16
ELECTRICAL CIRCUITS
2. BASIC CIRCUIT LAWS
Introduction
Circuit theory is the center of Electrical Engineering. It comes from the Electromagnetic
branch of Physics. Electromagnetics is the study of Electric charge, Electric fields,
Magnetic fields, and their interaction with each other, time and space. Basic circuit theory
has 2 principal variables: voltage and current. The concept of voltage comes from the
idea of an experiment of moving a test quantity of electric charge (think of electrons for
negative charge or protons for positive charge) about in an electric field. Since an electric
field exerts a force on the charge, force acting on a body moved over a distance results in
work. If that work is normalized to the quantity of charge used for our experiment then
that normalized work becomes voltage which in MKS units is joules per coulomb. Joules
and coulombs are the MKS units for work and charge. Note that if the test charge is
moved in any arbitrary manor that ends back at the starting point the work done against
the force exerted on the test charge is zero. This is because electric field, like gravity, is a
conservative field. Voltage is the potential for doing work. Current, the other principal
variable is in fact flowing charge. That charge flows in a medium called a conductor.
Conductors are typically metal such as copper, aluminum, silver and gold. These
particular metals are very good conductors, but not perfect. For the development of
circuit theory we will assume we have the perfect conductor for carrying current in the
circuit. Think of the conductor as a pipe and the current in the pipe is the total charge per
unit time that is flowing through a cross section of the conductor. Current in MKS units is
amps which is coulombs per second. Think of voltage as the pressure that is causing the
charge to flow.
Voltage Sources Current Sources and Resistors
Voltage sources, current sources and resistors are 2 terminal elements. We consider
voltage and current sources from the theoretical perspective, although a good physical
and simple example of a voltage source is a battery. There is no simple physical example
for a current source. It is the property of a voltage source that at its terminals it will flow
what ever current is necessary to maintain its voltage. The current will flow out from its
positive terminal and return to its negative terminal. Additionally, the voltage polarity is
plus at the positive terminal and minus at the negative terminal. It is the property of a
current source that at its terminals it will develop whatever voltage is necessary to
maintain it's source current. Additionally, the current flows from its output terminal and
returns to its input terminal. It is the property of a resistor that the voltage developed
across a resistor is proportional to the current flowing through it. An electric circuit is an
interconnection between voltage sources, current sources and resistors with current
carrying conductors (assumed perfect). It is the property of 2 terminal elements that
whatever current flows out one terminal must flow into the other terminal. It is the
property of all circuits that there must be a complete path in that whatever current goes
out must flow back. The symbol for the voltage source is Figure 1a, the symbol for the
current source is Figure 1b and the resistor is 1c
17
Figure 1a Voltage Source
1b Current source
1c Resistor
We consider a plot of current (Y axis) vs. voltage (X axis) for each of these devices.
Figure 2 shows the voltage source. It is plotted with a value of Va .
Figure 2 Voltage current relationships for the voltage source
Observe that the current can be anything but the voltage is constant, Va . Figure 3 gives
the voltage current relationships for the current source.
Figure 3 Voltage current relationships for the current source
In this case the plot shows that the voltage can be anything but the current is constant Ia .
Next we consider the voltage current relationship for the resistor. Figure 4 gives the plot.
