4.0 ac circuits - prepared by naz 1st half

8/13/2019 4.0 AC CIRCUITS - Prepared by NAZ 1st Half

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Vp

Ip

t

Vp

Ip

t

Vp

Ip

t

Resistor Capacitor Inductor



Impedance

52

Series Parallel

Zeq =Z1+Z2+Z3+…+Zn 1/Z eq =1/Z1+1/Z2+1/Z3+...+1/Zn



Impedance

53

The impedance, Z, of a circuit relates peakcurrent to peak voltage:

I V

Z p

p

(Units: Ohms)



Example 4.3.2

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A practical capacitor can be modeled by an ideal

capacitor in parallel with a resistor. Theresistance represents leakage losses in thecapacitor and is usually quite large. Find theimpedance of a practical capacitor at the radianfrequency ω =377 rad/s (60 Hz). C

1=10 -9 F, R

1=1M Ω

Impedance



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Z1=R 1||X C;R1=10 6 XC=1/jωC1=1/(377)(10 -9)=2.65 x 10 6 / jΩ

Z1= Write in polar form:

A= 10 6/√(12+0.377 2)=9.36 x 10 5

θ=tan -1(0.377/1) =-20.66 ° =-0.36 radZ(j377)= 9.36 10 ∠ − 0.36 Ω @



Example 4.3.3

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A Voltage Source V=100 ∠0° is supplied to 3

impedances in series; Z 1=5 ∠53° , Z 2=3 ∠90° ,Z3=1∠ − 90° .

Calculate Z eq , currents supplied and voltage at Z 2

element.

Impedance



Example 4.3.4

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A parallel circuit of several circuit elements contain

Z1=5.6 ∠30° , Z2=2.2 ∠ − 20° , Z3=3.3 ∠45° withV=120 V. Calculate Zeq and currents.

Impedance



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Zeq=(1/z1+1/z2+1/z3) -1

Zeq=[(1/5.6 ∠30°)+(1/2.2 ∠ − 20° )+(1/3.3 ∠45°)]-1

Zeq=[(0.18 ∠ − 30°) + (0.45 ∠20°)+(0.3 ∠ − 45° )]-1

Zeq=[(0.16-0.09j)+(0.42+0.15j)+(0.21-0.21j)] -1



Chapter 4 : AC Circuits 60

4.0 ac circuits - prepared by naz 1st half

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