Chapter 21, Problem 60P

To determine

(a)

The impedance of the circuit if the source frequency is $60\text{Hz}$

Answer to Problem 60P

Solution:

The impedance of the circuit if the source frequency is $60\text{Hz}$ will be $3.6\times {10}^3\text{Ω}$

Explanation of Solution

Given Info:

$\begin{array}{l}R=3.5k\Omega \\ C=3.0\mu F\end{array}$

Formula Used:

$Z=\sqrt{R^2+\frac{1}{4\pi f^2{c}^2}}$

Where,

$\begin{array}{l}R=3.5\text{kΩ}\\ C=3.0\text{μF}\\ f=60\text{Hz}\end{array}$

Calculation:

$\begin{array}{l}Z=\sqrt{R^2+\frac{1}{4\pi f^2{c}^2}}\\ =\sqrt{\left(3.5\times {10}^3\right)+\frac{1}{4{\pi }^2\left(60Hz\right){\left(3.0\mu F\right)}^2}}\\ =3.6\times {10}^3\Omega \end{array}$

To determine

(b)

The impedance of the circuit if the source frequency is $60,000\text{Hz}$

Answer to Problem 60P

Solution:

The impedance of the circuit if the source frequency is $60,000\text{Hz}$ will be $3.5\times {10}^3\text{Ω}$

Explanation of Solution

Given Info:

$\begin{array}{l}R=3.5k\Omega \\ C=3.0\mu F\end{array}$

Formula used:

$Z=\sqrt{R^2+\frac{1}{4\pi f^2{c}^2}}$

Where;

$\begin{array}{l}R=3.5\text{kΩ}\\ C=3.0\text{μF}\\ f=60,000\text{Hz}\end{array}$

Calculation:

$\begin{array}{l}Z=\sqrt{R^2+\frac{1}{4\pi f^2{c}^2}}\\ =\sqrt{\left(3.5\times {10}^3\right)+\frac{1}{4{\pi }^2\left(60000\text{Hz}\right){\left(3.0\text{μF}\right)}^2}}\\ =3.5\times {10}^3\text{Ω}\end{array}$