Figure 4 Voltage current relationship for the resistor
18
As seen in Figure 4 the voltage and current are proportional to each other for the resistor.
The convention for that relationship is given by Equation 1 also known as Ohms Law.
V  RI
(1)
Figure 5 illustrates the voltage and current polarities for the resistor. Figure 6 re-plots
Figure 4 to represent Equation 1.
Figure 5 V/I polarity for the resistor
Figure 6 Equation 1 resistor voltage current relationships
Observe the slope of the curve in the Figure is resistance which has the units of
Ohms = Volts/Amp.
Kirchhoff's Voltage and Current Laws
A simple electric circuit is a complete path interconnection, using perfect conducting
wire, of voltage sources, current sources or resistors. Figure 7 illustrates a voltage source
connected to a resistor.
Figure 7 Simple resistive circuit
19
By inspection we see that the voltage, Vr across the resistor is the battery voltage, Va .
Plugging this into Equation 1, Ohm's Law, and solving for the current Ir :
Ir 
Va
R
(2)
Recall in the Introduction that the work done in moving a quantity of charge around a
closed path in an electric field is always zero. Now look at Figure 1 and sum up the
voltages going around the closed current path. If we start at the negative terminal of the
battery traveling in the direction of the current we see a positive voltage rise of the
battery voltage Va and then a voltage drop across the resistor of Vr and the work done in
going around that complete path must be zero:
Va  Vr  0
(3)
This agrees with our previous observation that: Va  Vr . This simple observation is
restated as Kirchhof's Voltage Law: The sum of the voltage rises and voltage drops
around any closed loop in any circuit is always zero. Let's define a loop: In a circuit
a loop is a closed path through conductors or elements which returns to a starting
point. This law gives rise to loop equations for analysis of circuits. In general with loop
equations applied to a circuit the resistors and the voltage sources will be known and the
currents are the unknowns. The convention is illustrated by Figure 8: Assign unknown
loop currents all assumed in a clockwise direction. If the actual current is in a different
direction the answer will turn out negative. For a loop equation, put all the voltage source
magnitudes on the left side of the equation with a positive sign if the loop current is going
out of its positive terminal or with a negative sign if its going into its positive terminal. If
a voltage source is shared by 2 loops pick the sign it will get as if it only sees the current
for the loop presently being analyzed Put all the resistive voltage drops (Ohms Law) on
the right side of the equation. If a resistor is shared by 2 loops use the difference in loop
currents such that for the loop being analyzed the resistive voltage drop for that loop is
positive like Figure 7.
20
Figure 8 example circuit for loop equations
The loop equations obtained from Figure 8 are as follows:
Va  Vb  I1 R1   I1  I 2  R2
Vb   I 2  I1  R2  I 2 R4   I 2  I 4  R7   I 2  I 3  R5
0   I 3  I 2  R5  I 3  R3  R6 
0   I 4  I 2  R7  I 4  R8  R9  R10 
(4)
These Equations can be rearranged to a standard form and then converted into matrix
notation for matrix solution of simultaneous equations. However, one can go directly to
standard form. Pick loop currents all in a clockwise direction and put loop excitation
voltage on the left of the equal sign and voltage drops on the right. Notice that the
unknown array is:  I1 I 2 I3 I 4  . When writing the equations always write the voltage drop
terms in the same order given in this array. If an unknown current is not present include it
with a coefficient of zero and adjacent loop currents will generate terms that are negative.
The loop equations for Figure 8 written this way are:
Va  Vb  I1  R1  R2   I 2 R2  I 3 0  I 4 0
Vb   I1 R2  I 2  R2  R4  R5  R7   I 3 R5  I 4 R7
0  I1 0  I 2 R5  I 3  R3  R5  R6   I 4 0
0  I1 0  I 2 R7  I 3 0  I 4  R7  R8  R9  R10 
Now these equations can go directly into matrix form:
Va  Vb   R1  R2 
 Vb    R
2



