Question

A resistor of power \( P_1 \) at voltage \( V_1 \) is connected in series with another resistor of power \( P_2 \) at voltage \( V_2 \). The equivalent resistance will be:

• A. \( \frac{V_1^2}{P_1} + \frac{V_2^2}{P_2} \)
• B. \( \frac{P_1^2}{V_1} + \frac{P_2^2}{V_2} \)
• C. \( \frac{P_1 + P_2}{(V_1 + V_2)^2} \)
• D. \( \frac{P_1}{V_1^2} + \frac{P_2}{V_2^2} \)

Hint:

Power and resistance are related by the formula \( P = \frac{V^2}{R} \).

Answer 1

The Correct Option is A

Using the power formula \( P = \frac{V^2}{R} \), the equivalent resistance in series is derived as: \[ R_{\text{eq}} = \frac{V_1^2}{P_1} + \frac{V_2^2}{P_2} \] Thus, the correct answer is \( \mathbf{(A) \frac{V_1^2}{P_1} + \frac{V_2^2}{P_2}} \).