JEE Main 2026 — Current Electricity Question with Solution

The four parallel branches between the common left node $A$ and the common right node are all identical: each consists of a $27\text{ V}$ source in series with a $3\Omega$ resistor. In the third branch, the two cells $(14\text{ V}$ and $13\text{ V})$ aid each other, giving a net EMF of $14+13=27\text{ V}$.

Thevenin equivalent between $A$ and $B$:
Since all four branches have the same EMF and polarity in parallel:
$V_{th}=27\text{ V}$

The four $3\Omega$ resistors in parallel give:
$R_{th}=\frac{3}{4}=0.75\Omega$

Load branch (diagonal):
The diagonal connecting $A$ and $B$ consists of two $3\Omega$ resistors in series, so:
$R_L=3+3=6\Omega$

Current through the diagonal branch:
$I=\frac{V_{th}}{R_{th}+R_L}=\frac{27}{0.75+6}=\frac{27}{6.75}=4\text{ A}$

Voltage across $A$ and $B$:
$V_{AB}=I\cdot R_L=4\times 6=24\text{ V}$

Hence, the voltage is $24\text{ V}$ and the current is $4\text{ A}$, which corresponds to option $(2)24\text{ V},4\text{ A}$.