 0 
0

 
0
 0  
 R2
 R2  R4  R5  R7 
 R5
 R7
0
 R5
 R3  R5  R6 
0
0
  I1 
 
 R7
 I2 
  I3 
0
 
 R7  R8  R9  R10   I 4 
If numerical values are available for the resistors and the voltage sources then Matlab and
most pocket calculators can solve this via matrix inversion. A symbolic solution would
require inversion of a 4  4 matrix.
Kirchhof's voltage law is based upon the physical property of an electric field being
conservative. Kirchhof's current law is based upon a fundamental law of physics which is
that charge is always conserved. First a definition: In a circuit a node is the junction of
2 or more conductors. Now, Kirchhof's current law can be stated in 3 ways: Total
current into a node must equal total current out of a node; The sum of all currents
into a node is zero; or The sum of all currents out of a node is zero. With loop
equations the sum of the voltage drops around any closed loop in a circuit is set equal to
21
zero and loop currents were solved for. For node equations sum of all current in is set to
zero or sum of all out is set to zero or in equals out. With these node equations node
voltages are the unknowns and will be solved for. Which ever form of node equation is
used one should use the same for the complete set of equations for a given circuit.
Clearly, from the definition of a node, in a circuit there are an infinite number of nodes.
Nodes of interest would be one picked between two elements or the junction of 3 or more
conductors. To analyze a circuit with node equations, define a reference node as the
"ground" and assign the voltage of zero. Then, identify the remaining nodes of interest.
Assign unknown node voltages to those nodes. If only a voltage source separates one
node from another then the 2 nodes are related by either +/- the voltage source. Consider
an example, Figure 9:
Figure 9 example circuit for nodal analysis
Observe in Figure 9 the ground node at the negative terminal of the battery, the triangle
symbol. Figure 10 illustrates the assignment of the unknown nodes.
Figure 10 node example with assignment of unknown nodes
In writing node equations we use the current form of Ohm's Law: I 
sum all currents out of a node set equal to zero.
V
. We choose to
R
22
Va :
Vb :
Vc :
Vd :
Ve :
Vf :
Vg :
Vh :
Va  VIN Va  Vb Va  Vd


0
R1
R4
R2
Vb  Va Vb  Vc Vb  Ve


0
R4
R9
R7
Vc  Vb Vc  V f

0
R9
R12
Vd  Va Vd  Ve Vd  0


0
R2
R5
R3
Ve  Vd Ve  Vb Ve  V f Ve  Vg



0
R5
R7
R10
R8
V f  Ve V f  Vc V f  Vh


0
R10
R12
R13
Vg  0 Vg  Ve Vg  Vh


0
R6
R8
R11
Vh  Vg Vh  V f

0
R11
R13
These 8 equations can be manipulated into matrix notation for a matrix solution. If values
are available Matlab or most scientific calculators will solve that numerical problem. As
for a symbolic solution, the inversion of an 8  8 symbolically would be a formidable
task. As with loop equations node equations can be written directly in matrix notation.
The array of unknown node voltages is: Va Vb Vc Vd Ve V f Vg Vh  . As the
equation is being written if a node voltage is not present include it with a coefficient of
zero. Start with node Va and assume all the voltages to be subtracted from node Va are
zero and write an expression for the current going out from Va . Next subtract those
voltages that were assumed zero but divide by the appropriate resistor. Do the whole
process in the order that the node voltages appear in the array and move excitation
voltage to the other side of the equal sign also divided by the appropriate resistor. Using
that approach the ordered equations ready for matrix form are:
23
24
Va :
Vb :
Vc :
Vd :
Ve :
Vf :
Vg :
Vh :
 1
 1 
 1 
 1
1
1 
Va  
   Vb     Vc  0   Vd     Ve  0   V f  0   Vg  0   Vh  0   VIN  
 R1 R2 R4 
 R4 
 R2 
 R1 
 1
 1 
 1 
 1 
1
1 
Va     Vb 

   Vc     Vd  0   Ve     V f  0   Vg  0   Vh  0   0
 R4 
 R4 R7 R9 
 R9 
 R7 
 1 
 1
 1 
1 
Va  0   Vb     Vc  
  Vd  0   Ve  0   V f  
  Vg  0   Vh  0   0
 R12 
 R9 
 R9 R12 
 1
 1 
 1 
1
1 
Va     Vb  0   Vc  0   Vd 

   Ve     V f  0   Vg  0   Vh  0   0
 R2 
 R2 R3 R5 
 R5 
 1
Va  0   Vb  
 R7

 1 
 1
 1 
 1 
1
1
1 


  Vc  0   Vd     Ve  
  Vf  
  Vg     Vh  0   0

 R5 
 R5 R7 R8 R10 
 R10 
 R8 
 1 
 1
 1 
 1 
1
1 
Va  0   Vb  0   Vc  


  Vf 
  Vg  0   Vh  
0
  Vd  0   Ve  
 R12 
 R10 
 R10 R12 R13 
 R13 
 1
Va  0   Vb  0   Vc  0   Vd  0   Ve  
 R8

