What is the impedance of a 1.0-microfarad capacitor in series with a 500-ohm resistor at 100 kHz?

What is the impedance of a 1.0-microfarad capacitor in series with a 500-ohm resistor at 100 kHz?

• A. 500 - j159.15
• B. 500 + j159.15
• C. 500 - j318.3
• D. 500 + j318.3

Answer: (A)

To find the impedance of a capacitor in series with a resistor, it is essential to understand how to calculate the total impedance in the circuit. The impedance of a resistor remains purely real, while the impedance of a capacitor has an imaginary component due to the phase difference between voltage and current. First, we calculate the capacitive reactance (Xc) using the formula: \[ X_c = \frac{1}{2\pi f C} \] where: - \( f \) is the frequency (100 kHz or 100,000 Hz), and - \( C \) is the capacitance (1.0 microfarad or \( 1.0 \times 10^{-6} \) farads). Using these values, the calculation for capacitive reactance becomes: \[ X_c = \frac{1}{2\pi(100,000)(1.0 \times 10^{-6})} \] \[ X_c \approx \frac{1}{0.0006283} \approx 159.15 \; \Omega \] The negative sign indicates that this component is capacitive. Thus, the impedance of the capacitor is: \[ Z_c = -j159.15 \;

To find the impedance of a capacitor in series with a resistor, it is essential to understand how to calculate the total impedance in the circuit. The impedance of a resistor remains purely real, while the impedance of a capacitor has an imaginary component due to the phase difference between voltage and current.

First, we calculate the capacitive reactance (Xc) using the formula:

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

where:

• ( f ) is the frequency (100 kHz or 100,000 Hz), and

• ( C ) is the capacitance (1.0 microfarad or ( 1.0 \times 10^{-6} ) farads).

Using these values, the calculation for capacitive reactance becomes:

[ X_c = \frac{1}{2\pi(100,000)(1.0 \times 10^{-6})} ]

[ X_c \approx \frac{1}{0.0006283} \approx 159.15 ; \Omega ]

The negative sign indicates that this component is capacitive. Thus, the impedance of the capacitor is:

[ Z_c = -j159.15 ;