☑️19 - Superposition Theorem: Circuits with Dependent Sources 1
Summary
TLDRThis educational video script delves into the application of the superposition theorem for solving linear circuits involving dependent sources. The presenter guides through two examples, first calculating the voltage 'VX' using the theorem by deactivating one independent source at a time. The second example demonstrates finding 'V not' using a similar method, also introducing the principle of source transformation for circuit analysis. The script is a valuable resource for those learning to apply superposition in circuit problems.
Takeaways
- 🔌 The video script discusses the application of the Superposition Theorem in linear circuits involving dependent sources.
- 🔧 The Superposition Theorem is used when there are multiple independent sources in a circuit, and it involves considering each source one at a time while deactivating the others.
- 🔄 The first example given in the script illustrates how to find the voltage 'VX' by first considering the 20-volt source alone and then the 4-ampere current source alone.
- 🔍 In the first example, the dependent source is deactivated by treating it as an open circuit, and current distribution is used to set up equations.
- 📐 The script explains how to set up and solve equations involving voltage drops across resistors and the relationship between the currents in the circuit.
- 🔗 The concept of current division and the use of Ohm's law to find the value of unknown currents and voltages are demonstrated.
- 🔄 The second example in the script involves finding the voltage 'V naught' using the same principles of the Superposition Theorem.
- 🔄 The principle of source transformation is introduced, which allows for converting a voltage source in series with a resistor to a current source in parallel with the same resistor, and vice versa.
- 🔌 The script shows how to redraw circuits to simplify the analysis, such as combining resistors in parallel and transforming sources.
- 📘 The final values for 'VX' and 'V naught' are found by summing the effects of each independent source acting alone, as per the Superposition Theorem.
- 👋 The video concludes with a summary of the findings and a sign-off for the next video.
Q & A
What is the Superposition Theorem used for in the context of this script?
-The Superposition Theorem is used to find the voltage across a particular point (VX in the script) in a linear circuit with multiple independent sources by considering the effect of each independent source one at a time while deactivating the others.
How is the dependent source in the circuit described in the script?
-The dependent source in the circuit is described as a current source with a value of 0.1 VX, indicating that the current through this source is proportional to the voltage VX across it.
What is the first step in applying the Superposition Theorem to the circuit in the script?
-The first step is to consider the 20-volt source acting alone and deactivate the 4-ampere current source, then find the value of the current I in the circuit.
How is the current distribution determined when the 20-volt source is acting alone?
-The current distribution is determined by applying Ohm's Law to the circuit, considering the current I produced by the 20-volt source and the voltage drops across the resistors in the loop.
What is the equation derived from the loop with the 20-volt source acting alone?
-The equation derived is 20 = 20I + 4I1, where I is the current through the 20-ohm resistor and I1 is the current in the branch with the dependent source.
What is the relationship between I1 and I in the circuit?
-The relationship is given by the equation I1 = I + 0.1VX, indicating that the current I1 in the branch is the sum of the current I and 0.1 times the voltage VX.
How is the voltage VX across the 4-ohm resistor related to I1?
-The voltage VX across the 4-ohm resistor is equal to I1 times the resistance value, which is 4 ohms, as per Ohm's Law.
What is the principle of source transformation mentioned in the script?
-The principle of source transformation states that a voltage source in series with a resistor can be transformed into a current source in parallel with the same resistor, and vice versa, using Ohm's Law to find the equivalent source values.
How is the voltage source transformed into a current source in the script?
-The voltage source is transformed into a current source by using Ohm's Law (V = I * R), where the current I is calculated as the voltage across the source divided by the resistance value.
What is the final value of VX calculated using the Superposition Theorem in the script?
-The final value of VX is calculated to be 25 volts, obtained by summing the individual effects of the 20-volt source and the 4-ampere source acting alone.
What is the method used to find the value of V naught in the second example of the script?
-The method used to find the value of V naught in the second example is also the Superposition Theorem, by considering the effects of the 10-volt source and the 2-ampere source acting alone and then summing their individual contributions.
How are the resistors treated when the 2-ampere source is acting alone in the second example?
