What is the total resistance of a circuit comprising two 8MFD capacitors connected in series at 60HZ?

What is the total resistance of a circuit comprising two 8MFD capacitors connected in series at 60HZ?

• A. 500 ohms
• B. 600 ohms
• C. 663 ohms
• D. 750 ohms

Answer: (C)

To find the total impedance of capacitors in an AC circuit, such as the one with two 8 microfarads (MFD) capacitors connected in series at a frequency of 60 Hz, it is essential to first calculate the capacitive reactance (Xc) of each capacitor. The formula for capacitive reactance is given by: \[ X_c = \frac{1}{2\pi f C} \] where \( f \) is the frequency in hertz and \( C \) is the capacitance in farads. 1. Convert microfarads to farads: - 8 MFD = 8 × 10^-6 F 2. Calculate the reactance of one capacitor: \[ X_c = \frac{1}{2\pi (60) (8 × 10^{-6})} \] 3. Plugging the values in, we calculate: \[ X_c = \frac{1}{2\pi (60) (8 \times 10^{-6})} \approx 330.4 \text{ ohms} \] Since the capacitors are in series, the total capacitive reactance (X_total) is

To find the total impedance of capacitors in an AC circuit, such as the one with two 8 microfarads (MFD) capacitors connected in series at a frequency of 60 Hz, it is essential to first calculate the capacitive reactance (Xc) of each capacitor.

The formula for capacitive reactance is given by:

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

where ( f ) is the frequency in hertz and ( C ) is the capacitance in farads.

1. Convert microfarads to farads:

• 8 MFD = 8 × 10^-6 F

2. Calculate the reactance of one capacitor:

[ X_c = \frac{1}{2\pi (60) (8 × 10^{-6})} ]

3. Plugging the values in, we calculate:

[ X_c = \frac{1}{2\pi (60) (8 \times 10^{-6})} \approx 330.4 \text{ ohms} ]

Since the capacitors are in series, the total capacitive reactance (X_total) is