Circuit Theory
Chapter 9B
Sinusoidal Steady-State :
Node Voltage, Mesh
Current, Thevenin
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Sinusoidal Steady-State Analysis
Chapter 10
10.1
10.2
10.3
10.4
10.5
10.6
Basic Approach
Nodal Analysis
Mesh Analysis
Superposition Theorem
Source Transformation
Thevenin and Norton Equivalent Circuits
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10.1 Basic Approach (1)
Steps to Analyze AC Circuits:
1.
2.
3.
Transform the circuit to the phasor or frequency
domain.
Solve the problem using circuit techniques
(nodal analysis, mesh analysis,
superposition, etc.).
Transform the resulting phasor to the time
domain.
Time to Freq
Solve
variables in Freq
Freq to Time
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as circuits get more "complex"...
4
...we'll start writing matlab
scripts to solve them
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Complex Polar Helper Functions
download into your working folder
• ang.m: helps convert angle to rect
Here's how to use it:
--> 5*ang(45)
ans =
3.5355 + 3.5355i
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magPhs.m – prints in polar
--> magPhs(40 + 30j)
50.000 < 36.87
this just PRINTS in polar. Do not use in
calculations. Only at end to get polar 7
Find v(t) using NV
is=10cos(wt) A, vs = 100sin(wt) V
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% NVdemo.m
% solution to first NV demo in lecture 9b
% time domain values
w = 50000;
C = 9e-6;
L=100e-6;
R1=5;
R2=20;
% phasor domain
Is=10*ang(0);
Vs=100*ang(-90);
ZC = -j/(w*C);
ZL = j*w*L;
V = (Is + Vs/20)/(1/5 + 1/ZC + 1/ZL +1/20)
magPhs(V); % print out V in polar form
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10.2 Nodal Analysis (1)
Example 1
Using nodal analysis, find v1 and v2 in the circuit
of figure below.
Answer:
v1(t) = 11.32 sin(2t + 60.01) V
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v2(t) = 33.02 sin(2t + 57.12) V
%
%
%
w
C
L
NVdemo2.m
time domain
= 2;
= 0.2;
= 2;
% phasor domain
Is = 10*ang(-90);
ZC = -j/(w*C);
ZL = j*w*L;
%
%
%
%
%
Node eq at
-Is + V1/2
Node eq at
(V2-V1)/ZC
Vx = V1
V1:
+ (V1-V2)/ZC
V2:
+ V2/ZL + (V2-3*Vx)/4
%
%
%
%
%
Rearranging
V1( 0.5 + 1/ZC) - V2/ZC = Is
V1( -1/ZC -3/4) + V2(1/ZC + 1/ZL +1/4) = 0
A * Vunknown = B
1/R * V = I -> 1/Z *V = I -> Y*V = I
% matrix of admittances between nodes:
Y = [ .5 + 1/ZC , -1/ZC ; -1/ZC - 3/4, 1/ZC + 1/ZL + 1/4 ]
I = [ Is ; 0]
V = inv(Y)*I % V = Y\I
% solution vector of node voltages
magPhs(V) % Shows V1 and V2 in polar form
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Output from NVdemo2
Y=
0.5000 + 0.4000i 0.0000 - 0.4000i
-0.7500 - 0.4000i 0.2500 + 0.1500i
I=
0.0000 - 10.0000i
0
V=
9.8113 - 5.6604i
27.7358 - 17.9245i
11.327 < -29.98
33.024 < -32.87
(rectangular form)
(polar form)
Find the indicated currents expressed as cosine functions. Use
the node voltage analysis method first.
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i1(t) = cos(5t - 30°) A; i2(t) = 0.658cos(5t + 14.6°) A
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Mesh Analysis
Use mesh current analysis to find the
phasor voltages V1 and V2.
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V1 = 22.9∠110° V; V2 = 19.6∠-101° V
Solution to previous in
Freemat
% I*Z = V in phasor domain
Z = [ -2j, 5+3j ; -1,1]
V = [0; 10*ang(0) ]
I = inv(Z)*V
magPhs(I(1))
magPhs(I(2))
V1 = -2j*I(1)
magPhs(V1)
V2 = 5*I(2)
magPhs(V2)
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Output from previous slide
Z =
-0.0000 - 2.0000i
-1.0000 + 0.0000i
V =
0
10.0000 + 0.0000i
I =
-10.7692 - 3.8462i
-0.7692 - 3.8462i
11.435 < -160.35
3.922 < -101.31
V1 =
-7.6923 + 21.5385i
22.871 < 109.65
V2 =
-3.8462 - 19.2308i
19.612 < -101.31
5.0000 +
1.0000 +
3.0000i
0.0000i
(rectangular)
(polar)
(rectangular)
(polar)
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10.3 Mesh Analysis (1)
Example 2
Find Io in the following figure
using mesh analysis.
Answer: Io = 1.19465.44 A
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Find the indicated mesh
currents.
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I1 = 3.88 ∠ -153° A; I2 = 3.48 ∠ 176° A
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10.4 Superposition Theorem (1)
When a circuit has sources operating at
different frequencies,
• The separate phasor circuit for each
frequency must be solved
independently, and
• The total response is the sum of timedomain responses of all the individual
phasor circuits.
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10.4 Superposition Theorem (2)
Example 3
Calculate vo in the circuit of figure shown below
using the superposition theorem.
Vo = 4.631 sin(5t – 81.12) + 1.051 cos(10t – 86.24) V
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10.5 Source Transformation (1)
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10.5 Source Transformation (2)
Example 4
Find Io in the circuit of figure below using the
concept of source transformation.
Io = 3.28899.46 A
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10.6 Thevenin and Norton
Equivalent Circuits (1)
Thevenin transform
Norton transform
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Thevenin:
Find the Thevenin equivalent circuit at the
terminals F-G. Express all complex values
in your solution in both rectangular and
polar form.
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VTH = 250 - j200 V = 320 ∠ -38.7° V; ZTH = 100 + j20 Ω = 102 ∠ 11.3° Ω
10.6 Thevenin and Norton
Equivalent Circuits (2)
Example 5
Find the Thevenin equivalent at terminals a–b of
the circuit below.
Zth =12.4 – j3.2 
VTH = 18.97-51.57 V
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1) Use nodal analysis to determine which
impedance element has the lowest voltage
magnitude across its terminals.
3.3 kΩ resistor
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(voltage magnitude is 2.07 V, where other voltage magnitudes are 25.6 V and 13.7 V)
2)
Find the indicated mesh currents.
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I1 = 3.88 ∠ -153° A; I2 = 3.48 ∠ 176° A
3) Apply repeated source transformations to reduce this to an equivalent circuit at the
terminals G-H. The simplified circuit will consist of a voltage source in series with two
series-connected passive elements.
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Voltage source = 1140cos(60t + 68.4°); 200 Ω resistor; 74.1 μF capacitor
4) Find the Norton equivalent circuit at the terminals Q-R. Express all complex values in
your answer in both rectangular and polar form.
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IN = 0.769 - j1.15 A = 1.39∠-56.3° A; ZN = 0.523 - j1.27 W = 1.38∠-67.7° W
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