II An electric circuit, whether it's a simple lightbulb or a complex amplifier, has two input terminals that are connected to the two output terminals of the voltage source. The impedance between the two input terminals (often a function of frequency) is the circuit's input impedance. Most circuits are designed to have a large input impedance. To see why, suppose you need to amplify the output of a high-pass filter that is constructed with a $1.2\mathrm{k}\Omega$ resistor and a $15\mu \mathrm{F}$ capacitor. The amplifier you've chosen has a purely resistive input impedance. For a $60\mathrm{Hz}$ signal, what is the ratio $V_{\mathrm{R}\text{load}}/V_{\mathrm{R}\text{no load}}$of the filter's peak voltage output with (load) and without (no load) the amplifier connected if the amplifier's input impedance is (a) $1.5\mathrm{k}\Omega$and (b) $150\mathrm{k}\Omega$ ?

Step by step solution

01

Step1:definition of electric field:

A region associated with an electric charge distribution or a varying magnetic field in which forces caused by that charge or field act on other electric charges Sentence Examples of Electric Fields

02

find w:(part a)

$R_1=1.2\mathrm{k\Omega }\to 1200\mathrm{\Omega }$

$C=15\mu \mathrm{F}\to 1.5\times {10}^{-5}\mathrm{F}$

$f=60\mathrm{Hz}$

First, we find value for $\omega$

$\omega =2\pi f$

$\omega =2\pi \left(60\mathrm{Hz}\right)$

(Substitute values in equation.

$\omega =377\mathrm{Hz}$

Next, we find value for$XC$

$\begin{array}{r}{X}_C=\frac{1}{\omega C}\\ X_C=\frac{1}{2\pi fC}\end{array}$

$X_C=\frac{1}{\left(377\mathrm{Hz}\right)\left(1.5\times {10}^{-5}\mathrm{F}\right)}$

(Substitute values in equation.)

$X_C=176.8\mathrm{\Omega }$

03

Step2:Find value XC(part a)

$R_2=1.5\mathrm{k\Omega }:$

$R_{eq}={\left(\frac{1}{R_1}+\frac{1}{R_2}\right)}^{-1}$

$R_{eq}={\left(\frac{1}{1.2\mathrm{k\Omega }}+\frac{1}{1.5\mathrm{k\Omega }}\right)}^{-1}$

$R_{eq}={\left(\frac{1}{1200\mathrm{\Omega }}+\frac{1}{1500\mathrm{\Omega }}\right)}^{-1}$

$R_{eq}={\left(\frac{5+4}{6000\mathrm{\Omega }}\right)}^{-1}$

$R_{eq}={\left(\frac{9}{6000\mathrm{\Omega }}\right)}^{-1}$

$R_{eq}=666.66\mathrm{\Omega }$

04

Step4:Find 150kΩ(part b)

$\frac{V_{R_{\text{load}}}}{V_{R_{\text{noload}}}}:$

$\frac{V_{R_{\text{load}}}}{V_{R_{\text{noload}}}}=\frac{R_{eq}}{R}\frac{\sqrt{R^2+X_C^2}}{R_{eq}^2+X_C^2}$

$\frac{V_{R_{\text{load}}}}{V_{R_{\text{noload}}}}=\frac{666.66\mathrm{\Omega }}{1200\mathrm{\Omega }}\frac{\sqrt{(1200\mathrm{\Omega }{)}^2+(176.8\mathrm{\Omega }{)}^2}}{(666.66\mathrm{\Omega }{)}^2+(176.8{)}^2}$

$\frac{V_{R_{\text{load}}}}{V_{R_{\text{noload}}}}=0.9997$

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