JEE Main 2026 — Current Electricity Question with Solution

Using the principle of superposition, we analyse the circuit with one source active at a time.

Case 1: Only the $10\text{ V}$ source is active (the $5\text{ V}$ source is replaced by a short circuit).

With the $5\text{ V}$ source shorted, the top and bottom nodes of the bridge are directly connected by a wire, forming a Wheatstone bridge.

Checking for balance:
Left arm ratio: $\frac{4}{4}=1$
Right arm ratio: $\frac{2}{2}=1$

The ratios are equal, so the bridge is balanced. Hence, no current flows through the middle branch connecting the two nodes, and the contribution to $I_2$ from the $10\text{ V}$ source is $0\text{ A}$.

Equivalent resistance of the balanced bridge:
$(4+2)\Omega \parallel (4+2)\Omega =6\Omega \parallel 6\Omega =3\Omega$

Total resistance including the series $1\Omega$ resistor:
$R_{eq}=3+1=4\Omega$

Current drawn from the $10\text{ V}$ source:
$I_{1(10V)}=\frac{10}{4}=2.5\text{ A}$

This current splits equally between the two $6\Omega$ branches, so the current through the bottom $4\Omega$ resistor is:
$I_{3(10V)}=\frac{2.5}{2}=1.25\text{ A}$

Case 2: Only the $5\text{ V}$ source is active (the $10\text{ V}$ source is replaced by a short circuit).

Now the $5\text{ V}$ source is applied across the top and bottom nodes, and the $1\Omega$ resistor lies in the middle branch between the left and centre nodes.

For the left branch ($4\Omega$ and $4\Omega$ in series), the potential at the midpoint is $\frac{5}{2}=2.5\text{ V}$.
For the right branch ($2\Omega$ and $2\Omega$ in series), the potential at the midpoint is also $\frac{5}{2}=2.5\text{ V}$.

Since both midpoints are at the same potential, no current flows through the $1\Omega$ middle branch. Hence, the contribution to $I_1$ from the $5\text{ V}$ source is $0\text{ A}$.

Equivalent resistance seen by the $5\text{ V}$ source:
$(4+4)\Omega \parallel (2+2)\Omega =8\Omega \parallel 4\Omega =\frac{8\times 4}{8+4}=\frac{8}{3}\Omega$

Total current supplied by the $5\text{ V}$ source:
$I_{2(5V)}=\frac{5}{8/3}=\frac{15}{8}=1.875\text{ A}$

Current through the left branch (which includes the bottom $4\Omega$ resistor):
$I_{3(5V)}=\frac{5}{8}=0.625\text{ A}$

Applying superposition:
$I_1=I_{1(10V)}+I_{1(5V)}=2.5+0=2.5\text{ A}$
$I_2=I_{2(10V)}+I_{2(5V)}=0+1.875=1.875\text{ A}$
$I_3=I_{3(10V)}+I_{3(5V)}=1.25+0.625=1.875\text{ A}$

Hence, the correct option is $(1)I_1=2.5\text{ A},I_2=1.875\text{ A},I_3=1.875\text{ A}$.