GATE 2024 EE Q.12: solution and answer

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

1. Definition. A junction (principal node) is a point where three or more circuit elements meet. All points joined only by ideal wire are one electrical node, so collapse every bare wire first.

2. Collapse the ideal wires. Label the top row $A,B,T,T^{\prime }$, the middle row $D,O,E$ and the bottom row $F,G,H,H^{\prime }$. The bare wire from $O$ to $E$ makes the centre dot $O$ and the tee $E$ one node. The bare wires from $T$ to $T^{\prime }$ and from $H$ to $H^{\prime }$ merge those pairs in the same way.

3. Count the elements at each candidate node, top and middle rows. $B$ has 2 resistors and 1 capacitor, so 3. $\{T,T^{\prime }\}$ has a resistor, an inductor and the outer capacitor, so 3. $D$ has the source, a capacitor and an inductor, so 3. $\{O,E\}$ has 2 resistors, 1 capacitor and 2 inductors, so 5.

4. Bottom row. $G$ has the source and 2 resistors, so 3. $\{H,H^{\prime }\}$ has a resistor, an inductor and the outer capacitor, so 3.

5. Discard the simple nodes. Corner $A$ (resistor and source), corner $F$ (inductor and source) and the point between the capacitor and the resistor in the upper middle branch each join only 2 elements. They are nodes but not junctions. With the six nodes above, the circuit has $9$ nodes in all.

6. Count the junctions: $B,\{T,T^{\prime }\},D,\{O,E\},G,\{H,H^{\prime }\}$, which is $6$ junctions. The answer is (A).

7. Check with topology. The circuit has $13$ elements (5 resistors, 3 capacitors, 3 inductors, 2 sources) and $9$ nodes, so the number of independent loops is $13-9+1=5$. Merging the series pairs leaves $10$ principal branches between the $6$ junctions, and $10-6+1=5$ again. That matches the 5 meshes visible in the drawing: four window panes plus the strip on the outer right.