Ch2 Circuit Elements
Definition : An Electric Source is a device that is capable of converting nonelectric
energy to electric energy and vice versa.
Example : Battery
Discharging battery convert chemical energy to electric energy
Charging Battery convert electric energy to chemical energy
Four double-A rechargeable batteries )AA(
Electrical generator
An electrical generator is a device that converts mechanical energy to electrical energy
The source of mechanical energy may be a reciprocating or turbine steam
engine, water falling through turbine or waterwheel, an internal combustion
engine, a wind turbine or any other source of mechanical energy
Portable Electric Generator
Electric Generator
Theses souses can either deliver power like the battery when operate devices like toys,
radio, mobile phone, car electric instrument,….etc. or the electric generator that
deliver power to houses , cities, countries. Or absorb power like charging the battery
or operating electric drill.
We are going to use ideal sources to model practical sources as will be shown next.
There are two type of sources that we will disuse
Voltage Source
Current Source
We are going to use ideal sources to model practical sources as will be shown next.
Those ideal sources do not exist as practical devices , they are idealized model of
the actual devices
Ideal Voltage Source : is a circuit element that maintaining a prescribed voltage across
its terminals regardless of the current flowing in those terminals
Example : if we consider the 1.5 volt dry battery you from the market as an ideal, then
you will get a 1.5 volt across the battery regards of what is connected across it or in
another way the battery will supply a steady 1.2 V regardless what current flowing
through it
i
1.2 V
We should know that can not be possible since if the current i is large due to some load
as we will see later when we discuss Ohms law, the power deliver by the small battery
will be very large
We also will classified sources as Independent and Dependent sources
Independent source establishes a voltage or a current in a circuit without relying on a
voltage or current elsewhere in the circuit
Dependent sources establishes a voltage or a current in a circuit whose value depends
on the value of a voltage or a current elsewhere in the circuit
We will use circle to represent Independent source and diamond shape to represent
Dependent sources
Independent source
Dependent sources
Independent and dependent voltage and current sources can be represented as
+
-
5V
Independent voltage source
3A
Independent current source
+
4 ix V
4 vx A
-
were ix is some current
through an element
Dedependent voltage source
Voltage depend on current
were vx is some voltage
across an element
Dedependent current source
Current depend on voltage
The dependent sources can be also as
+
4 vx V
-
were vx is some current
through an element
Dedependent voltage source
Voltage depend on voltage
7 ix A
were ix is some voltage
across an element
Dedependent current source
Current depend on current
Example 2.1 for each of the following connections establish which interconnections
are permissible and which violate the constrains by the ideal source
Connection is valid
Connection is valid
Connection is not permissible
Connection is valid
Connection is not permissible
Example 2.2 for each of the following connections establish which interconnections
are permissible and which violate the constrains by the ideal source
Connection is not permissible
Connection is valid
Connection is valid
Connection is not permissible
2.2 Electrical Resistance (Ohm’s)
Georg Simon Ohm, a German physicist who is famous for defining the fundamental
relationship among voltage, current, and resistance through Ohm's law
Ohm's law states that, in an electrical circuit, the current passing through a conductor, from
one terminal point on the conductor to another terminal point on the conductor, is directly
proportional to the potential difference (i.e. voltage drop or voltage) across the two
terminal points and inversely proportional to the resistance of the conductor between the
two terminal points
were
V = the voltage in volts
(V)
I = the current in amperes (A)
R = the resistance in ohms (W)
Most materials exhibit measurable resistance to current. The amount of resistance
depend on the material.
This is similar in some way when water flow in a pipe. If there is a dirt or material on
the pipe it will impede ( ‫ )يعيق‬the flow of water.
For purpose of circuit analysis, we must reference the current in the resistor to the
terminal voltage. For the passive sign convention
i
+
v
v = Ri
R
-
Otherwise we introduce a minuses sign similar to what we did when we calculated power
+
v
R
i
v = - Ri
Example
2A
+
+
v
3W
v = (3)(2) = 6 V
2A
+
-2 A
+
-
v = -(3)(2) = -6 V
-
-
v
3W
v
3W
v = (3)(- 2) = -6 V
v
-
3W
-2 A
v = -(3)(- 2) = 6 V
The reciprocal of resistance (‫ )مقاومه‬is conductance (‫ )موصل‬and have the symbol G
1
R
for (mho) is spelling backward
An 8 W resistor is equivalent to 1/8= 0.125 S or
W
and have the unit S for (siemens) or
for ohm
W
G=
We may calculate the power at the terminal of a resistor in several way
First method : we use the defining equation as follows:
i
+
v
R
p = vi
For the passive sign convention
-
Otherwise we introduce a minuses sign
+
v
R
i
p = -vi
Second method : we express the power at the terminal of resistor in terms of the current
and resistor as follows:
v
+
-
p  vi  (iR )i  Ri 2
R
i
For the non passive sign convention we have
v
+
-
p  -vi  -(-iR )i  Ri 2
R
i
Third method : we express the power at the terminal of resistor in terms of the voltage
and resistor as follows:
+
v
-
p  vi
R
i
v
v2
v ( ) 
R
R
For the non passive sign convention we have
+
v
R
p  -vi  -v ( -v )  v
R
R
-
2
i
Fourth method : we express the power at the terminal of resistor in terms of the voltage
or current and conductance G as follows:
+
v
-
p  vi  v (Gv )  Gv 2
R
i
p  vi
i
i2
 ( )i 
G
G
For the non passive sign convention we will have identical relation as the passive sign
convention
+
v
R
-
p  Gv 2
i
i2
p
G
Example 2.3 In each circuit find the followings:
v a , p8W ?
