JEE Main 2026 — Physics Electromagnetism

The given circuit consists of four parallel branches connected between the left and right common vertical wires.

Let's find the equivalent electromotive force (EMF) and resistance for each branch:

For the top three active branches, each contains three $5\text{ V}$5 V cells and three $3\ \Omega$3 Ω resistors in series.

Equivalent EMF of each active branch, $E = 5 + 5 + 5 = 15\text{ V}$E=5+5+5=15 V.

Equivalent resistance of each active branch, $R = 3 + 3 + 3 = 9\ \Omega$R=3+3+3=9 Ω.

Since these three identical active branches are connected in parallel, we can replace them with a single equivalent voltage source:

Equivalent EMF, $E_{eq} = 15\text{ V}$Eeq​=15 V (as they are identical and in parallel).

Equivalent internal resistance, $R_{eq} = \dfrac{9}{3} = 3\ \Omega$Req​=39​=3 Ω.

The fourth (bottom) branch is connected between terminals $A$A and $B$B and acts as the load for this equivalent source. It contains three $3\ \Omega$3 Ω resistors in series.

Resistance of the branch $AB$AB, $R_{AB} = 3 + 3 + 3 = 9\ \Omega$RAB​=3+3+3=9 Ω.

Now, the circuit is simplified to an equivalent source of $15\text{ V}$15 V with internal resistance $3\ \Omega$3 Ω connected across a load of $9\ \Omega$9 Ω.

The current $I$I flowing between terminals $A$A and $B$B (through the bottom branch) is given by Ohm's law:

$$I = \dfrac{E_{eq}}{R_{eq} + R_{AB}}$$I=Req​+RAB​Eeq​​

$$I = \dfrac{15}{3 + 9} = \dfrac{15}{12}\text{ A}$$I=3+915​=1215​ A

$$I = 1.25\text{ A}$$I=1.25 A

Answer: $1.25$1.25