 1
 1 
1
1 

  V f  0   Vg  
  Vh  
0
 R11 

 R6 R8 R11 
 1 
 1
 1 
1 
Va  0   Vb  0   Vc  0   Vd  0   Ve  0   V f  

  Vg  
0
  Vh 
 R11 
 R13 
 R13 R11 
Putting this set of equations in matrix form:
25

VIN

0

0
0

0

0
0

0
 1 
 
 R1  











 1
 1 
1
1   1 

0
0
0
0
0
 
  
 
 R2 
 R1 R2 R4   R4 

 1
 1 

1
1   1 
  1 

   
0
0
0
0

 
 R4 
 R4 R7 R9   R9 
 R7 

 1 
 1
 1 
1 


0
0
0
 

 0


0
R9 
R9 R12 
R12 




 1 
 1
1
1   1 
0
0

   
0
0
0
  

 R2 
 R2 R3 R5   R5 

 1 
 1 
 1
 1 
1
1
1   1 
0
0


0
 
 
 
 

 

 R7 
 R5 
 R5 R7 R8 R10   R10 
 R8 


 1 
 1
 1 
 1 
1
1 
0
0





 0




0
R
R
R
R
R

12

10
10
12
13




 R13 


 1 
 1
1
1   1 
0
0
0
0
0


 

 


 R8 
 R6 R8 R11   R11 

 1 
 1
 1 
1

0
0
0
0






0
 R11 
 R13 
 R11 R13








 Va 
 
 Vb 
 V 
 c 
 Vd 
 V 
 e 
 V f 
 
 Vg 
 Vh 








26
This approach will generate the circuit node equations but the question is raised as to which
approach, node or loops, should be used? Both should be considered and the simplest
selected. Consider Figure 9 set up for loop equations.
Figure 11 circuit configured for loop equations
Clearly in this case with 5 loop currents but 8 node voltages, the circuit should be analyzed
with loops.
Resistance and Conductance
Observe in the 2 forms of Ohm's Law:
V  IR
(5a)
I
V
R
(5b)
Also observe in the loop equations we see the coefficients of the unknown currents being
sums of resistors and in the node equations, the coefficients of the unknown currents being
sums of the inverse of resistors. Thus in Equation 5b we define a new element, conductance,
G with units of Siemens, Amps per Volt:
G
1
R
(6)
Thus Equation 5b can be rewritten:
I  VG
(7)
The symbol for conductance G is the same as resistance.
We observed in loop equations that the coefficients of the unknown current for that loop is a
sum of resistors and for node equations the coefficient of the unknown voltage is a sum of
1
or conductance. This hint suggests that resistors in a circuit can be combined to simplify
R
the circuit. Series resistors combine by adding them and parallel resistors combine by
27
changing to conductance, adding the conductance and changing the total conductance
back to resistance. Equations 8 and 9 illustrate.
N
RTOTAL   Ri
N resistors in series:
(8)
i 1
N resistors in parallel:
1
RTOTAL
N

