GATE 2024 EE Q.13: solution and answer

Official answer: ASource: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. Take the bottom wire as the reference node. The $2\Omega$ resistor is in series with the ideal $10\text{A}$ current source, so it carries exactly $10\text{A}$ into node A. A resistor in series with an ideal current source does not change that current.

2. Write KCL at node A, with node voltage $V_A$ and currents entering equal to currents leaving: $$\frac{10-V_A}{\alpha }+10=\frac{V_A}{1}$$

3. Multiply through by $\alpha$: $$10-V_A+10\alpha =\alpha V_A\Rightarrow V_A(1+\alpha )=10(1+\alpha )\Rightarrow V_A=10\text{V}$$ This holds for every $\alpha >0$.

4. The current through the $\alpha \Omega$ resistor is also the current through the $10\text{V}$ source: $$I=\frac{10-V_A}{\alpha }=\frac{10-10}{\alpha }=0$$

5. Check: all $10\text{A}$ from the current source flows down the $1\Omega$ resistor, so the voltage across it is $10\text{A}\times 1\Omega =10\text{V}$. That equals the source voltage, so there is no voltage across $\alpha$ to drive a current. The same holds in the limit $\alpha =0$: the source then fixes $V_A=10\text{V}$ directly, the $1\Omega$ resistor draws $10\text{A}$, the current source supplies all of it, and the voltage source again carries nothing.

6. Power delivered by the voltage source: $P=10\text{V}\times 0=0\text{W}$, whatever the value of $\alpha$. Answer (A).