Q 12-03-238JEE MainJEE Main 2020 (4 Sep, Shift 2)Medium
In the circuit diagram shown, the value of current $i_1$ flowing from $A$ to $C$ through the two $4\ \Omega$ resistors along $AC$ is:
Answer: (D) $1\ \text{A}$
The network is symmetric about the line $AC$ and also about the line $BD$, so $B$, $D$ and the centre $O$ are all at the same potential (midway between $A$ and $C$). No current flows in the two $5\ \Omega$ resistors.
The path $A \to O \to C$ is then just $4 + 4 = 8\ \Omega$ directly across the $8$ V cell:
$$i_1 = \frac{8}{8} = 1\ \text{A}$$
(The other two paths, $ABC$ and $ADC$, carry $2$ A each, so the cell supplies $5$ A in all.)
Solution by Sreeraj P, M.Sc Physics