Question

A signal generator having a source resistance of 50\(\Omega\) is set to generate a 1 kHz sinewave. Open circuit terminal voltage is 10 V peak-to-peak. Connecting a capacitor across the terminals reduces the voltage to 8 V peak-to-peak. The value of this capacitor is \(\underline{\hspace{1cm}}\) \(\mu\text{F}\). (Round off to 2 decimal places.)

Hint:

For AC sources with internal resistance, terminal voltage changes due to loading follow the impedance voltage-divider formula.

Answer 1

Correct Answer: 2.3

The source has internal resistance \(R_s = 50\Omega\). Open-circuit voltage (peak): \[ V_{oc} = \frac{10}{2} = 5\text{ V} \] With capacitor connected, the terminal peak voltage becomes: \[ V = \frac{8}{2} = 4\text{ V} \] Voltage division gives: \[ \frac{V}{V_{oc}} = \frac{1}{\sqrt{1 + (\omega C R_s)^2}} \] Substituting values: \[ \frac{4}{5} = \frac{1}{\sqrt{1 + (2\pi 1000 \cdot 50 C)^2}} \] Solving: \[ \sqrt{1 + (314000 C)^2} = 1.25 \] \[ (314000 C)^2 = 0.5625 \] \[ C = 2.36 \times 10^{-6}\text{ F} \] Thus, \[ C \approx 2.36\ \mu F \] which lies in the range 2.30 to 2.50 µF.