GATE 2024 EC Q.62: solution and answer

Official answer: 2.5Source: Official answer key, GATE 2024 EC (IISc Bengaluru)

1. Maximum power transfer theorem: the load takes maximum power when $R_L=R_{th}$, the Thevenin resistance seen from the terminals of $R_L$. The circuit has a dependent source, so find $R_{th}$ from the terminal V–I relation, not by combining resistors.

2. Take the bottom wire as the reference. Node A (the left vertical wire) is then at $50\text{V}$ and node B (the right vertical wire) is at $V_L$, the load voltage. Let $V=V_A-V_B=50-V_L$.

3. Top branch (2 Ω and 3 Ω in series): its current from A to B is $I_1=\frac{V}{5}$. This current enters the 3 Ω at its left ($-$) end and leaves at its right ($+$) end, so the left end is the higher one and $V_x=-3I_1=-\frac{3V}{5}$.

4. Middle branch (KVL): the dependent source has $-$ on the node-A side, so its right terminal is at $V_A+V_x$. The current through the 2 Ω towards B is $$I_2=\frac{V_A+V_x-V_B}{2}=\frac{V-\frac{3V}{5}}{2}=\frac{V}{5}$$

5. KCL at node B: the current delivered to the load is $$I=I_1+I_2=\frac{2V}{5}=\frac{50-V_L}{2.5}$$

6. Compare this with the Thevenin form $I=\frac{V_{th}-V_L}{R_{th}}$. Open circuit ($I=0$) gives $V_{th}=50\text{V}$, and short circuit ($V_L=0$) gives $I_{sc}=20\text{A}$, so $$R_{th}=\frac{V_{th}}{I_{sc}}=\frac{50}{20}=2.5\Omega$$

7. Check with a test source: replace the 50 V source with a short, so node A is at 0, and apply $V_t$ at B. The top branch now carries $\frac{V_t}{5}$ from B to A, entering the 3 Ω at its $+$ end, so $V_x=+\frac{3V_t}{5}$. The dependent source's right terminal is at $\frac{3V_t}{5}$, so the middle branch draws $\frac{V_t-3V_t/5}{2}=\frac{V_t}{5}$. Then $I_t=\frac{2V_t}{5}$ and $R_{th}=\frac{V_t}{I_t}=2.5\Omega$, which agrees.

8. So $R_L=R_{th}=2.5\Omega$. The maximum power would be $\frac{V_{th}^2}{4R_{th}}=\frac{2500}{10}=250\text{W}$, but the question does not ask for it.