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Lecture #5
EGR 272 – Circuit Theory II
Phasor Analysis
“I have found the equation that will enable us
to transmit electricity through alternating
current over thousands of miles. I have
reduced it to a simple problem in algebra.”
Charles Proteus Steinmetz
The use of complex numbers to solve ac circuit problems – the so-called phasor method considered in
this chapter – was first done by German-Austrian mathematician and electrical engineer Charles
Proteus Steinmetz in a paper presented in 1893. He is noted also for the laws of hysteresis and for his
work in manufactured lighting.
Steinmetz was born in Breslau, Germany, the son of a government railway worker. He was deformed
from birth and lost his mother when he was 1 year old, but this did not keep him from becoming a
scientific genius. Just as his work on hysteresis later attracted the attention of the scientific
community, his political activities while he was at the University at Breslau attracted the police. He
was forced to flee the country just as he had finished the work for his doctorate, which he never
received. He did electrical research in the United States, primarily with the General Electric
Company. His paper on complex numbers revolutionized the analysis of ac circuits, although it was
said at the time that no one but Steinmetz understood the method. In 1897 he also published the first
book to reduce ac calculations to a science.
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(Electric Circuit Analysis, 2nd Edition, by Johnson, Johnson, and Hilburn, p. 307)
Lecture #5
EGR 272 – Circuit Theory II
Read: Chapter 9 and Appendix B in Electric Circuits, 6th Edition by Nilsson
Development of “phasor analysis”
AC circuit analysis (or phasor analysis) is very similar to DC analysis, but the
justification for the method of analysis is somewhat complicated.
Recall from a study of differential equations that the forced response to a circuit will
be in the form of the input (+ derivatives of the input). So if the input is sinusoidal,
the output must also be sinusoidal as shown below.
Input:
Vpcos(wt)
AC
Circuit
Output:
Acos(wt) + Bsin(wt)
= Kcos(wt + q)
Note that the input and the output on differ by the magnitude and the phase.
Recall that polar numbers are represented by a magnitude and a phase angle.
(More on this shortly)
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Lecture #5
EGR 272 – Circuit Theory II
Phasor – a complex number in polar form representing either a sinusoidal voltage or
a sinusoidal current.
Symbolically if a time waveform is designated at v(t), then the corresponding phasor
is designated by V.
Examples: Convert to the other notation (time waveform or phasor)
Time Domain
v(t) = 30cos(10t) V
i(t) = 4cos(400t - 15) mA
v(t) = 100sin(2t + 40) V
Phasor Domain
V  1030  ,
w  10 rad/s
Solving complex equations
Two equations can be generated from a single complex equation by equating the real
and imaginary parts of the equation.
Example:
A + jB = C + jD (one complex equation)
so A = C
(real parts of the equation)
and
B=D
(imaginary parts of the equation)
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Lecture #5 EGR 272 – Circuit Theory II
Further justification for “phasor analysis”
Suppose that we replaced a real sinusoidal forcing function v(t) = Vpcos(wt + q) with
a complex forcing function V(s) whose real part equals v(t).
Define V(s) = Vpej(wt+q) = complex forcing function.
Recall Euler's Identity: e±jA = cos(A) ± jsin(A)
So V(s) = Vpej(wt+q) = Vp[cos(wt + q ) + jsin(wt + q )]
And if the real part of V(s) is designated Re[V(s)] then
Re[V(s)] = Vpcos(wt + q) = v(t)
Replacing sinusoidal forcing functions with complex forcing functions will result in
a cancellation of exponential terms leaving only phasor terms.
Example: Use KVL on the circuit below to show how the phasor equation can be
developed.
+
+
v1 (t)
+
v(t) = Vpcos(wt+q)
_
v2 (t)
-
Since v1(t) and v2(t) can only vary from v(t) in magnitude and phase
(as seen earlier) then v1(t) and v2(t) could be expressed as:
v1 (t) = V1cos(wt+a)
v2 (t) = V2cos(wt+b)
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Lecture #5
EGR 272 – Circuit Theory II
Before we get into the details of phasor analysis, we need a method of representing
components in AC circuits. We will introduce a new term called impedance.
Complex Impedance
Z = impedance or complex impedance (in )
Z 
V
phasor voltage

phasor current
I
I
+
Z
V
_
Note that the relationship above is similar to Ohm’s Law.
Now we will define impedance for resistors, inductors, and capacitors.
Resistors:
V
RIq
If i(t) = Ipcos(wt + q), find v(t) and show that
ZR 

 R0  R
I
Iq
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Lecture #5
EGR 272 – Circuit Theory II
Inductors:
If i(t) = Ipcos(wt + q), find v(t) and show that
ZL  jwl  j2 fL  wl90
Capacitors:
1
1
-j
1
Z




  90
If v(t) = Vpcos(wt + q), find i(t) and show that C
jwC
j2 fC
wC
wC
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Lecture #5
EGR 272 – Circuit Theory II
Example: Find I(t) in the circuit below using phasor analysis.
i(t)
+
100cos(500t) V
2H
400
_
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Lecture #5
EGR 272 – Circuit Theory II
KVL and KCL in AC circuits:
KVL and KCL are satisfied in AC circuits using phasor voltages and currents.
They are not satisfied using the magnitudes of the voltages or the currents.
Example: For the previous example, show that:
V  VL  VR
V  VL  VR
(KVL is satisfied using phasors)
(KVL is NOT satisfied using magnitudes only)
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Lecture #5
EGR 272 – Circuit Theory II
Phasor diagrams:
KVL and KCL are only satisfied by phasors as indicated above, but it can be
difficult to see exactly why certain voltages combine or cancel in some circuits.
A phasor diagram is a vector diagram that illustrates graphically exactly how
the phasors combine. Phasor diagrams will help the student to get a better feel
for what happens when AC voltages or AC currents combine.
Example: For the previous example, show that V  VL  VR
using a phasor diagram.
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Lecture #5
EGR 272 – Circuit Theory II
AC Circuit Analysis Procedure:
1) Draw the phasor circuit (showing voltage and current sources as phasors and
using complex impedances for the components).
2) Analyze the circuit in the same way that you might analyze a DC circuit.
3) Convert the final phasor result back to the time domain.
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Lecture #5
EGR 272 – Circuit Theory II
Example: Analyze the circuit below using phasor analysis. Specifically,
A) Find the total circuit impedance
B) Find the total current, i(t)
C) Use current division to find i1(t) and i2(t)
D) Show that KCL is satisfied using current phasors, but not current magnitudes.
i (t)
2
i(t)
4H
+
i (t)
1
20 uF
120cos(100t)
300
200
_
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Lecture #5
EGR 272 – Circuit Theory II
Example: (continued)
E) Show that KCL is satisfied using a phasor diagram.
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