GATE 2024 EE Q.60: solution and answer

Official answer: 1.64 to 1.70Source: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. Neither node is labelled in the figure, so name them here. Node A is the junction of the two left-hand 1 kΩ resistors, which is also the anode of $D2$. Node B is the cathode of $D1$, where the series 1 kΩ from $D2$ and the lower right 1 kΩ to ground also meet. Assume both ideal diodes are ON, so each acts as a short circuit, and check this at the end. Work in mA and kΩ.

2. $D1$ ON ties node B to its anode supply: $V_B=3\text{V}$.

3. $D2$ ON puts its cathode at $V_A$, so the current through $D2$ and its series 1 kΩ resistor, from A towards B, is $I_{D2}=\frac{V_A-3}{1\text{k}\Omega }$.

4. KCL at node A. The current arriving from the +10 V supply equals the current into the lower 1 kΩ plus the current into $D2$: $$\frac{10-V_A}{1}=\frac{V_A}{1}+\frac{V_A-3}{1}\Rightarrow 3V_A=13\Rightarrow V_A=\frac{13}{3}\approx 4.333\text{V}$$

5. Current through $D2$: $I_{D2}=\frac{13}{3}-3=\frac{4}{3}\approx 1.333\text{mA}$. It is positive (anode to cathode), so $D2$ really is ON.

6. Current from node B to ground through the lower right 1 kΩ: $\frac{V_B}{1\text{k}\Omega }=3\text{mA}$.

7. KCL at node B. $D2$ and $D1$ together must supply those 3 mA: $I+\frac{4}{3}=3\Rightarrow I=\frac{5}{3}\approx 1.667\text{mA}$. $I$ is positive and flows downward from anode to cathode, as the arrow shows, so $D1$ really is ON. Both assumptions hold, and for ideal diodes this consistent state is the solution.

8. Rounded to two decimal places, $I\approx 1.67\text{mA}$.