Question

A 1 kHz sine-wave generator having an internal resistance of 50Ω generates an open-circuit voltage of 10 Vp-p. When a capacitor is connected across the output terminals, the voltage drops to 8 Vp-p. The capacitance of the capacitor (in microfarads) is _________.

Hint:

To calculate the capacitance, use the voltage divider rule and the formula for the impedance of a capacitor in an AC circuit.

Answer 1

Correct Answer: 2.35

We are given the following information: - Frequency \( f = 1 \, kHz = 1000 \, Hz \)
- Internal resistance \( R = 50 \, \Omega \)
- Open-circuit voltage \( V_{oc} = 10 \, V_{p-p} \)
- Voltage with capacitor \( V_C = 8 \, V_{p-p} \)
Step 1: Use voltage divider formula
The voltage across the capacitor \( V_C \) when it is connected in parallel with the internal resistance can be found using the voltage divider formula: \[ \frac{V_C}{V_{oc}} = \frac{Z_C}{R + Z_C}, \] where \( Z_C \) is the impedance of the capacitor and \( R \) is the internal resistance. Step 2: Impedance of the capacitor
The impedance of the capacitor \( Z_C \) is given by: \[ Z_C = \frac{1}{j 2\pi f C}. \] Substitute the values for frequency and impedance into the voltage divider formula: \[ \frac{8}{10} = \frac{\frac{1}{j 2\pi \times 1000 \times C}}{50 + \frac{1}{j 2\pi \times 1000 \times C}}. \] Step 3: Solve for capacitance
After solving the equation, we find the capacitance: \[ C \approx 2.39 \, \mu F. \] Thus, the capacitance of the capacitor is approximately \( 2.39 \, \mu F \).