Question

A Wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances \[ (R_1=R_2=R_3=R_4). \] When \(R_3\) resistance is heated, its resistance value increases by \(10%\). The potential difference \((V_a-V_b)\) after \(R_3\) is heated is _______ V. \includegraphics[width=0.5\linewidth]{30.png}

• A. \(0\)
• B. \(0.95\)
• C. \(2\)
• D. \(1.05\)

Hint:

In Wheatstone bridge problems, always find midpoint potentials using the voltage divider rule.

Answer 1

The Correct Option is B

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 \(R_3\) \[ R_3'=1.1R \] Supply voltage across the bridge: \[ V=40\,\text{V} \]
Step 3: Calculate potentials at midpoints Potential at point \(a\): \[ V_a=40\cdot\frac{R_2}{R_1+R_2} =40\cdot\frac{R}{2R}=20\,\text{V} \] Potential at point \(b\): \[ V_b=40\cdot\frac{R_4}{R_3'+R_4} =40\cdot\frac{R}{1.1R+R} =40\cdot\frac{1}{2.1} \approx 19.05\,\text{V} \]
Step 4: Potential difference \[ V_a-V_b=20-19.05=0.95\,\text{V} \] Final Answer: \[ \boxed{0.95\text{ V}} \]