EIT Review – Electric Circuits
Electric circuits are used in the generation, transmission and
consumption of electric power and energy.
Electric circuits are used in the encoding, decoding, storage,
retrieval, transmission and processing of information.
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Interpreting the circuit diagram
An electric circuit is an interconnection of electric circuit elements. (Circuit elements are also called
devices or components.)
Each circuit element has at least two terminals, i.e. places where that element can be connected to other
circuit elements. (Terminals are sometimes called leads.) The parts of the circuit where terminals are
connected together are called nodes. (Nodes are also called vertices.)
This circuit is an interconnection of two terminal elements.
○ The shape of the element indicates it behavior.
○ Each element is characterized by a parameter, represented either as a value with units or as a variable.
○ Each element has two terminals that are connected to nodes of the network. The element is said to be
incident to the nodes ate which its terminals are connected.
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Element Equations
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Current and Voltage
Two quantities are identified for each two terminal circuit element: the
element current and the element voltage.
• Current and voltage each have a direction as well as value.
• Changing the direction corresponds to multiplying the value by –1
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Voltage and Current Waveforms
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Steady State and Transient Parts of a Complete Response
DC Circuits:
○ All of the inputs are constant voltages and currents.
○ The circuit is at steady state.
○ All of the voltages and currents are constant functions of time and can
be represented using real numbers
AC Circuits:
○ All of the inputs are sinusoidal voltages and currents having the same
frequency.
○ The circuit is at steady state.
○ All of the voltages and currents are sinusoidal functions of time at the
input frequency and can be represented using complex numbers
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The Passive Convention
When the reference direction of a particular current points from the + toward the  of the
polarity of a particular voltage, that current and voltage are said to adhere to the passive
convention.
• ia and va adhere to the passive convention
• ib and vb adhere to the passive convention
• ia and vb adhere to the passive convention
• ib and va adhere to the passive convention
The “element equations”, e.g. Ohm's law, use the passive convention:
va = R ia
and
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vb = va = R ia
The Passive Convention
Specification of power and energy use of the passive convention.
Power and Energy:
p  v i and w   pdt
power
uses the passive convention
supplied by
No
supplied to
Yes
absorbed by
Yes
received by
Yes
delivered by
No
delivered to
Yes
In cases a and b, the element receives 24 W. In cases c and d, the element supplies 24 W.
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Kirchhoff's Laws
KCL: The algebraic sum of the currents entering
any node is zero.
i1  i 2  i3  i 4  0, i 4  i5  0  i 4  i5 ,
i1  i 2  i3  i5  0,...
KVL: The algebraic sum of the voltages in any
loop are zero.
v1 v4 v5 v6  0, v2 v3  0,...
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Informal Analysis of DC circuits
Problem Determine the voltage and current of each of the circuit
elements in this circuit.
Next, apply KCL at the bottom right node to get
i3 i4 0.251.00.250.75 A
Next, apply KVL to the loop consisting of the voltage source
and the 60 W resistor to get
v2 15  0  v2  15 V
Apply Ohm’s law to each of the resistors to get
v2 15
i 2    0.25 A, v3 10 i3 100.75  7.5 V,
60 60
Solution We can label the circuit as follows:
v3 10 i3 100.75  7.5 V
and
v4  20i 4  201  20 V
Next, apply KCL at the bottom left node to get
i1 i3 i2 0.750.251.0 A
Next, apply KVL to the loop consisting of the 0.75 A current
source and three resistors to get
v6  v4 v3 v2 20(7.5) 15 12.5 V
Apply KCL at the top right node to get
i4 0.250.75  0  i4 1.0 A
Finally, apply KVL to the loop consisting of the 0.25 A current
source and the 20 W resistor to get
v5  v4   20  20 V
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Informal Analysis of DC circuits
Problem Determine the voltage and current of each of the circuit
elements in this circuit.
Next, apply KCL at the bottom right node to get
i3 i4 0.251.00.250.75 A
Next, apply KVL to the loop consisting of the voltage source
and the 60 W resistor to get
v2 15  0  v2  15 V
Apply Ohm’s law to each of the resistors to get
v2 15
i 2    0.25 A, v3 10 i3 100.75  7.5 V,
60 60
Solution We can label the circuit as follows:
v3 10 i3 100.75  7.5 V
and
v4  20i 4  201  20 V
Next, apply KCL at the bottom left node to get
i1 i3 i2 0.750.251.0 A
Next, apply KVL to the loop consisting of the 0.75 A current
source and three resistors to get
v6  v4 v3 v2 20(7.5) 15 12.5 V
Apply KCL at the top right node to get
i4 0.250.75  0  i4 1.0 A
Finally, apply KVL to the loop consisting of the 0.25 A current
source and the 20 W resistor to get
v5  v4   20  20 V
http://people.clarkson.edu/~jsvoboda/Syllabi/ES250/dcCkts/More%20Ad%20Hoc%20Analysis.pdf
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Dependent Sources
•
•
•
•
Controlling and controlled elements
Gain
Units of the gain
Power
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Node Voltages
Node (Voltage) Equations
1. Express everything in terms of the node voltages.
2. Apply KCL at all nodes except for the reference node.
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Example:
Apply KCL at node 1 to get
v1 v1  v 2

