Question

Certain resistors connected in parallel and the equivalent resistance is \( X \). If one of the resistances is removed the equivalent resistance is \( Y \). What is the conductance value of removed resistance?

• A. \( \frac{XY}{Y-X} \)
• B. \( \frac{XY}{X-Y} \)
• C. \( \frac{X-Y}{XY} \)
• D. \( \frac{Y-X}{XY} \)

Hint:

For parallel resistances, use \( \frac{1}{R_{\text{eq}}} \) sum rule and carefully rearrange when one resistance is removed.

Answer 1

The Correct Option is D

For resistors in parallel: \[ \frac{1}{X} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots \] When one resistor is removed: \[ \frac{1}{Y} = \frac{1}{X} - \frac{1}{R} \] Rearranging: \[ \frac{1}{R} = \frac{1}{X} - \frac{1}{Y} = \frac{Y-X}{XY} \] So, the conductance value of removed resistance: \[ G = \frac{1}{R} = \frac{Y-X}{XY} \]