GATE 2020 IN Q.7: solution and answer

Official answer: ASource: Official answer key, GATE 2020 IN (IIT Delhi)

1. Read the orientations. Call the junction of the diode, the Zener and the $100\Omega$ resistor the node, at voltage $V_N$. The series diode has its anode towards the $50\Omega$ resistor, so it conducts only when $v_s=10\sin (100\pi t)$ is above $V_N$. The Zener has its cathode at the node and its anode at ground. Taking it as ideal: it carries no current for $0<V_N<5\text{V}$, it breaks down and holds $V_N=5\text{V}$ when reverse current is forced into it, and it would conduct forward only if $V_N<0$.

2. Diode off ($v_s<V_N$). The diode cuts the source off from the node. Only the 5 V battery and the $100\Omega$ resistor drive the node. The Zener cannot conduct below $5\text{V}$, so there is no path for current, and $V_N=5\text{V}$. That is exactly the breakdown voltage, with zero Zener current, so the assumption is consistent.

3. Diode on ($v_s>5\text{V}$). Suppose the Zener stayed off. KCL at the node would give $\frac{v_s-V_N}{50}=\frac{V_N-5}{100}$. The left side is positive because the diode conducts, so $V_N>5\text{V}$. The Zener cannot block more than $5\text{V}$, so this is a contradiction. The Zener is therefore in breakdown and holds $V_N=5\text{V}$. All of the source current $\frac{v_s-5}{50}$ flows down through the Zener.

4. Negative half-cycle, or $0\le v_s<5\text{V}$. The diode's anode is below its cathode at $5\text{V}$, so the diode is reverse-biased and step 2 applies: $V_N=5\text{V}$. Forward-biasing the Zener would need $V_N<0$. With the diode off, nothing can pull the node below the battery's $5\text{V}$.

5. So $V_N=5\text{V}$ at every instant. Both ends of the $100\Omega$ resistor are at $5\text{V}$. By Ohm's law, $$I_{100}=\frac{5-V_N}{100}=\frac{5-5}{100}=0$$

6. Power: $P_{100}=I_{100}^2\times 100=0\text{W}$ at every instant, so the average power is also $0\text{W}$. Answer: (A).