Question

In a Wheatstone bridge arrangement (P, Q, R and S), the resistors P and Q are nearly equal. The bridge is balanced when R = 500 \(\Omega\). On interchanging P and Q, the value of R for balancing is 505 \(\Omega\). The value of S will be:

• A. 500 \(\Omega\)
• B. 502.5 \(\Omega\)
• C. 505 \(\Omega\)
• D. 5 \(\Omega\)

Hint:

Use \(S = \sqrt{R_1 R_2}\) for the two balancing values.

Answer 1

The Correct Option is B

Step 1: Idea:
Since P and Q are nearly equal, the two balancing values of R are almost the same. The unknown S is then close to the average of the two.

Step 2: Write the two balance equations:
First, $P S = 500\,Q$. After interchange, $Q S = 505\,P$.

Step 3: Eliminate the ratio:
Multiply the two equations and cancel $PQ$: $S^2 = 500 \times 505$. Taking the root gives $S = 502.49$ ohm.

Step 4: Cross check with the mean:
The arithmetic mean is $(500+505)/2 = 502.5$ ohm. The two values agree to the first decimal.

Step 5: Match:
The nearest option is 502.5 ohm, the second printed option.

Final Answer:
S is about 502.5 ohm, option 2. \[ \boxed{S \approx 502.5\ \Omega} \]