Question

If resistance between \(A\) and \(C\) is increased by \(10%\) through heating, then calculate \(|V_A - V_B|\).

• A. \(\dfrac{10}{21}\)
• B. \(\dfrac{5}{21}\)
• C. \(\dfrac{20}{21}\)
• D. \(\dfrac{5}{7}\)

Hint:

In symmetric resistor networks, changing resistance in only one branch breaks balance — use {voltage division} separately for each branch.

Answer 1

The Correct Option is C

To solve this problem, we need to analyze the given circuit and understand the effect of increasing the resistance between \(A\) and \(C\) by \(10\%\).

1. Initially, the resistances in the Wheatstone bridge are all \(R\). Given the symmetry and equal resistances, the bridge is balanced.
2. If the bridge is balanced, no current flows through the galvanometer branch \(BD\), and the points \(A\) and \(B\) are at the same potential. Thus, \(V_A = V_B\).
3. When the resistance \(R_{AC}\) increases by \(10\%\), it becomes \(1.1R\). This imbalance causes current to flow through \(BD\), resulting in a potential difference between \(A\) and \(B\).
4. Using Kirchhoff's rules or Thevenin equivalents, the potential difference between \(A\) and \(B\) can be derived from the altered circuit.

Calculating \(|V_A - V_B|\) after the resistance change:

1. Total voltage across the bridge is \(40V\).
2. The effective resistance in one arm (with increase) is \(1.1R + R = 2.1R\).
3. The other arm remains \(2R\).
4. The potential at \(A\) and \(B\) can be found using voltage division:

Voltage division in arm with increased resistance:

V_A = \frac{1.1R}{2.1R + 2R} \times 40 = \frac{1.1}{4.1} \times 40

Voltage division in symmetrical arm:

V_B = \frac{R}{2R + 2.1R} \times 40 = \frac{1}{4.1} \times 40

Calculate \(|V_A - V_B|\):

|V_A - V_B| = \left| \frac{1.1}{4.1} \times 40 - \frac{1}{4.1} \times 40 \right| = \frac{0.1}{4.1} \times 40 = \frac{4}{4.1} \approx \frac{20}{21}

Thus, the correct answer is \(\frac{20}{21}\).