20
8
25
We will simplify this equation by doing two things:
1. Multiplying each side by to eliminate fractions.
2. Move the terms that don’t involve node voltages to the right side of the
equation.
The result is
Emphasize and label the nodes:
33 v1  8 v 2  400
Next, apply KCL at node 2 to get
v2 v2  24 v1  v2


9
14
25
Simplify this equation:
126v1  701v2  5400
Solving these simultaneous equations, perhaps using MATLAB, the node
voltages are v1 = -10.7209 V and v2 = 5.7763 V.
v 3  24 V
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Mesh Currents
Mesh (Current) Equations
1. Express everything in terms of the mesh currents.
2. Apply KVL to all meshes.
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Mesh (Current) Equations
1. Express everything in terms of the mesh currents.
2. Apply KVL to all meshes.
Example:
Notice that
i3  2 A
Apply KVL to mesh 1 to get
25i1  i3   9i1  i 2   8i1  0
Substituting
Solution: Label the label the mesh currents.
Then, label the element currents in terms of
the mesh currents:
i3  2 A
and doing a little algebra gives
42 i 1  9 i 2  50
Next, apply KVL to mesh 2 to get
14  i 2  i 3   24  9  i1  i 2   0
Substituting and doing a little algebra gives
9 i1  23 i 2  24  14  2   4
Solving these simultaneous equations, perhaps using MATLAB, the
mesh currents are i1 = 1.3401 A andi2 = 0.6983 A.
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Circuit Equivalence
Series and Parallel Resistors - Voltage and Current Division
Equivalence: Replacing some circuit elements by an equivalent element does not
change the current or voltage of the remaining circuit elements.
Successful application of circuit equivalences involves recognizing opportunities
to simplify the a circuit without changing the quantity of interest.
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Voltage Division
Example
The input to this circuit is the voltage of the
independent voltage source. The output is the
voltage measured by the meter.
The output is proportional to the input.
Determine the value of the constant of
proportionality.
Solution
Using equivalent resistance and voltage division in the
right part of the circuit
va 
20 20
1
vs  vs
20   20 20
3
Using voltage division in the left part of the circuit
3
1
 12 
vo  
 10va   10 vs  2vs
12

8
5
3


So vo is proportional to vs and the constant of
proportionality is 2 V/V.
http://people.clarkson.edu/~jsvoboda/Syllabi/ES250/vDiv/VdivVCVSworksheet.pdf
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Source Transformations
Equivalence: Replacing some circuit elements by an equivalent element does
not change the current or voltage of the remaining circuit elements.
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Superposition
The response of a linear circuit to several inputs working together is equal to the sum of the responses
to each input working separately.
Circuit inputs and outputs
 Inputs are usually the voltages of independent voltage sources and the currents of independent current
sources.
 Outputs can be any current or voltage.
 The circuit designer designates the input and output of the circuit.
 A circuit can have more than one input or output.
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Superposition
The response of a linear circuit to several inputs working together is equal to the sum of the responses to each
input working separately.
Example
This circuit has one output, vo , and three inputs, v1, i2 and v3.
Express the output as a linear combination of the inputs.
Solution 1
Writing and solving node equations gives
vo 
v1
5
 8i2 
v3
5
Solution 2
vo 
10
1
v1  v1
40 10
5
vo 
40 10
i2  8 i2
40  10
http://people.clarkson.edu/~jsvoboda/Syllabi/ES250/Thev/SuperpositionHOsoln.pdf
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vo 
10
v3    15 v3
40 10
Sunday, January 23, 2011
11:42 AM
Thevenin Equivalent Circuits
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Example
We want to find the Thevenin equivalent circuit for this circuit:
Solution
voc 2 V
i sc 
3
A
4
Finally, here's the Thevenin equivalent circuit:
http://people.clarkson.edu/~jsvoboda/Syllabi/ES250/Thev/ThevVCVS_HOsoln.pdf
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Rt  
8

