For a 50 Hz circuit with a 25 µF capacitor, what is the capacitive reactance in ohms after conversion?

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Multiple Choice

For a 50 Hz circuit with a 25 µF capacitor, what is the capacitive reactance in ohms after conversion?

To determine the capacitive reactance of a capacitor in a circuit, the formula used is: \[ X_C = \frac{1}{2 \pi f C} \] where: - \( X_C \) is the capacitive reactance in ohms, - \( f \) is the frequency of the circuit in hertz (Hz), and - \( C \) is the capacitance in farads (F). In this scenario, the frequency \( f \) is 50 Hz and the capacitance \( C \) is 25 µF, which is equal to \( 25 \times 10^{-6} \) F. Now, we can substitute these values into the formula: 1. Calculate \( 2 \pi f \): - \( 2 \pi (50) = 314.16 \) (approximately) 2. Now, use this to calculate \( X_C \): - \( X_C = \frac{1}{314.16 \times 25 \times 10^{-6}} \) 3. Calculate the denominator: - \( 314.16 \times 25 \times 10^{-6} = 0.007854 \) 4.

To determine the capacitive reactance of a capacitor in a circuit, the formula used is:

[ X_C = \frac{1}{2 \pi f C} ]

where:

  • ( X_C ) is the capacitive reactance in ohms,

  • ( f ) is the frequency of the circuit in hertz (Hz), and

  • ( C ) is the capacitance in farads (F).

In this scenario, the frequency ( f ) is 50 Hz and the capacitance ( C ) is 25 µF, which is equal to ( 25 \times 10^{-6} ) F. Now, we can substitute these values into the formula:

  1. Calculate ( 2 \pi f ):
  • ( 2 \pi (50) = 314.16 ) (approximately)
  1. Now, use this to calculate ( X_C ):
  • ( X_C = \frac{1}{314.16 \times 25 \times 10^{-6}} )
  1. Calculate the denominator:
  • ( 314.16 \times 25 \times 10^{-6} = 0.007854 )
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