A wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances ( R₁ = R₂ = R₃ = R₄$R_{1} = R_{2} = R_{3} = R_{4}$ ). When R₃$R_{3}$ resistance is heated to some temperature, its resistance value has gone up by 10%. The potential difference ( Vₐ - Vb$V_{a} - V_{b}$ ) (after R₃$R_{3}$ is heated) is ____ V.
A standard wheatstone bridge circuit powered by a 40V source.
A.1.05$1.05$
B.0$0$
C.0.95$0.95$
D.2$2$
Solution & Explanation
Core Logic
A standard wheatstone bridge circuit powered by a 40V source.
Initially, R₁ = R₂ = R₃ = R₄ = R$R_{1} = R_{2} = R_{3} = R_{4} = R$. The bridge is balanced, and VA = VB$V_A = V_B$.
After heating, R₃$R_3$ increases by 10%, meaning the new R₃' = R + 0.1R = 1.1R$R_3' = R + 0.1R = 1.1R$.
The other arms remain R$R$.
The battery provides a voltage of V = 40 V$V = 40\text{ V}$.
Step 1: Calculate Potentials at nodes A and B
Let the lower node (D) be at 0 V$0\text{ V}$ and the upper node (C) be at 40 V$40\text{ V}$.
The potential at node A (between R₃'$R_3'$ and R₄$R_4$) can be found via the voltage divider rule across the right branch (R₃'$R_3'$ and R₄$R_4$).
Wait, examining the circuit labels carefully:
Assume the branches are: Left branch contains R₁$R_1$ and R₂$R_2$. Right branch contains R₃$R_3$ and R₄$R_4$.
Node B is between R₁$R_1$ and R₂$R_2$. Node A is between R₃$R_3$ and R₄$R_4$.
From the provided solution:
VA = (V)/(2)$$V_A = \frac{V}{2}$$
This implies the branch containing node A consists of two equal resistances R$R$, which contradicts the textual convention if R₃$R_3$ changed. Looking at the provided solution image logic, the branch with B$B$ has the changed resistance 1.1R$1.1R$.
Let's follow the PDF's mathematical layout exactly.
VA = (V)/(2)$$V_A = \frac{V}{2}$$ (Meaning branch A has equal resistors R$R$ and R$R$)
VB = (V)/(R + 1.1R) × R = (V)/(2.1)$$V_B = \frac{V}{R + 1.1R} \times R = \frac{V}{2.1}$$
(Note: The PDF solution computes VB = V / 2.1$V_B = V / 2.1$, implying the resistance below node B is R$R$, and above is 1.1R$1.1R$)
The closest match among the options is 0.95 V$0.95\text{ V}$.
Pattern Recognition
For a slightly unbalanced Wheatstone bridge due to a fractional change Δ R/R = x$\Delta R/R = x$, the potential difference Δ V$\Delta V$ is approximately V · (x)/(4)$V \cdot \frac{x}{4}$ for small x$x$. Here x=0.1$x=0.1$, so Δ V ≈ 40 × 0.1 / 4 = 1 V$\Delta V \approx 40 \times 0.1 / 4 = 1 \text{ V}$. Exact calculation gives 0.95 V$0.95\text{ V}$.
Chapter Mix
Class 12 Physics: Current Electricity
A standard wheatstone bridge circuit powered by a 40V source.
Keywords:#wheatstone bridge is initially at room temperature#JEE Main 2026 Evening Q31#Current Electricity JEE Main 2026#Wheatstone Bridge JEE Main 2026#Wheatstone bridge#resistances#potential difference#battery 40V
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