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.

• 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

The problem involves analyzing a modified Wheatstone bridge circuit, where the resistance of one arm is changed due to heating. Let's solve the problem step-by-step.

1. Initially, the bridge is balanced since all resistances are equal: \(R_1 = R_2 = R_3 = R_4 = R\).
2. The condition for a balanced Wheatstone bridge is: \(\frac{R_1}{R_2} = \frac{R_3}{R_4}\). As all resistances are equal, this condition is satisfied.
3. When \(R_3\) is heated, its resistance increases by 10%: \(R_3 = R + 0.1R = 1.1R\).
4. Now, calculate the potential difference across \((V_a - V_b)\). Using the voltage divider rule, the potentials can be derived based on the resistances:
   • In the left arm: \(V_a = \frac{R_2}{R_1 + R_2} \times 40\text{V}\)
   • In the right arm: \(V_b = \frac{R_4}{R_3 + R_4} \times 40\text{V}\)
5. Substitute the values:
   • Left arm: \(V_a = \frac{R}{R + R} \times 40 = \frac{1}{2} \times 40 = 20\text{V}\)
   • Right arm: \(V_b = \frac{R}{1.1R + R} \times 40 = \frac{R}{2.1R} \times 40 = \frac{40}{2.1} \approx 19.05\text{V}\)
6. The potential difference: \((V_a - V_b) = 20 - 19.05 = 0.95\text{V}\).

Therefore, the potential difference \((V_a - V_b)\) after heating \(R_3\) is 0.95 V. This confirms that the correct answer is \(0.95\text{V}\).