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

Concept:
The circuit is a Wheatstone bridge–like network
supplied by a \(40\text{ V}\) source between points \(C\) and \(D\).

All resistances are initially equal to \(R\).
Heating increases the resistance of branch \(AC\) by \(10%\).
Potential difference between two nodes is found using voltage division
.

Step 1: Modified Resistance Values
\[ R_{AC} = 1.1R \] All other resistances remain equal to \(R\).
Step 2: Potentials at Points \(A\) and \(B\)
Potential at \(A\):
Point \(A\) lies on the branch \(C \rightarrow A \rightarrow D\). Total resistance of this branch: \[ R_{CAD} = 1.1R + R = 2.1R \] Using voltage division: \[ V_A = 40 \times \frac{R}{2.1R} = \frac{40}{2.1} \] Potential at \(B\):
Branch \(C \rightarrow B \rightarrow D\) has equal resistances: \[ R_{CBD} = R + R = 2R \] Thus, \[ V_B = 40 \times \frac{R}{2R} = 20 \]
Step 3: Calculate Potential Difference
\[ |V_A - V_B| = \left| \frac{40}{2.1} - 20 \right| \] \[ = \left| \frac{400 - 420}{21} \right| = \frac{20}{21} \] \[ \boxed{|V_A - V_B| = \frac{20}{21}} \]