Chapter 21, Problem 66P

To determine

(a) To determine:

Voltage drop R for a $130\text{mV}$ signal of frequency at $60\text{Hz}$

Answer to Problem 66P

Solution:

Voltage drop R for a $130\text{mV}$ signal of frequency at $60\text{Hz}$ will be $V=30\text{mV}$

Explanation of Solution

Given:

$\begin{array}{l}R=520\Omega \\ C=1.2\text{μF}\\ f=60\text{Hz}\\ V=130\text{mV}\end{array}$

Formula used:

$V_R=IR=\frac{V_{in}}{Z}R=\frac{V_{in}R}{\sqrt{R^2+1/{\left(2\pi fC\right)}^2}}$

$V=$ Voltage

$I=$ Current

$C$= capacitor

Calculation:

Voltage drop R for a $130\text{mV}$ signal of frequency at $60\text{Hz}$

$\begin{array}{l}{V}_R=IR=\frac{V_{in}}{Z}R=\frac{V_{in}R}{\sqrt{R^2+1/{\left(2\pi fC\right)}^2}}\\ =\frac{\left(130mV\right)\left(520\Omega \right)}{\sqrt{{\left(520\Omega \right)}^2+1/{\left[2\pi \left(60Hz\right)\left(1.2\times {10}^{-6}F\right)\right]}^2}}=30\text{mV}\end{array}$

To determine

(b)

Voltage drop R for a $130\text{mV}$ signal of frequency at $60\text{kHz}$

Answer to Problem 66P

Solution:

Voltage drop R for a $130\text{mV}$ signal of frequency at $60\text{kHz}$ will be $V=130\text{mV}$

Explanation of Solution

Given:

$\begin{array}{l}R=520\Omega \\ C=1.2\text{μF}\\ f=60\text{Hz}\\ V=130\text{mV}\end{array}$

Formula used:

$V_R=IR=\frac{V_{in}}{Z}R=\frac{V_{in}R}{\sqrt{R^2+1/{\left(2\pi fC\right)}^2}}$

$V=$ Voltage

$I=$ Current

$C$= capacitor

Calculation:

The phase angle will be $-{12.6}^0$

Voltage drop R for a $130\text{mV}$ signal of frequency at $60\text{kHz}$

$\begin{array}{l}{V}_R=IR=\frac{V_{in}}{Z}R=\frac{V_{in}R}{\sqrt{R^2+1/{\left(2\pi fC\right)}^2}}\\ =\frac{\left(130mV\right)\left(520\Omega \right)}{\sqrt{{\left(520\Omega \right)}^2+1/{\left[2\pi \left(6000Hz\right)\left(1.2\times {10}^{-6}F\right)\right]}^2}}=130\text{mV}\end{array}$