Circuit Theorems
1
Current Sources
A current source is said to be the dual of a voltage source.
Remember that a voltage source supplies a fixed voltage
and the current from it will vary depending upon the load.
A current source will supply a fixed current to the
branch to which it is connected, while the terminal
voltage will vary depending upon the load.
IL
I
RL
RS
2
Find VS and II for
the given circuit
I  I1  10 mA
VS  V1  I 1 R1  10 mA 20 k 
 200V
3
Determine VS, I1
and I2
Vs  E  12V
VR E 12V
I2 
 
3 A
R R 4
Applying Kirchoff' s current law
I  I1  I 2
I1  I  I 2  7 A  3 A  4 A
4
IL
RS
E
If the value of Rs is very
small in comparison to
the load RL then it may be
ignored, and the voltage
source in the shaded area
may be considered as an
“IDEAL” voltage source
5
IL
E
I
Rs
RS
RL
If the value of Rs is very large in comparison
to the load RL then it may be ignored, and the
current source in the shaded area may be
considered as an “IDEAL” current source
6
Source Conversions
If the internal resistance is included with either voltage or current
sources then that source may be converted to the other. Source
conversions are equivalent at their external terminals.
a
a
c
RS
I
E
RS
RS
E  IRS
b
.
b
7
RS

E

IL
2
RL
6V
4
Convert the voltage source to a
current source
E 6V
IS 

 3A
RS 2
RS is the same for the current source
Determine IL for both circuits
From these results you should
see that the LOAD RL does not
know if the source is a voltage
or current
8
Convert the current source to
a voltage source
VS  I S  RS  9mA  3k  27V
RS is the same as for the
current source
a
Determine IL for both circuits
RS
RL
These results show that the
LOAD RL does not know if the
source is a voltage or current.
b
9
Current sources in parallel may be added together
10
Reduce the parallel current sources to a single current source
R1
R2
R3
R3
11
Reduce to a single current source and find IL
12
13
14
Invalid situation.
15
The current through, or voltage across,
an element in a linear network is equal
to the algebraic sum of the currents or
voltages produced independently by each
source.
16
When removing a current
source it must be replaced by
an open circuit
When replacing a voltage
source it must be replaced by
a short circuit
17
When determining the power delivered to a
resistive element the total current through or
the total voltage across the element must be
used. NOT the simple sum of power levels
established by each source.
(The currents due to each source acting
individually may not all be in the same
direction through the element, and have to
be added algebraically.)
18
Using superposition find the current through the 6Ω resistor
19
Replacing the voltage source by
a short circuit gives
The current through the resistor
due to the current source is
obviously 0A due to the short
circuit
Replacing the current source by
an open circuit gives
The current through the resistor
due the voltage source gives
30V/6Ω = 5A
Thus the total current through the resistor is the sum from both
sources i.e. 0A + 5A = 5A
20
Using the superposition theorem determine
the current through the 4Ω resistor
21
Replace the 48V battery by a S/C
the current I 3 can now be calculated
RT  R1  R2 //R3  24 Ω  12 Ω// 4Ω
 24 Ω  3Ω  27 Ω
54V
I
 2A
27
R2
12 
24 A
I 3  I 
 2A