+
1 A
8 W
va
v a  (1)(8)  8 V
pW  8i 2  8(1) 2  8 W
-
OR
p W
v a2 (8) 2
 
8 W
8
8
i d , p 25W ?
+
-
25 W
50 V
-50
 -2 A
25
 25i d2  25(-2) 2  100 W
id 
p 25W
id
OR
p 25W
502

 100 W
25
2.3 Construction of a Circuit Model
This course will be focus on circuit analysis (i.e., finding voltages , currents and
powers of circuit elements)
However you would need to construct a model for the electric device as much as
analyzing it.
We are going to develop a circuit model based on the behavior of the circuit components
and interconnections
2.4 Kirchhoff’s Law
The objective of this course is to find (or solve) for voltages and currents in
every element in the circuit
Example consider the following circuit:
i1
i3
2W
+
i 2 + v3 -
v1 +
5V
+
-
3W
v2
6W
-
+
v4
5W
-
Suppose we want to find the current i1 ?
In this circuit we have 7 unknowns , namely
i1 ,v1 i 2 ,v 2 i 3 ,v 3
and v 4 ( note i 3 is the same as i 4 )
i1
i3
2W
+
i 2 + v3 -
v1 +
5V
+
-
3W
v2
-
6W
+
v4
5W
-
To solve for the 7 unknown we need 7 equations
Ohms law can provide us with 4 equations, namely
v 1 = 2 i1
v 2 = 6 i 2 v3 = 3 i3 v4 = 5 i3
However Ohms equations can not be sufficient to solve for the 7 unknown,
we need still 3 equations , what are these equations ?
Gustav Kirchhoff Russian scientist who first stated them in 1848 in a published paper
and they are named after him as
Kirchhoff Current Law (KCL)
Kirchhoff Voltage Law (KVL)
Kirchhoff's Current Law ( KCL):
The algebraic sum of all the currents at any node in a circuit equals zero.
First we have to define a node.
A node is a point where two or more circuit elements meet
i1
i3
i2
Kirchhoff's Current Law ( KCL):
The algebraic sum of all the currents at any node in a circuit equals zero.
i1
i2
i3
The algebraic signify a sign on the current that is positive or negative. Since the current
is a reference quantity by direction. Then we can state the following
Current entering the node is positive and current leaving the node is negative
OR
Current entering the node is negative and current leaving the node is positive
Example
i1
i2
i3
Current entering the node is positive and leaving the node is negative
i1  ( - i 2 )  i 3
0
 i1 - i 2

i3
0
Current entering the node is negative and leaving the node is positive
(-i1 ) 
i2
 (-i 3 )  0
 i1 - i 2

i3
0
Note the algebraic sign is regardless if the sign on the value of the current
Suppose
i1
 -1 A
i2
2 A
i3
-1 A
3 A
3A
Now KCL
i1 - i 2
2A

i3
0
 ( -1 ) - 2  3  0
Since electric current is a rate flow of charges then Kirchhoff's Current Law is similar to
the flow of water from different direction to a water valve ( ‫(صمام‬
Water
Valve
i1
Now we go back to our circuit
i3
2W
+
i 2 + v3 -
v1 +
5V
v2
+
-
3W
+
6W
-
Ohms law can provide us with 4 equations, namely
v 1 = 2 i1
-
v 2 = 6 i 2 v3 = 3 i3 v4 = 5 i3
We have 4 nodes
Node 1
i3
2W
+
5V
Node 1
Node 2
i1
v1 -
+
v2
+
-
5W
v4
i2
3W
+
v3
v4
-
2W
Node 2
-
+
6W
short
Node 3
Which can be redrawn as
5W
5V
+
-
6W
-
3W
Node 3
short
Node 4
Same Node
5W
Node 4
Now if we apply KCL to each node
we will have the followings
Node 1
2W
+
Node 1
Node 1
i1
i1
5V
+
-
5V
2W
v1 -
+
v2
+
-
i1
2W
Node 3
3W
i2
v3 -
+
+
6W
5W
v4
-
-
short
short
Node 3
i1 - i1
0
i3
i2
6W
i3 -i3
 i3  i3
0
5W
Nothing new !
