JEE Main 2026 Physics (28 Jan Shift 2) | Unbalanced Wheatstone Bridge Problem
Detailed step-by-step solution for JEE Main 2026 (28 January, Shift 2) Physics question covering Current Electricity and Unbalanced Wheatstone Bridges.
📌 QUESTION:
Q.37 A Wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances (R1 = R2 = R3 = R4). When R3 resistance is heated to some temperature, its resistance value has gone up by 10%. The potential difference (Va - Vb) (after R3 is heated) is _________ V.
[Diagram details: A Wheatstone bridge circuit with four resistors R1, R2, R3, R4 connected to a 40 V DC source. Points a and b are the output terminals of the bridge.]
Options:
1. 1.05
2. 0.95
3. 2
4. 0
⏱️ TIMESTAMPS:
0:00 - Question Reading & Introduction
0:35 - Circuit Diagram & Initial Setup
1:31 - Calculating Voltage at Node A
2:07 - Calculating Voltage at Node B
2:38 - Finding the Potential Difference (Va - Vb) & Final Answer
📝 STEP-BY-STEP SOLUTION:
1. Let the top node be 40V and the bottom node be 0V.
2. Initially R1 = R2 = R3 = R4 = R.
3. After heating, R3 increases by 10%, so R3 = 1.1R.
4. Calculate Voltage at Node A (Va) using the left branch:
(40 - Va) / R = (Va - 0) / R
40 - Va = Va
2Va = 40 then Va = 20V
5. Calculate Voltage at Node B (Vb) using the right branch:
(40 - Vb) / 1.1R = (Vb - 0) / R
(40 - Vb) / 1.1 = Vb
40 - Vb = 1.1Vb
40 = 2.1Vb then Vb = 40 / 2.1 ≈ 19.047V
6. Find the Potential Difference (Va - Vb):
Va - Vb = 20 - (40 / 2.1)
Va - Vb = (42 - 40) / 2.1 = 2 / 2.1 ≈ 0.95 V
Correct Option: 2 (0.95)
#JEE2026 #JEEPhysics #CurrentElectricity #WheatstoneBridge #PhysicsUnlocked #JEEMains