Question

A high-pass filter consists of a $1.59mF$capacitor in series with a $100\Omega$resistor. The circuit is driven by an $AC$source with a peak voltage of $5.00V$.

a. What is the crossover frequency $f_c$?

b. What is $V_{\mathrm{R}}$when $f=\frac{1}{2}{f}_{\mathrm{c}},f_{\mathrm{c}}$, and $2f_{\mathrm{c}}$?

Short answer

(a) Crossover frequency , $f_c=1000\mathrm{Hz}$

(b) Value of$V_R=3.54\mathrm{V}$

Step-by-step solution

AC Source

The flow of electrical charge that occasionally reverses direction is understood as electricity (ac).

As mentioned in Induced Electric Fields, an alternating emf generated in an exceedingly very station produces an ac.

An ac circuit is one during which the ac source varies frequently, often sinusoidally.

Find Crossover frequency

Given values

$C=1.59\times {10}^{-6}\mathrm{F}$(farad)

$R=100\Omega$(ohm)

${\epsilon }_o=5.00\mathrm{V}$(voltage)

we use formula

$w=\frac{1}{RC}$

Derive from the formula,

$w=\frac{1}{RC}$

$2\pi f_c=\frac{1}{RC}$

$f_c=\frac{1}{2\pi RC}$

localid="1650213173701" $f_c=\frac{1}{2\pi \times \left(100\right)\Omega \times \left(1.59\times {10}^{-6}\right.)F}$

$f_c=1000\mathrm{Hz}$

Find value of voltage

$V_R=\frac{{\epsilon }_{o}R}{\sqrt{R^2+\frac{1}{w^2{C}^2}}}$

$V_R=\frac{{\epsilon }_{o}R}{R^2+\frac{R^2{C}^2}{C^2}}$

Substitute the values,

$V_R=\frac{5.00\mathrm{V}}{1.41}$

$V_R=3.54\mathrm{V}$

$f=f_c\Rightarrow V_R^{\text{'}}=\frac{{\epsilon }_o}{\sqrt{2}}$