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:
Carefully track the placement of equivalent resistance terms when substituting into a bridge balance condition to avoid inverting the fraction incorrectly.

Step 2: Key Formula or Approach:
For parallel resistors, equivalent resistance S_eq = (Product)/(Sum). When this appears in a denominator of the form R/S_eq, substitution flips it to R × (Sum)/(Product).

Step 3: Detailed Explanation:
Since S_eq sits in the denominator of the balancing condition R/S_eq, replacing S_eq with Product/Sum naturally inverts the fraction to Sum/Product. This ensures the product term S₁S₂ remains in the overall denominator of the final expression. Misplacing this inversion is a common exam error; consciously noting the flip prevents algebraic mistakes under time pressure.

Step 4: Final Answer:
The product term correctly stays in the overall denominator.