Question

In a Wheatstone's bridge, three resistance 'Q', 'R' and 'S' are connected in three arms and first arm is formed by three resistances '\(P_{1}\)', '\(P_{2}\)' and '\(P_{3}\)' in parallel. The condition for the bridge to be balanced is

• A. \(\frac{P_{1}P_{2}P_{3}/(P_{1}P_{2}+P_{2}P_{3})}{Q}=\frac{S}{R}\)
• B. \(\frac{P_{1}P_{2}P_{3}/(P_{2}P_{3}+P_{1}P_{3}+P_{1}P_{2})}{Q}=\frac{S}{R}\)
• C. \(\frac{P_{1}P_{2}P_{3}/(P_{2}P_{3}+P_{1}P_{3}+P_{1}P_{2})}{Q}=\frac{R}{S}\)
• D. \(\frac{P_{1}P_{3}+P_{1}P_{2}}{P_{1}P_{2}P_{3}Q}=\frac{S}{R}\)

Hint:

Replace the three parallel resistors by their equivalent and apply the balance ratio.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The arm AB holds \(P_1\), \(P_2\), \(P_3\) in parallel. Their equivalent is \(P=\dfrac{P_1P_2P_3}{P_1P_2+P_2P_3+P_1P_3}\).

Step 2: Bridge arms
The other arms are BC = \(Q\), CD = \(R\) and DA = \(S\). The galvanometer joins B and D.

Step 3: Balance condition
At balance B and D are at the same potential, so \(\dfrac{P}{Q}=\dfrac{S}{R}\).

Step 4: Final form
\(\dfrac{P_1P_2P_3/(P_2P_3+P_1P_3+P_1P_2)}{Q}=\dfrac{S}{R}\)

Final Answer:
The balance condition uses the parallel equivalent over Q equal to S over R. \[ \boxed{\text{(B) }\dfrac{P_1P_2P_3/(P_2P_3+P_1P_3+P_1P_2)}{Q}=\dfrac{S}{R}} \]