 1.5 A
R2  R3
12   4 16
22
Replace the 54V battery with a S/C
the current I3 can now be calculated
R T  R3  R1 // R2  4  24  // 12 
 4  8  12 
I 3 
48V
 4A
12
The total current th rough the 4 resistor is
I 3  I 3 I 3  4 A  1.5 A  2.5 A
23
Any two-terminal dc network can be replaced
by an equivalent circuit consisting of a voltage
source and a series resistor
The Thevenin voltage ETh is the open
circuit voltage between the two
terminals under consideration
The Thevenin resistance RTh is the resistance
looking into the two terminals of the network
with all voltage sources acting as short circuits
and current sources acting as open circuits.
24
The network behind
terminals ab may be
replaced by its
Thevenin equivalent.
The voltage ETH is the open
circuit voltage between a and b
The resistance RTh is the
resistance looking into the
network from a b with E1
short circuit.
25
Produce Thevenin circuit for the shaded network
☺
E1
1 Find the Thevenin resistance
RTH looking into a b (remember
to S/C E1)
RTH  R1 // R2  3 // 6  2
2 Find VTH (same as V O/C at terminals a b )
Use voltage divider rule
VTH  Vo / c
R2
6
 E1 
 9V 
 6V
R2  R1
6  3
26
The equivalent Thevenin circuit is as shown
To the left of aa
27
.
Find the the Thevenin equivalent circuit for the shaded area
28
i.e. the Thevenin circuit looking into the terminals at a b
29
Now applying the theorem
find RTh (S/C the supply)
It should be seen that due
to the S/C that R1 and R3
are in parallel as are R2 and
R4
30
RTh  R1 // R3   R2 // R4 
 2  3  5
31
Applying KVL to loop shown and ETh polarities as assumed
0  VR1  VR 2  ETh
Use voltagedivider rule to findVR1 andVR 2
6
VR1  72V 
 48V
9
VR 2
12
 72V 
 54V
16
insert in eqn.above 0  48V  54V  ETh
T hus ETh  6V
32
Substituting the Thévenin equivalent circuit for
the network external to the resistor RL
33
Any two-terminal, linear dc network can be
replaced by an equivalent circuit consisting of
a current source and a parallel resistor
The Norton current IN is the short
circuit current between the two
terminals under consideration.
The resistance RN is the resistance
looking into the network with
voltage sources acting as short
circuits and current sources
acting as open circuits.
34
The network behind
terminals ab may be replaced
by its equivalent Norton
Circuit.
The current IN is the short
circuit current between a and b
The resistance RN is the
resistance looking into the
network between terminals
a and b with E1 short
circuit.
35
Produce Norton circuit for the shaded network
1 Find the Norton resistance
RTH looking into a b (remember
to S/C E1)
RN  R1 // R2  3 // 6  2
Note that the Norton resistance is the same as the Thevenin resistance
2 Find IN (same as S/C current between terminals a b )
IN  IS /C
E1 9V


 3A
R1 3
36
Produce the equivalent Norton circuit for the shaded network
37
Produce the equivalent Norton circuit for the shaded network
Convert to
current source
IS 
7V
 1.75 A
4
RS = R1
38
Network redrawn with the
voltage source converted to
a current source.
Add the two currents to obtain the Norton current
i.e. 8A – 1.75A = 6.25Awhich will flow in the same
direction as the 8A.
To find the Norton resistance look into a b with the
current sources O/C (use product over sum) i.e. 2.4Ω.
39
Note that conversion between Norton
and Thevenin equivalent circuits may
be carried out by using the source
conversion methods introduced earlier.
Also remember that RN = RTH
40
The Maximum Power Transfer Theorem
A load will receive maximum power
from a linear dc network when its
total resistance is exactly equal to
the Thevenin resistance of the
network as seen by the load
41
Defining the conditions for maximum power to a load using the Thévenin
equivalent circuit.
IL
RTh
ETh
RL
Maximum power will be delivered to the load
RL when its resistance is equal to RTh
42
Defining the conditions for maximum power to a
load using the Norton equivalent circuit.
Remember that RTh = RN
I
IN
RN
RL
Maximum power will be delivered to the load
RL when its resistance is equal to RN
43
RTh
PL
9
IL

RL
VL
ETh

Power developed in the load
ETh
IL 
RTh  RL
2
 ETh 
 RL
PL  I RL  
 RTh  RL 
2
L
A tabulation of
PL
versus RL gives the
table as shown on the
next slide.
44
RL
PL
0.1
4.35
0.2
8.51
0.5
19.94
1
36.00
2
59.50
3
75.00
4
85.21
5
91.84
6
96.00
7
98.44
8
99.65
9
100.00
10
99.72
11
99.00
12
97.96
13
96.69
14
95.27
15
93.75
16
92.16
17
90.53
18
88.89
19
87.24
20
85.61
Tabulated values of RL and PL
Max.
Plot of PL against RL
45
RTh
9
ETH
IL
RL
60V
PL

VL

ETh2 RTh
PL 
4 RTh2
Power developed in the load
ETh
IL 
RTh  RL
2
 ETh 
 RL
PL  I RL  
 RTh  RL 
2
L
Expanding the last expression
with RL = RTH gives
( watts , W )
and
ETh2
PL max 
4 RTh
46
Calculate the value of R for maximum transfer of
power and the magnitude of this power.
For max transfer
of power R must =
RTh as shown
R
RTh
RTh = (R1 // R2) + R3 = 2+8 = 10 Ω
VTh = Voc = VR2 = 4V (using voltage divider rule)
PL max
ETh2
42


 0.4W
4 RTh 4  10
47