Download Example 2.7 for the circuit shown apply KVL to each designated

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
represent Dependent sources
Independent source
Dependent sources
shape to
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
7 ix A
-
were vx is some current
through an element
Dedependent voltage source
Voltage depend on voltage
were ix is some voltage
across an element
DE dependent current source
Current depend on current
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
le 2.7 for the circuit shown apply KVL to each designated path in the circuit
path b
-v a v 3 v 5  0
le 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
le 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
le 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
-i o  i1 - 6  0
Combinning this with the KCL equation
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
DEPENDENT VOLTAGE AND CURRENT SOURCES
•
A linear dependent source is a voltage or current source that depends linearly on
some other circuit current or voltage.
Example: You watch a certain voltmeter V1 and manually adjust a voltage source Vs to
be 2 times this value. This constitutes a voltage-dependent voltage source.
+
Circuit A
V1
2V1
-
+
-
Circuit B
This is just a manual example, but we can create such dependent sources
electronically.
We will create a new symbol for dependent sources.
DEPENDENT VOLTAGE AND CURRENT SOURCES
•
We can have voltage or current sources depending on voltages or currents
elsewhere in the circuit.
•
Here, the voltage V provided by the dependent source (right) is
proportional to the voltage drop over Element X. The dependent source
does not need to be attached to the Element X in any way.
Element X
+
VX
-
+
-
V  A V VX
•
A diamond-shaped symbol is used for dependent sources, just as a
reminder that it’s a dependent source.
•
Circuit analysis is performed just as with independent sources.
THE 4 BASIC LINEAR DEPENDENT SOURCES
Parameter being sensed
Constant of proportionality
Output
Voltage-controlled voltage source
… V = Av Vcd
Current-controlled voltage source …
Current-controlled current source
… I = Ai Ic
Voltage-controlled current source …
+
_
Av Vcd
+
_
Rm Ic
V = Rm Ic
I = Gm Vcd
Ai Ic
Gm Vcd
NODAL ANALYSIS WITH DEPENDENT SOURCES
R1
R3
Va
R5
Vb
Vc
R2
+
-
VSS
+
-
AvVc
R4
R6
ISS
Va - VSS Va - A v Vc Va - Vb


0
R1
R2
R3
Vb - Va Vb Vb - Vc


0
R3
R4
R5
Vc - Vb Vc

 ISS
R5
R6
NODAL ANALYSIS WITH DEPENDENT SOURCES
R1
Va
ISS
Vb
I2
R3
R2
R4
+
-
Rm I2
Va - Vb
ISS 
R1
Vb - Va
Vb - Rm I2
 I2 
0
R1
R3
Vb
I2 
R2
INTUITION
• A dependent source is like a “krazy resistor”, except its voltage
(current) depends on a different current (voltage).
• If the controlling voltage or current is zero, the dependent voltage
source will have zero voltage (and the dependent current source will
have zero current).
• There must be independent sources in the circuit for nonzero
voltages or currents to happen!
• Math explanation: Node voltage analysis leads to
Ax = b
The A matrix is nonsingular for real circuits, and is made of resistor
values and dependent source parameters. The b vector is made of
independent voltage & current source values. If b is zero and A is
nonsingular, the solution must be zero!