-When the 2-ampere source is acting alone, the 20-ohm and 4-ohm resistors are treated as being in parallel, and their combined resistance is calculated to simplify the circuit before finding the current and voltage values.
What is the final value of V naught calculated in the second example of the script?
-The final value of V naught calculated in the second example is 11.2 volts, which is the sum of the individual effects of the 10-volt source and the 2-ampere source acting alone.
Outlines
🔌 Application of Superposition Theorem in Linear Circuits
The first paragraph introduces the application of the Superposition Theorem in linear circuits with dependent sources. The example provided involves calculating the voltage VX in a circuit with two independent sources and one dependent source. The process involves deactivating one independent source at a time and analyzing the circuit to find the current distribution and the resulting voltage across the resistors. The equations derived from the circuit analysis are used to solve for the unknown current and voltage values, demonstrating the step-by-step application of the theorem.
🔧 Circuit Analysis with Superposition and Source Transformation
In the second paragraph, the script continues with the circuit analysis, focusing on finding the voltage VX when only the four-ampere current source is active. The concept of source transformation is introduced to simplify the circuit by converting the voltage source in series with a resistor into an equivalent current source in parallel with the resistor. The analysis involves calculating the equivalent resistance of parallel resistors and using the transformed source to find the voltage across a resistor, eventually determining the value of VX.
🔍 Further Exploration of Superposition and Source Transformation
The third paragraph delves deeper into the application of the Superposition Theorem and source transformation in a different circuit configuration. It discusses the process of deactivating one source and analyzing the circuit with the remaining source. The principle of source transformation is applied again to convert a current source in parallel with a resistor into a voltage source in series, facilitating the calculation of the current and voltage in the circuit. The paragraph concludes with the calculation of the voltage across a resistor due to the active source.
🛠 Calculation of Voltage Using Superposition and Ohm's Law
This paragraph presents a methodical approach to calculating the voltage V naught in a circuit with multiple sources using the Superposition Theorem. The script explains the process of deactivating one source and using Ohm's law to find the equivalent current source. It also discusses the identification of series and parallel resistors and the assignment of currents to find the voltage across specific resistors. The paragraph concludes with the calculation of the voltage V naught due to the active source.
🔗 Final Calculations and Conclusion of Circuit Analysis
The final paragraph wraps up the circuit analysis by considering the original circuit with both independent sources active. It uses the previously calculated values of V naught from the superposition scenarios to find the total voltage across the resistor when both sources are active. The paragraph demonstrates the use of source transformation to simplify the circuit and calculate the current and voltage across the dependent source. The video concludes with the final value of V naught and a sign-off, inviting viewers to the next video.
Mindmap
Keywords
💡Linear Circuits
💡Dependent Sources
💡Superposition Theorem
💡Independent Sources
💡Current Distribution
💡Ohm's Law
💡Source Transformation
💡Parallel Resistors
💡Circuit Deactivation
💡Voltage Division
💡Circuit Analysis
Highlights
Introduction to using the superposition theorem for linear circuits involving dependent sources.
Explanation of how to apply the superposition theorem by considering one independent source at a time and deactivating the others.
Step-by-step approach to solving for VX in a circuit with two independent sources and a dependent source.
Technique of deactivating the current source and treating it as an open circuit for the first part of the example.
Current distribution method to find the value of I and I1 in the circuit.
Derivation of equations to relate the current I, I1, and VX in the circuit.
Calculation of VX by substituting derived equations to find the voltage across the 4-ohm resistor.
Finding the value of VX when only the 20-volt source is active (VX Prime).
Deactivation of the 20-volt source and activation of the 4-ampere source to find another value of VX (VX Prime Prime).
Use of the principle of source transformation to simplify the circuit for the second example.
Conversion of a voltage source in series with a resistor to a current source in parallel with the same resistor.
Application of Ohm's law to find the equivalent current source for the transformed circuit.
Assignment of currents and the use of source transformation to find the value of V naught in the second example.
Solving for the current I and the voltage V naught when the 10-volt source is active (V naught Prime).
Approach to find the current and voltage distribution when the 2-ampere source is active (V naught Prime Prime).
Final calculation of V naught by combining the effects of both independent sources.