3
Op Amps
Ideal Op Amp
Op Amp IC
5 Terminal Op Amp
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Op Amp Circuits
Determine the values of the node voltages; v1, v2 and vo ; of this circuit.
Solution
0.25v1
2010
3

v1
2010
3
vo v2
0  v1 0.125 V125 mV
v1

v2
2010 20103
3

v2
2010 20103
3
0  vo 2v2 25 mV
0  v2 v1 125mV
http://people.clarkson.edu/~jsvoboda/Syllabi/ES250/oaCkts/ioaHOsoln.pdf
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Capacitors
Inductors
Capacitors act like open circuits in dc circuits.
Inductors act like short circuits in dc circuits.
The capacitor voltage is continuous unless
the capacitor current is unbounded.
The inductor current is continuous unless the
inductor voltage is unbounded.
Circuits containing capacitors and inductors are represented by differential equations:
After the switch closes, this circuit is represented by the differential equation
 10 R  d
 1 
20
i t   
for t  0
 i t   
 i t  
CL
dt
CL
 3 R 10 L  dt
d2
2
When
1) R = 10 W, L = 0.4 H and C = 0.25 mF
2) The circuit is at steady state before the switch closes.
The capacitor voltage, v(t), can be shown to be
vt1616.525e2.5t cos 9.682t 165.5 V for t 0
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First Order Circuits
i t   I sc  i 0  I sc eat for t  0
vt Voc v0 Voc eat for t 0
where
a
where
1
a
Rt C
and the initial condition, v(0), is the capacitor
voltage at time t = 0.
L
and the initial condition, i(0), is the inductor
current at time t = 0.
http://people.clarkson.edu/~jsvoboda/Syllabi/ES250/foc/FOC_HOsoln.pdf
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Rt
First -Order Circuits
http://people.clarkson.edu/~jsvoboda/Syllabi/ES250/foc/FOCsteps.pdf
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First-Order Circuit Example
This diagram represents a circuit for t ≥ 0 . Given the initial condition i(0) = 2.4 A and the
inductance, L = 6 H, represent the inductor current , i(t), as a function of t for t  0.
Solution
Simplify the circuit using source transformations and equivalent resistance:
Now recognize that Is c = 0.8 A and Rt = 18 Ω. Then
a
Rt 18
1
 3
L 6
s
Finally
i t   0.82.40.8e3t  0.81.6e3t for t  0
http://people.clarkson.edu/~jsvoboda/Syllabi/ES250/foc/FOC_HOsoln.pdf
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AC and DC Circuits
DC Circuits:
○ All of the inputs are constant voltages and currents.
○ The circuit is at steady state.
○ All of the voltages and currents are constant functions of time and
can be represented using real numbers
AC Circuits:
○ All of the inputs are sinusoidal voltages and currents having the
same frequency.
○ The circuit is at steady state.
○ All of the voltages and currents are sinusoidal functions of time at
the input frequency and can be represented using complex numbers
Voltage and Current Waveforms
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Phasors
Impedances
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Time and Frequency Domains
Time Domain
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Frequency Domain
DC Circuits (in the Time Domain)
AC Circuits (in the Frequency Domain)
i1  i 2  i
v1 
v2 
R1
R1  R2
I1  I 2  I
v
V1 
v
V2 
R2
R1  R2
i2 
R2
R1  R 2
R1
R1  R 2
Z1  Z 2
V
Z2
Z1  Z 2
V
V1  V 2  V
v1  v 2  v
i1 
Z1
i
I1 
i
I2 
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Z2
Z1  Z 2
Z1
Z2  Z2
I
I
AC and DC Circuits:
DC Circuit:


KCL and Ohm's law: v c  4 i 2  4.5  4 i 2  4.5
KVL:  10 i 2  2.5 v c  v c  0
v c   45 
 1 4 v c  18
1.5 10   i    0    i   6.75

  2 


 2 
v o  2.5 v c  112.5 V
AC Circuit:


KCL and Ohm's law: Vc   j 4 I 2  4.50   j 4 I 2  j18
KVL:  j10 I 2  2.5 Vc  Vc  0
Vc   j18
Vc  11.2590
I      I   

 2  0 
 2  1.68750
Vo  2.5 Vc  28.12590 V
j4 
1
1.5  j10


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