(a) In Figure 18-18, find the value of C so that $X_{C}$ is approximately ten times less than R at an input frequency of 10 kHz.
(b) If a 5 V sine wave with a dc level of 10 V is applied, what are the output voltage magnitude and the phase shift?

Question

(a) In Figure 18-18, find the value of C so that [latex]X_{C}[/latex] is approximately ten times less than R at an input frequency of 10 kHz.
(b) If a 5 V sine wave with a dc level of 10 V is applied, what are the output voltage magnitude and the phase shift?

Accepted Answer

(a) Determine the value of C as follows:

[latex]X_{C}[/latex] = 0.1 R = 0.1(680 Ω) = 68 Ω

[latex]C= \frac{1}{2\pi fX_{C}}= \frac{1}{2\pi (10 \ kHz )(68 \ \Omega )}= 0.234 \ \mu F [/latex]

Using the nearest standard value of C = 0.22[latex]\mu F [/latex],

[latex]X_{C}= \frac{1}{2\pi fC}= \frac{1}{2\pi (10 \ kHz )(0.22 \ \mu F)}= 72 \ \Omega [/latex]

(b) Determine the magnitude of the sinusoidal output using the voltage-divider formula.

[latex]V_{out}= \left(\frac{R}{\sqrt{R^{2} + X^{2}_{C}} } ight) V_{in}= \left(\frac{680 \ \Omega }{\sqrt{(680 \ \Omega)^{2} + (72 \ \Omega)^{2}} } ight)5V= 4.97 \ V[/latex]

The phase shift is

[latex]\phi = an ^{-1}\left(\frac{X_{C}}{R} ight)= an ^{-1}\left(\frac{72 \ \Omega}{680 \ \Omega} ight) = 6.04°[/latex]

At f = 10 kHz. which is a decade above the critical frequency, the sinusoidal output is almost equal to the input in magnitude, and the phase shift is very small. The 10 V dc level has been filtered out and does not appear at the output.

Question 18.6: (a) In Figure 18-18, find the value of C so that Xc is appro...

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