Thevenin's Theorem with Dependent Sources
Cases in Thevenin's Theorem:
■ CASE 1: If the network has no dependent sources, we turn
off all independent sources. RTH is the input resistance of the
network looking between terminals A and B.
■ CASE 2: If the network has dependent sources, we turn off
all independent sources. As with superposition, dependent sources are not to be
turned off because they are controlled by circuit variables. We apply a voltage
source at terminals A and B and determine the resulting current Io. Then,
RTH = Vo/Io . Alternatively,
we may insert a current source Io at terminals A-B and find the terminal voltage Vo. Again RTH = Vo/Io.
Either of the two approaches will give the same result. In either approach we may
assume any value of and Vo and Io. For example, we may
use Vo = 1V or Io = 1A or even use unspecified values of
Vo or Io.
It often occurs that RTH takes a negative value.
In this case, the negative resistance (V = -iR) implies that the circuit is
supplying power. This is possible in a circuit with dependent sources.
Thevenin’s theorem is very important in circuit analysis. It
helps simplify a circuit. A large circuit may be replaced by a single
independent voltage source and a single resistor. This replacement technique is
a powerful tool in circuit design.
As mentioned earlier, a linear circuit with a variable load
can be replaced by the Thevenin equivalent, exclusive of the load. The
equivalent network behaves the same way externally as the original circuit. Consider
a linear circuit terminated by a load RL, . The
current IL through the load and the voltage VL across the
load are easily determined once the Thevenin equivalent of the circuit at the load’s
terminals is obtained.
Suppose that R1 =5Ω , R2
=3Ω and IS =2A .
Solution
The circuit has both independent and dependent sources. In
these cases, we need to find the open circuit voltage to
determine the Thevenin Equivalent Circuit
Open Circuit Voltage
Open circuit voltage means the voltage across the terminals
of the network without connecting any extra element or connection:
Since there is no connection, the current of R2 is
zero. To solve the circuit lets write KVL for the left hand side loop assuming IR1 defined
from left to right:
+Vx+R1×IR1+Vx=0
→2Vx=−R1×IR1
→Vx=−R1×IR12
But what is IR1? R1 is in series with the current source; they have only one node shared and there is no other element connected there. This means that all current of IS must pass through R1 . Therefore, IR1=IS=2A .
If we apply this to the equation above, we have
→2Vx=−R1×IR1
→Vx=−R1×IR12
But what is IR1? R1 is in series with the current source; they have only one node shared and there is no other element connected there. This means that all current of IS must pass through R1 . Therefore, IR1=IS=2A .
If we apply this to the equation above, we have
Vx=−R1×IR12→Vx=−5V.
Since no current passing through R2 we can easily see that VOC=Vx . If it is not clear, you could find this by applying KVL to the right hand side loop:
−Vx+R2×IR2+VOC=0
→−Vx+R2×0+VOC=0
→VOC=Vx=−5V
Since no current passing through R2 we can easily see that VOC=Vx . If it is not clear, you could find this by applying KVL to the right hand side loop:
−Vx+R2×IR2+VOC=0
→−Vx+R2×0+VOC=0
→VOC=Vx=−5V
Calculating RTH
The only thing left is to calculate RTh which
can be easily found by
RTH=VOC / ISC=−5−53=3Ω
RTH=VOC / ISC=−5−53=3Ω
Thevenin's Equivalent Network
VTh=VOC=−5V
RTh=3Ω
RTh=3Ω
Reflection/Learnings:
This week, We discussed about Thevenin's Theorem with Dependent Sources. I didnt include this in the previous section because I wanted to explain it in more detail. Actually, In Thevenin's Theorem, There were two cases. Case 1 are for circuits that doesn't have any dependent sources while Case 2 are for circuits that contains at least one dependent source. To solve for problems with dependent sources, We first turn off all independent sources by replacing all voltage sources as a closed branch and replacing all current sources with an open branch. Dependent sources cannot be turned off because they are controlled by circuit variables and then we simply remove one load in a circuit and instead of replacing it with an open circuit and measuring the open circuit voltage, We add an additional voltage or current source. The values of the added sources completely depends on you. You can assume any value for it as long as it is not 0.
Video(s):
Thevenin's Theorem with Dependent Source
Thevenin's Theorem with one Dependent Source
Electricity is often called wonderful, beautiful; but it
is so only in common with the other forces of nature. The beauty of electricity
or of any other force is not that the power is mysterious, and unexpected,
touching every sense at unawares in turn, but that it is under law, and that
the taught intellect can even govern it largely. The human mind is placed
above, and not beneath it, and it is in such a point of view that the mental
education afforded by science is rendered super-eminent in dignity, in
practical application and utility; for by enabling the mind to apply the
natural power through law, it conveys the gifts of God to man.
--Michael Faraday
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