For only 2 resistors R1 , R2 in parallel:
i 1
1
Ri
(9)
RTOTAL 
R1 R2
R1  R2
(10)
Another resistance transformation that is useful in combining resistors is the "Wye/Delta"
conversion. Figure 12 illustrates the Wye and the Delta resistor connection in which either
form could appear in a circuit. The points labeled A, B, C are connection points within a
circuit. Often in a task of combining resistance if a Wye configuration could be converted to
a Delta or visa versa then the application of Equations 8 and 9 could proceed to combine
resistance to the maximum possible. Equation 11 gives the transformations
Figure 12 the Delta and Wye resistance forms
Rc 
R1 R2  R1 R3  R2 R3
R3
Ra 
R1 R2  R1 R3  R2 R3
R2
Rb 
R1 R2  R1 R3  R2 R3
R1
Ra Rc
R1 
Ra  Rb  Rc
R2 
Rb Rc
Ra  Rb  Rc
R3 
Ra Rb
Ra  Rb  Rc
(11)
28
Note that all elements have an internal resistance or conductance and it is the slope of their
V/I characteristic curve. Examine Figure 4 for the resistor and the slope is conductance, in
Figure 5 the alternate plot and the slope is resistance. Now consider Figure 2, the V/I plot of
the perfect voltage source, the slope is conductance and it is infinite which means the
internal resistance of a perfect voltage source is zero. In the same way Figure 3 indicates
that the internal resistance of a perfect current source is infinite.
Voltage and Current Dividers
In circuit analysis a common occurrence is a voltage source driving a series connection of
resistors and the voltage drop across one of them is desired. Figure 13 illustrates a general
representation of the Voltage Divider and Equation 12 gives an expression for the desired
voltage.
Figure 13 Voltage Divider configuration
VOUT  VIN
Ri
n
R
k 1
(12)
k
A similar configuration can occur called the Current Divider, whereby input current flows to
a parallel connection of resistors and the current flowing through one of them is desired.
Figure 14 illustrates the situation and Equation 13 provides the result.
Figure 14 Current Divider configuration
29
I OUT  I IN
 1 
 
 Ri 
n
 1 
 

k 1  Rk 
(13)
Thevenin and Norton Equivalent Circuit Law
The Thevenin and Norton Circuit Law enables a very complicated circuit to be replaced
either by a single voltage source in series with resistor or a single current source in parallel
with the same resistor. Figure 15 illustrates the situation.
Figure 15 Thevenin and Norton equivalence
To obtain either form, Thevenin or Norton, in the original circuit ANALYTICALLY place a
short circuit between A and B. Then solve for the current in the short: I SC . Remove the
short and solve for the open circuit voltage, VOC , between A and B. Remember, if A and B
connect to any additional circuitry in Figure 15 to the right of the complicated circuit it must
be disconnected prior to this analysis. The Thevenin and Norton elements are as follows:
I NORTON  I SC , VTHEVENIN  VOC , REQ 
VOC
I SC
(16)
Note that REQ can also be found analytically by, in the complicated circuit, replacing all
current sources with open circuits and all voltage sources with short circuits. Then, combine
all resistance seen looking into A and B using the "Series, Parallel and Delta-Wye" resistance
combination rules.
The End, Superposition and Controlled Sources
The concept of Superposition is very simple, as electrical circuits are linear systems thus, if a
circuit has more than one source exciting it, one can remove all the sources except the one of
30
interest and replace them with their internal resistance (open for current sources and shorts
for voltage sources) and solve for the response of that source. If this is done for each of the
sources and all the responses are added up the total response is obtained.
A controlled source is either a current source or a voltage source whose output value is a
constant times a sense voltage or current elsewhere in the circuit. We consider four types:
1.
2.
3.
4.
Voltage controlled voltage
Voltage controlled current
Current controlled current
Current controlled voltage
For types 1 and 3 the control constant or gain is unitless while for type 2 the gain has units of
siemens and for type 4 it has units of ohms. NOTE: WHEN ANALYZING CIRCUITS
WITH THESE SOURCES, IN GENERAL, NEITHER SUPERPOSITION NOR
THEVENIN CAN BE APPLIED TO THEM. Figure 16 illustrates the common symbols
used for controlled sources. The letter beneath each symbol is the PSpice designator for that
element.
Figure 16 symbols for controlled sources
Example 1
In the following Figure with a current controlled current source, use node equations to
solve for VOUT .
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Figure 17 example with current controlled source
This circuit has but one unknown node, VOUT . We choose to set the current in equal current
out.
 V  VOUT 
Vin  VOUT VOUT

   in

R1
R2
R1


Solving for VOUT :
VOUT 
Vin
1   
R1
 1    R2
 Vin 
1
1
 R1  1    R2
1    
R1
R2



This treatment of the basic circuit laws has been very abbreviated and is meant as a review.
For a more in depth material the reader is referred to a typical text used in a sophomore EE
circuits course such as "Electric Circuits" by Nilsson.