A capacitor of 5 microfarads is connected to an A.C source of frequency 50Hz. The capacitive reactance is:

March 30, 2026Physics

ScanSolve AI Answer

March 30, 2026

Answer

636.62 \Omega

Step 1: Convert the capacitance from microfarads to farads. $C = 5 \muF = 5 \times 10^{-6} F$

Step 2: Use the formula for capacitive reactance. $X_C = \frac{1}{2 \pi f C}$

Step 3: Substitute the given values into the formula. $X_C = \frac{1}{2 \pi (50 Hz) (5 \times 10^{-6} F)}$ $X_C = \frac{1}{2 \pi (250 \times 10^{-6})}$ $X_C = \frac{1}{0.001570796}$ $X_C \approx 636.62 \Omega$

The capacitive reactance is approximately 636.62 $\Omega$ .

$636.62 \text{ \Omega}$

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Quick Answer

Convert the capacitance from microfarads to farads. C = 5 = 5 × 10^-6 F Step 2: Use the formula for capacitive reactance.

Asked on March 30, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

ScanSolve AI AnswerAnswered on March 30, 2026

Step 1: Convert the capacitance from microfarads to farads. C = 5 = 5 × 10^-6 F Step 2: Use the formula for capacitive reactance. X_C = (1)/(2 f C) Step 3: Substitute the given values into the formula. X_C = (1)/(2 (50 Hz)) (5 × 10^-6 F) X_C = (1)/(2 (250 × 10^-6)) X_C = (1)/(0.001570796) X_C ≈ 636.62 The capacitive reactance is approximately 636.62 . 636.62