JEE Physics 2026 Question with Solution

Given: All four arms of the Wheatstone bridge initially have equal resistance, so $R_1=R_2=R_3=R_4=R$. After heating, $R_3' = 1.1R$. The supply voltage across the bridge is $40 \, \text{V}$.

Find: The potential difference $(V_a-V_b)$ after $R_3$ is heated.

Concept: In a Wheatstone bridge, the potential difference between the midpoints is zero when the bridge is balanced:

$$\frac{R_1}{R_2}=\frac{R_3}{R_4}$$

Any change in one resistance unbalances the bridge, producing a potential difference between the midpoints.

Step 1: Initial condition Initially,

$$R_1=R_2=R_3=R_4=R$$

Hence, the bridge is balanced and:

$$V_a=V_b$$

Step 2: After heating $R_3$

$$R_3'=1.1R$$

Step 3: Calculate potentials at midpoints using the voltage divider rule Potential at point $a$:

$$V_a=40\cdot\frac{R_2}{R_1+R_2}$$ $$V_a=40\cdot\frac{R}{2R}=20 \, \text{V}$$

Potential at point $b$:

$$V_b=40\cdot\frac{R_4}{R_3'+R_4}$$ $$V_b=40\cdot\frac{R}{1.1R+R}$$ $$V_b=40\cdot\frac{1}{2.1}\approx 19.05 \, \text{V}$$

Step 4: Potential difference

$$V_a-V_b=20-19.05=0.95 \, \text{V}$$

Therefore, the potential difference is $0.95 \, \text{V}$ and the correct option is B.