current and final state (KVL around the loop). $$i=\frac{v_C-V_Z}{R}$$ Current flows only while $v_C>5\text{V}$, so the capacitor settles at $v_C(\infty )=V_Z=5\text{V}$, not at 0 V. With $\tau =RC=(10\times 10^3)(10\times 10^{-6})=0.1\text{s}$, $$v_C(t)=5+5e^{-t/\tau }\text{V},i(t)=\frac{5}{10\text{k}\Omega }{e}^{-t/\tau }=0.5e^{-10t}\text{mA}$$

4. Charge that passes through the Zener. All of it flows through the series loop: $$Q=C[v_C(0^{+})-v_C(\infty )]=10\mu \text{F}\times (10-5)\text{V}=50\mu \text{C}$$

5. Zener energy = constant voltage × charge. The Zener voltage stays at $V_Z$ the whole time, so $$W_Z={\int }_0^{\infty }{V}_{Z}idt=V_{Z}Q=5\text{V}\times 50\mu \text{C}=250\mu \text{J}=0.250\text{mJ}$$

6. Energy balance check. Energy released by the capacitor: $\frac{1}{2}C(10^2-5^2)=\frac{1}{2}(10\times 10^{-6})(75)=375\mu \text{J}$. Energy lost in the resistor: ${\int }_0^{\infty }{i}^{2}Rdt=\frac{1}{2}C(\Delta V{)}^2=\frac{1}{2}(10\times 10^{-6})(5^2)=125\mu \text{J}$. Then $375-125=250\mu \text{J}$, which matches $W_Z$. Answer: 0.250 mJ.