JEE Main 2026 — Physics Electromagnetism

2026mcqmedium

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

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

A
$I_1 = 2.5$ A, $I_2 = 1.875$ A, $I_3 = 1.875$ A
B
$I_1 = 1.875$ A, $I_2 = 2.5$ A, $I_3 = 1.875$ A
C
$I_1 = 1.875$ A, $I_2 = 1.875$ A, $I_3 = 2.5$ A
D
$I_1 = 2.5$ A, $I_2 = 2.5$ A, $I_3 = 1.875$ A

Did you get this right?

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: $\dfrac{4}{4} = 1$

Right arm ratio: $\dfrac{2}{2} = 1$

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:

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

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

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

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 $\dfrac{5}{2} = 2.5\text{ V}$ .

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

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:

$(4 + 4)\,\Omega \parallel (2 + 2)\,\Omega = 8\,\Omega \parallel 4\,\Omega = \dfrac{8 \times 4}{8 + 4} = \dfrac{8}{3}\,\Omega$

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

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

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

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

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 $(1)\ I_1 = 2.5\text{ A},\ I_2 = 1.875\text{ A},\ I_3 = 1.875\text{ A}$ .

Your note

JEE Main Electromagnetism in other years

Electromagnetism 2025 Electromagnetism 2024 Electromagnetism 2023 Electromagnetism 2022 Electromagnetism 2021 Electromagnetism 2020 Electromagnetism 2019 Electromagnetism 2018 Electromagnetism 2017 Electromagnetism 2016 Electromagnetism 2015 Electromagnetism 2014

More JEE Main Electromagnetism questions

A solenoid has a core made of material with relative permeability $400$. The magnetic field produced in the interior of solenoid is $1.0$ T. The magnetic intensity in SI units is $\alpha \times 10^5$. The value of $\alpha$ is ______. (Free space permeability $\mu_0=4\pi \times 10^{-7}$ SI units.)

2026easy

Two identical small bar magnets each of dipole moment $3\sqrt{5}$ J/T are placed at a center to center separation of $10$ cm, with their axes perpendicular to each other as shown in figure. The value of magnetic field at the point P midway between the magnets is $\alpha \times 10^{-3}$ T. The value of $\alpha$ is ______. ($\mu_0=4\pi \times 10^{-7}$ Tm/A) ![JEE Main 2026 Physics, Magnetic Properties of Matter — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Magnetic_Properties_of_Matter/69d8074781239f6567c4f3e1/question_1__q_69d8074781239f6567c4f3e1__cdn-question-pool.getmarks.app__04APRS2__q48_d1__b2e00d1c52_final_ppt_sync.jpg)

2026hard

The heat generated in 1 minute between points $A$ and $B$ in the given circuit, when a battery of 9 V with internal resistance of $1 \Omega$ is connected across these points is $\_\_\_\_$ J. ![JEE Main 2026 Physics, Current Electricity — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Current_Electricity/6983c17d8bfdbb2435639562/question_1__q_6983c17d8bfdbb2435639562__cdn-question-pool.getmarks.app__21S1_q_49_1__47430bbfa5_final_ppt_sync.png)

2026medium

Two known resistances of $R \Omega$ and $2 R \Omega$ and one unknown resistance $X \Omega$ are connected in a circuit as shown in the figure. If the equivalent resistance between points $A$ and $B$ in the circuit is $X \Omega$, then the value of $X$ is $\_\_\_\_$ $\Omega$. ![JEE Main 2026 Physics, Current Electricity — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Current_Electricity/6983c1818bfdbb2435639596/question_1__q_6983c1818bfdbb2435639596__cdn-question-pool.getmarks.app__21S2_q_26_1__e476cb4a94_final_ppt_sync.png)

2026medium

A cylindrical conductor of length 2 m and area of cross-section $0.2 \mathrm{~mm}^{2}$ carries an electric current of 1.6 A when its ends are connected to a 2 V battery. Mobility of electrons in the conductor is $\alpha \times 10^{-3} \mathrm{~m}^{2} / V. s$. The value of $\alpha$ is : (electron concentration $=5 \times 10^{28} / \mathrm{m}^{3}$ and electron charge $\left.=1.6 \times 10^{-19} \mathrm{C}\right)$

2026medium

View all Electromagnetism questions

Keep practicing Electromagnetism

Work through every JEE Main Electromagnetism PYQ, year by year.

Browse all Physics topics

Options

$I_1 = 2.5$ A, $I_2 = 1.875$ A, $I_3 = 1.875$ A

$I_1 = 1.875$ A, $I_2 = 2.5$ A, $I_3 = 1.875$ A

$I_1 = 1.875$ A, $I_2 = 1.875$ A, $I_3 = 2.5$ A

$I_1 = 2.5$ A, $I_2 = 2.5$ A, $I_3 = 1.875$ A

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: $\dfrac{4}{4} = 1$

Right arm ratio: $\dfrac{2}{2} = 1$

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:

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

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

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

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 $\dfrac{5}{2} = 2.5\text{ V}$ .

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

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:

$(4 + 4)\,\Omega \parallel (2 + 2)\,\Omega = 8\,\Omega \parallel 4\,\Omega = \dfrac{8 \times 4}{8 + 4} = \dfrac{8}{3}\,\Omega$

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

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

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

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

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 $(1)\ I_1 = 2.5\text{ A},\ I_2 = 1.875\text{ A},\ I_3 = 1.875\text{ A}$ .

JEE Main 2026 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism

JEE Main 2026 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism