Refer to the figure given below. The values of I_1, I_2 and I_3 are . JEE Main 2026 Physics, Current Electricity — question figure

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

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

With the $5\text{ V}$5 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: $\dfrac{4}{4} = 1$44​=1

Right arm ratio: $\dfrac{2}{2} = 1$22​=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 $I_2$I2​ from the $10\text{ V}$10 V source is $0\text{ A}$0 A.

Equivalent resistance of the balanced bridge:

$$(4 + 2)\,\Omega \parallel (4 + 2)\,\Omega = 6\,\Omega \parallel 6\,\Omega = 3\,\Omega$$(4+2)Ω∥(4+2)Ω=6Ω∥6Ω=3Ω

Total resistance including the series $1\,\Omega$1Ω resistor:

$$R_{eq} = 3 + 1 = 4\,\Omega$$Req​=3+1=4Ω

Current drawn from the $10\text{ V}$10 V source:

$$I_{1(10V)} = \dfrac{10}{4} = 2.5\text{ A}$$I1(10V)​=410​=2.5 A

This current splits equally between the two $6\,\Omega$6Ω branches, so the current through the bottom $4\,\Omega$4Ω resistor is:

$$I_{3(10V)} = \dfrac{2.5}{2} = 1.25\text{ A}$$I3(10V)​=22.5​=1.25 A

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

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

For the left branch ($4\,\Omega$4Ω and $4\,\Omega$4Ω in series), the potential at the midpoint is $\dfrac{5}{2} = 2.5\text{ V}$25​=2.5 V.

For the right branch ($2\,\Omega$2Ω and $2\,\Omega$2Ω in series), the potential at the midpoint is also $\dfrac{5}{2} = 2.5\text{ V}$25​=2.5 V.

Since both midpoints are at the same potential, no current flows through the $1\,\Omega$1Ω middle branch. Hence, the contribution to $I_1$I1​ from the $5\text{ V}$5 V source is $0\text{ A}$0 A.

Equivalent resistance seen by the $5\text{ V}$5 V source:

$$(4 + 4)\,\Omega \parallel (2 + 2)\,\Omega = 8\,\Omega \parallel 4\,\Omega = \dfrac{8 \times 4}{8 + 4} = \dfrac{8}{3}\,\Omega$$(4+4)Ω∥(2+2)Ω=8Ω∥4Ω=8+48×4​=38​Ω

Total current supplied by the $5\text{ V}$5 V source:

$$I_{2(5V)} = \dfrac{5}{8/3} = \dfrac{15}{8} = 1.875\text{ A}$$I2(5V)​=8/35​=815​=1.875 A

Current through the left branch (which includes the bottom $4\,\Omega$4Ω resistor):

$$I_{3(5V)} = \dfrac{5}{8} = 0.625\text{ A}$$I3(5V)​=85​=0.625 A

Applying superposition:

$$I_1 = I_{1(10V)} + I_{1(5V)} = 2.5 + 0 = 2.5\text{ A}$$I1​=I1(10V)​+I1(5V)​=2.5+0=2.5 A

$$I_2 = I_{2(10V)} + I_{2(5V)} = 0 + 1.875 = 1.875\text{ A}$$I2​=I2(10V)​+I2(5V)​=0+1.875=1.875 A

$$I_3 = I_{3(10V)} + I_{3(5V)} = 1.25 + 0.625 = 1.875\text{ A}$$I3​=I3(10V)​+I3(5V)​=1.25+0.625=1.875 A

Hence, the correct option is $(1)\ I_1 = 2.5\text{ A},\ I_2 = 1.875\text{ A},\ I_3 = 1.875\text{ A}$(1) I1​=2.5 A, I2​=1.875 A, I3​=1.875 A.