A signal generator having a source resistance of 50 Ω 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

The correct answer is 2.38.

To find the value of the capacitor connected across the terminals of the signal generator, we can model the setup as a series AC circuit containing the generator's internal source voltage, its source resistance, and the connected capacitor.

First, let's identify the given parameters:
- Source resistance:
$R_s=50 \Omega$
- Frequency of the signal:
$f=1 \mathrm{kHz}=1000 \mathrm{Hz}$
- Open circuit terminal voltage (which is equal to the source voltage):
$V_s=10 V$
- Terminal voltage after connecting the capacitor:
$V_c=8 V$

The terminal voltage is the voltage across the capacitor, which can be found using the voltage divider formula for an RC circuit:

$V_c=V_s\times \frac{X_c}{\sqrt{{R_s}^2+{X_c}^2}}$

where $X_c$ is the capacitive reactance.

Substituting the given values into the equation:

$8=10\times \frac{X_c}{\sqrt{{50}^2+{X_c}^2}}$

Divide both sides by 10:

$0.8=\frac{X_c}{\sqrt{2500+{X_c}^2}}$

Now, square both sides of the equation to eliminate the square root:

$0.64=\frac{{X_c}^2}{2500+{X_c}^2}$

Multiply both sides by $2500+{X_c}^2$:

$0.64(2500+{X_c}^2)={X_c}^2$

$1600+0.64{X_c}^2={X_c}^2$

Rearrange the equation to solve for ${X_c}^2$:

$1600={X_c}^2-0.64{X_c}^2$

$1600=0.36{X_c}^2$

${X_c}^2=\frac{1600}{0.36}\approx 4444.44$

Taking the square root:

$X_c\approx 66.67 \Omega$

The capacitive reactance is related to the capacitance by the formula:

$X_c=\frac{1}{2\pi fC}$

Rearranging the formula to solve for the capacitance $C$:

$C=\frac{1}{2\pi f{X}_c}$

Substitute the values of $f$ and $X_c$:

$C=\frac{1}{2\times \pi \times 1000\times 66.67}$

$C\approx 2.387\times {10}^{-6} F$

Converting to microfarads ($\mu F$):

$C\approx 2.38 \mu F$