Node 4
3W
i2
6W
i1 - i 2 - i 3
Node 3
i3
 i1  i1
Node 2
3 W i3
Node 4
Same Node
Nothing new !
Node 2
i3
Node 2
i1
0
i1
i3
-i1  i 2  i 3  0
Same as node 2
Node 4
i1
i3
2W
+
i 2 + v3 -
v1 +
5V
v2
+
-
3W
6W
-
+
v4
5W
-
Now we have 5 equations, namely
Ohms law can provide us with 4 equations, namely
v 1 = 2 i1
v 2 = 6 i 2 v3 = 3 i3 v4 = 5 i3
KCL provide us with 1 equations, namely
i1 - i 2 - i 3
0
We have now 5 equations , we still need two more equations
Kirchhoff Voltage Law (KVL) will provide us with the other two equations
as will be shown nex
Kirchhoff Voltage Law (KVL)
The algebraic sum of all the voltages around any closed path in a circuit equals zero.
First we have to define a closed path
a
b
c
+
-
f
e
d
A closed path or a loop is defined as starting at an arbitrary node, we trace closed path in a
circuit through selected basic circuit elements including open circuit and return to the
original node without passing through any intermediate node more than once
abea
Kirchhoff Voltage Law (KVL)
The algebraic sum of all the voltages around any closed path in a circuit equals zero.
First we have to define a closed path
a
b
c
+
-
f
e
d
A closed path or a loop is defined as starting at an arbitrary node, we trace closed path in a
circuit through selected basic circuit elements including open circuit and return to the
original node without passing through any intermediate node more than once
abea
bceb
Kirchhoff Voltage Law (KVL)
The algebraic sum of all the voltages around any closed path in a circuit equals zero.
First we have to define a closed path
a
b
c
+
-
f
e
d
A closed path or a loop is defined as starting at an arbitrary node, we trace closed path in a
circuit through selected basic circuit elements including open circuit and return to the
original node without passing through any intermediate node more than once
abea
bceb
cdec
Kirchhoff Voltage Law (KVL)
The algebraic sum of all the voltages around any closed path in a circuit equals zero.
First we have to define a closed path
a
b
c
+
-
f
e
d
A closed path or a loop is defined as starting at an arbitrary node, we trace closed path in a
circuit through selected basic circuit elements including open circuit and return to the
original node without passing through any intermediate node more than once
abea
bceb
cdec
aefa
Kirchhoff Voltage Law (KVL)
The algebraic sum of all the voltages around any closed path in a circuit equals zero.
First we have to define a closed path
a
b
c
+
-
f
e
d
A closed path or a loop is defined as starting at an arbitrary node, we trace closed path in a
circuit through selected basic circuit elements including open circuit and return to the
original node without passing through any intermediate node more than once
abea
bceb
cdec
aefa
abcdefa
Kirchhoff Voltage Law (KVL)
The algebraic sum of all the voltages around any closed path in a circuit equals zero.
2W
+
3W
v1 -
+
+
5V
+
-
v2
-
6W
v3 +
v4
5W
-
The "algebraic" correspond to the reference direction to each voltage in the loop.
Assigning a positive sign to a voltage rise ( - to + )
Assigning a negative sign to a voltage drop (  to - )
OR
Assigning a positive sign to a voltage drop (  to - )
Assigning a negative sign to a voltage rise ( - to  )
Example
2W
+
3W
v1 -
+
+
5V
v2
+
-
-
We apply KVL as follows:
Loop 1
v1 v 2
Loop 2
v 3 v 4 -v 2
-5  0
0
6W
v3 +
v4
-
5W
Example
2W
+
3W
v1 -
+
+
5V
v2
+
-
-
6W
v3 +
v4
5W
-
We apply KVL as follows:
Loop 1
Loop 2
Loop 3
v1 v 2
-5  0
v 3 v 4 -v 2  0
v1 v 3 v 4 - 5  0
However we notice that KVL on Loop 3 is only the summation of Loop 1 and Loop 2
Therefore only two KVL equations are valid
Loop 1
Loop 2
v1 v 2 - 5  0
v 3 v 4 -v 2  0
i1
Now we go back to our circuit
i3
2W
+
i 2 + v3 -
v1 +
5V
v2
+
-
6W
-
Now we have 7 equations, namely
Ohms law provide us with 4 equations, namely
v 1 = 2 i1
v 2 = 6 i 2 v3 = 3 i3 v4 = 5 i3
KCL provide us with 1 equations, namely
i1 - i 2 - i 3
0
KVL provide us with 2 equations, namely
v1 v 2 - 5  0
v 3 v 4 -v 2  0
Now we can solve the 7 equations to obtain the 7 variables
i1 ,v1 i 2 ,v 2 i 3 ,v 3
3W
and v 4
+
v4
-
5W
We can think of KVL as a gas that distribute pressure gas to different houses. When it leaves
the station it is full of gas. Then it start delivering the gas to each house depend on the size of
the house
House 1
House 2
House 3
Gas Station
House 4
House 6
House 5
Example 2.6 for the circuit shown apply KCL to each node a, b, c, and d.
