Question

In a Wheatstone's bridge, three resistances $P$, $Q$ and $R$ are connected in the three arms and the fourth arm is formed by two resistances $S_1$ and $S_2$ connected in parallel. The condition for the bridge to be balanced is

• A. $\frac{P}{Q} = \frac{2R}{S_1 + S_2}$
• B. $\frac{P}{Q} = \frac{R(S_1 + S_2)^2}{S_1S_2}$
• C. $\frac{P}{Q} = \frac{R(S_1 + S_2)}{S_1S_2}$
• D. $\frac{P}{Q} = \frac{R(S_1S_2)}{S_1 + S_2}$

Hint:

Take note of how the terms are placed to avoid mistakes under pressure. Since $S_{\text{eq}}$ is in the denominator of the balancing condition ($\frac{R}{S_{\text{eq}}}$), and $S_{\text{eq}} = \text{Product}/\text{Sum}$, substituting it naturally flips the fraction to $\text{Sum}/\text{Product}$. This ensures that the product term $S_1S_2$ remains in the overall denominator!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the balancing condition of a standard Wheatstone bridge circuit network. Three branches have standalone resistors $P$, $Q$, and $R$, while the final balancing branch consists of a sub-circuit where two resistors ($S_1$ and $S_2$) are placed in parallel.

Step 2: Key Formula or Approach:
1. The general balance condition for a standard Wheatstone bridge configuration with arms balanced in order is: $$\frac{P}{Q} = \frac{R}{S_{\text{eq}}}$$ Where $S_{\text{eq}}$ is the net equivalent resistance of the fourth branch. 2. For two resistors linked together in parallel, the equivalent formula is: $$\frac{1}{S_{\text{eq}}} = \frac{1}{S_1} + \frac{1}{S_2} \implies S_{\text{eq}} = \frac{S_1S_2}{S_1 + S_2}$$

Step 3: Detailed Explanation:
Let's plug the parallel equivalent formula for $S_{\text{eq}}$ directly into the general balancing condition of the bridge network: $$\frac{P}{Q} = \frac{R}{\left(\frac{S_1S_2}{S_1 + S_2}\right)}$$ When a fraction sits in the denominator, its reciprocal moves to the numerator as a multiplier: $$\frac{P}{Q} = \frac{R(S_1 + S_2)}{S_1S_2}$$

Step 4: Final Answer:
The correct balance formula corresponds to option (C).