ISRO EC 2025 Q.34: solution and answer

Official answer: CSource: ISRO's final answer key of 15 May 2026

1. The left node is held at $75\text{V}$ by the ideal source. Let the middle node (across the $20\Omega$) be at $V$ and the right node (across the $12\Omega$) be at $W$.

2. The $6\text{A}$ source points left, so it carries $6\text{A}$ from node $W$ to the $75\text{V}$ node. KCL at $W$ (currents leaving): $\frac{W-V}{8}+\frac{W}{12}+6=0$, that is $\frac{V-W}{8}=\frac{W}{12}+6$.

3. Multiply by $24$: $3(V-W)=2W+144$, so $W=\frac{3V-144}{5}$.

4. KCL at the middle node (current in through $5\Omega$ equals current out through $20\Omega$ and $8\Omega$): $\frac{75-V}{5}=\frac{V}{20}+\frac{V-W}{8}$.

5. Multiply by $40$: $8(75-V)=2V+5(V-W)$, so $600=15V-5W$.

6. Substitute $5W=3V-144$: $600=15V-3V+144$, so $12V=456$ and $V=38\text{V}$.

7. Check: $W=(114-144)/5=-6\text{V}$. The $8\Omega$ current is $(38+6)/8=5.5\text{A}$ and the $12\Omega$ current is $-0.5\text{A}$, so $5.5=-0.5+6$ holds. At the middle node, $7.4\text{A}$ from the $5\Omega$ equals $1.9+5.5$ through the $20\Omega$ and $8\Omega$.