Question

Phasors and complex impedance 1. A resistor R and capacitor Care connected in series with an AC voltage source with frequency f and maximum voltage Vo. a. Find the complex impedance (in the form Z = R +jX). If the impedance is written in polar form (Z = Zej®), find expressions for Z and Ø. Write your answers in terms of the variables R, C, and (=21f). b. If the voltage source is described by the phasor V = V.ejut, and the current is described by I = 1, ej(wt-o), find lo in terms of R, C, 0, and V.. [Use I = V/Z and the properties of complex numbers.] C. Find the amplitude of the voltage across the capacitor, Vc, in terms of R, C, o, and Vo d. If R = 5.0 and C = 770 pF, find the frequencies for which Vc = 0.90 Vand Vc= 0.10 V.. [Note that the voltage across the capacitor is much larger for the lower frequency. If a load was placed across the capacitor, a sizeable signal would be transferred to the load only for lower frequencies – therefore connecting across a capacitor in an R-C circuit is known as a low-pass filter.]

Answer 1

Impedence of resistor = R

Impedence of capacitor = - j [1/ ( $\omega$ C ) ]

where R is resistance of resistor, $\omega$ is angular frequency that is given by $\omega$ = 2$\pi$f , f is frequency AC voltage , C is capacitance, and j is complex number $\sqrt{-1}$ .

When resistance and capacitance are in series , then impedence Z = R - j [ 1/ ( $\omega$ C ) ]

Impedence Z in polar form is written as

$Z = \left | Z \right | e^{j\phi }$

where | Z | is given as

................................(1)

12 1 R2 + W²02

where phase angle $\phi$ is given by

$\phi = tan^{-1}\left ( \frac{-1}{(\omega C)R} \right )$ ........................(2)

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Current I = ( V / Z )

$I_{o} e^{j(\omega t -\phi )} = \frac{V_{o} e^{j\omega t}}{\left | Z \right | e^{j\phi }} = \frac{V_{o}}{\left | Z \right |} e^{j (\omega t -\phi )}$

Hence by substituting |Z| from eqn.(1) , we get

$I_{o} = \frac{V_{o}}{\left | Z \right |} = \frac{V_{o}}{\sqrt{R^{2}+ \frac{1}{\omega ^{2} C^{2}}}}$ ...........................(3)

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Amplitude of voltage across capacitor = I_o X_C

Where X_C is the impedence of capacitor that is given by , X_C = 1 / ( $\omega$ C )

By substituting I_o from eqn.(3) , amplitude of voltage across capacitor is given by

$V_{C} = \frac{V_{o}}{\sqrt{R^{2}+ \frac{1}{\omega ^{2} C^{2}}}} \times \frac{1}{\omega C}$ ................................(4)

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Let us substitute , $\alpha$ = 1/($\omega$C) in eqn.(4) , then eqn.(4) is written as

$\frac{V_{C}}{V_{o}} = \frac{\alpha }{\sqrt{R^{2} + \alpha ^{2}}}$ ...........................(5)

if V_C = 0.9 V_o , then from eqn.(5) we get

$\frac{\alpha ^{2}}{ R^{2} + \alpha ^{2}} = 0.81$

if R = 5 $\Omega$ , the we get $\alpha$ = 10.324 from above eqn.

Hence $\alpha$ = 1/($\omega$C) = 1/( 2$\pi$f C) = 10.324 ..................(6)

if capacitance of capacitor is 770 $\mu F$ , then from above eqn.(6) , we get frequency f = 20 Hz.

f V_C = 0.1 V_o , then from eqn.(5) we get

$\frac{\alpha ^{2}}{ R^{2} + \alpha ^{2}} = 0.01$

if R = 5 $\Omega$ , the we get $\alpha$ = 0.503 from above eqn.

Hence $\alpha$ = 1/($\omega$C) = 1/( 2$\pi$f C) = 0.503 ..................(7)

if capacitance of capacitor is 770 $\mu F$ , then from above eqn.(7) , we get frequency f = 411 Hz.