How much capacitive reactance would a 60 cycle circuit with an 88.464 microfarad capacitor have, rounded to the nearest whole number?

How much capacitive reactance would a 60 cycle circuit with an 88.464 microfarad capacitor have, rounded to the nearest whole number?

• A. 20 Ohm
• B. 25 Ohm
• C. 30 Ohm
• D. 35 Ohm

Answer: (C)

To find the capacitive reactance, you can use the formula: \[ X_C = \frac{1}{2\pi f C} \] where: - \( X_C \) is the capacitive reactance in ohms. - \( f \) is the frequency in hertz (60 Hz for this problem). - \( C \) is the capacitance in farads (88.464 microfarads, which is \( 88.464 \times 10^{-6} \) farads). Substituting the given values into the formula: 1. Calculate \( 2\pi f \): \[ 2\pi \times 60 \approx 376.99 \text{ radians/second} \] 2. Now plug in the values to find \( X_C \): \[ X_C = \frac{1}{376.99 \times 88.464 \times 10^{-6}} \] 3. Calculate \( 376.99 \times 88.464 \times 10^{-6} \) first: \[ 376.99 \times 88.464 \times 10^{-6} \approx 0

To find the capacitive reactance, you can use the formula:

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

where:

• ( X_C ) is the capacitive reactance in ohms.

• ( f ) is the frequency in hertz (60 Hz for this problem).

• ( C ) is the capacitance in farads (88.464 microfarads, which is ( 88.464 \times 10^{-6} ) farads).

Substituting the given values into the formula:

1. Calculate ( 2\pi f ):

[

2\pi \times 60 \approx 376.99 \text{ radians/second}

]

2. Now plug in the values to find ( X_C ):

[

X_C = \frac{1}{376.99 \times 88.464 \times 10^{-6}}

]

3. Calculate ( 376.99 \times 88.464 \times 10^{-6} ) first:

[

376.99 \times 88.464 \times 10^{-6} \approx 0