GATE 2018 EC Q.53: solution and answer

Official answer: 7.9 to 8.1Source: Official answer key, GATE 2018 EC (IIT Guwahati)

1. Symmetry. The network is its own mirror image about the vertical line through $A$ and $B$, and the source is connected across $A$ and $B$. So $V_C=V_D$ and $V_E=V_F$.

2. Branches with no current. Each $R_2$ joins two nodes at the same potential ($C$–$D$ and $E$–$F$), so each carries zero current and can be removed. Equipotential nodes can also be merged: call $C,D$ node $U$ and $E,F$ node $L$.

3. The diagonals. Each $R_3$ runs from the upper pair to the lower pair ($C$ to $F$ and $D$ to $E$), so after merging, both sit between $U$ and $L$. It makes no difference whether the crossing is a junction. The upper half of each diagonal is plain wire, so the crossing point on the $C$–$F$ diagonal is at $V_C$ and the one on the $D$–$E$ diagonal is at $V_D$. These are equal, so joining them changes nothing.

4. Series–parallel reduction. Between $A$ and $U$: $R_1\parallel R_1=0.5\Omega$. Between $U$ and $L$: $(R_1\parallel R_1)\parallel (R_3\parallel R_3)=0.5\parallel 1.5=\frac{0.5\times 1.5}{0.5+1.5}=0.375\Omega$. Between $L$ and $B$: $R_1\parallel R_1=0.5\Omega$.

5. Equivalent resistance. $$R_{eq}=0.5+0.375+0.5=1.375\Omega =\frac{11}{8}\Omega$$

6. Source current (Ohm's law). $$I=\frac{11\text{V}}{11/8\Omega }=8.00\text{A}$$ Check: $V_U=11-8(0.5)=7\text{V}$ and $V_L=8(0.5)=4\text{V}$. The $U$–$L$ drop is $3\text{V}=8\times 0.375$, which agrees.