Note there is no connection dot at the center of the diagram (i.e, no node).
The 4 W branch crosses the branch containing the ideal current ia
Applying KCL to nodes a, b, c, and d , we have
Node
Node
Node
Node
a
b
c
d
i1  i 4 - i 2 - i 5  0
i 2  i 3 - i1 - i b - i a
i b i 3 -i 4 -i c  0
i5 i a ic  0
0
Example 2.7 for the circuit shown apply KVL to each designated path in the circuit
Example 2.7 for the circuit shown apply KVL to each designated path in the circuit
Example 2.7 for the circuit shown apply KVL to each designated path in the circuit
path a
-v 1 v 2 v 4 -v b -v 3  0
Example 2.7 for the circuit shown apply KVL to each designated path in the circuit
path b
-v a v 3 v 5  0
Example 2.7 for the circuit shown apply KVL to each designated path in the circuit
path c
v b -v 4 -v c -v 6 -v 5  0
Example 2.7 for the circuit shown apply KVL to each designated path in the circuit
path d
-v a -v 1 v 2 -v c v 7 -v d
0
Example 2.7 for the circuit shown apply KVL to each designated path in the circuit
path a
-v 1 v 2 v 4 -v b -v 3  0
path b
path c
path d
-v a v 3 v 5  0
v b -v 4 -v c -v 6 -v 5  0
-v a -v 1 v 2 -v c v 7 -v d
0
Example 2.8 for the circuit shown use Kirchcoff’s laws and Ohm’s law to find
io ?
Solution
We will redraw the circuit and assign currents and voltages as follows
io
Since io is the current in the 120 V source , therefore there is only two unknown currents
in the circuit namely:
io and i1
io
Therefore we need two equations relating io and i1
Applying KCL to the circuit nodes namely a,b and c will give us the following
Node a
io -io
0
 io io
Node b
-i o  i1 - 6  0
Node c
-i1  i o  6  0
Nothing new !
The same as node b
Therefore KCL provide us with only one equation relating
-i o  i1 - 6  0
We need another equation to be able to solve for io
That equation will be provided by KVL as shown next
io and i1
namely
io
We have three closed loops
However only one loop that you can apply KVL to it namely abca
Since the other two loops contain a current source namely 6 A and since we can not relate
the voltage across it to the current through it , therefore we can not apply KVL to that loop
Applying KVL around loop abca clockwise direction assigning a positive sign to
voltage drops ( + to - ) and negative sign to voltage rise , we have
- 120  10i o  50i 1  0
Combinning this with the KCL equation
-i o  i1 - 6  0
we have two equations and unknowns which can be solved simultaneously to get
io= -3 A
i1= 3 A
2.5 Analysis of a circuit containing dependent sources
For the circuit shown we want apply Kirchhof’s and Ohm law to find vo ?
Solution
We are going to device a strategy for solving the circuit
Let io be the current flowing on the 20 W resistor
Since vo=20
io
→
therefore we seek io
KCL will provide us with one equation relating io with iΔ namely
Node b
i   5i  - i o  0
 6i  - i o  0
We need two equations to be able to solve for io
KVL and Ohm’s law will provide us with the additional equation
relating io with iΔ as will be shown next
The circuit has three closed loops
However only the loop abca is the one that you can apply KVL to it
Since the other two loops contain a current source that, you will not be
able to write the voltage across it in terms of the current (i.e, you will
not be able to write KVL around the loop the one that you can apply
KVL to it )
Therefore
-500  5i   20i o  0
 5i   20i o  500
Therefore we have two equations relating io with iΔ namely
KCL at node b
KVL around loop abca
6i  - i o  0
5i   20i o  500
Two equations and two unknowns io and iΔ , we can solve simultaneously and get
io = 24 A and iΔ = 4 A
Therefore vo=20 io = (20)(24) = 480 V