Conclusion of the video with the final value of V naught being 11.2 volts.
Invitation to the next video and sign-off, indicating the end of the educational content.
Transcripts
so still under linear circuits involving
dependent sources we are going to use
superposition theorem to find the value
of VX in this question now this happens
to be our first example let's try to
solve this example together
now looking at this circuit you realize
that we have two independent sources we
have 20 volts and then four ampere and
source and then we have a dependent
source which is 0.1 VX carrying Source
now with superposition to your remote we
do is we consider one of the independent
sources at a time and then we deactivate
the other later we consider the other
one and then we deactivate the initial
one so you use superposition theorem
whenever you have more than one
independent source
so in the first step we are going to
consider the 20 volts and then
deactivate the four ampers current
source and then we try to find the value
of ax so first let's
20 volts
act alone
so because this is a carrying Source
it's going to be an open circuit
so when 20 volts is acting alone like we
said we are going to deactivate the four
ampere scaling Source now the next thing
we are going to do is to do current
distribution so we have current I
produced by the 20 volts this current
flows through this direction and then at
this point
we have current i1
flowing in this Branch now
considering this Loop
we have the source voltage to be 20
volts and that is equal to the sum of
the voltages dropped across the loop so
we have
20 times I because we have I flowing
through this resistor and then plus we
have
four times i1
now let's call this equation one
now you should realize that we have this
current Source 0.1 VX moving in the
anti-clockwise direction and then it's
going to combine with
I at this note and then they both flow
through this Branch therefore
therefore we have
i1 to be equal to
I plus 0.1
VX so this is the value of or this is
the expression for i1
again
so the currents I produced by the 20
volts flows through this 20 ohms
resistor and then as it flows through
the resistor some of the voltage is
being dropped across the resistor now
the left voltage the voltage that is
left is said to be the voltage VX and
that is the voltage across this four
ohms resistor therefore we have VX
to be equal to
the voltage across this resistor which
is i1 times the value of the resistor so
we have vx to be equal to
for i1 let's call this equation two now
we are going to put equation two
we are going to put equation 2
into
i1 so for i1 we have i1 equals
I
plus 0.1
times
VX which is 4i1
and that becomes
I
plus 0.1 times 4 is 0.4 and then i1 now
we are going to transpose 0.4 i1 to the
left hand side
so that's going to be
i1 minus 0.4 i1 which is
0.6 i1 equals I
so let's call this equation three
now we are going to put equation three
we're going to put equation three
into equation one so that is 20 equals
2 sorry 20 times
20 equals 20 times
in place of I we have 0.6 i1 so
0.6 i1
plus 4 i1
now 20 times
um 0.6 becomes 12 so you have 12 i1 plus
4 I 1 equals 20. now this becomes 16 i1
so we divide through by 16. by 16 and
then you have i1 to be equal to now 20
over 16 gives
1.25 so we have I want to know one point
two five amperes
now if that is the case then we can find
the value of VX
we can find the value of a x which is
equal to 4 times i1 so that is 4 times
i1 1.25
and that is equal to
5 volts
now since VX is as a result of 20 volts
acting alone we are going to call this
VX Prime so that is
5 volts again we are going to deactivate
20 volt and then make
four amperes at a loan
so for four amperes I think alone
you see let's
for amps act alone
so let's redraw the circuit so we are
going to have
20 votes becoming short circuits
we have this 20
ohms here
this is VX
and then we have the current source
for amperes
we have this resistor
for ohms
and then we have the current
Source dependent
Korean sauce
all right
so let's try to find the value of VX now
whenever you go through a loop without
passing through any other circuit
element except two resistors then it
means that those two resistors are
connected in in parallel so 20 ohms and
then 4 Ohms are connected in parallel
so we can have the parallel combination
you can have the parallel combination
that is 20 parallel four that is 20
times 4 over
20 plus 4. now this becomes 80 over 24
8 goes here three times eight goes here
ten times so that is 10.
over 3 ohms
so we can redraw the circuits such that
we have this to be
10 over 3 ohms we have the current
source
which is four amperes
and then we have the
dependent carrying source
so this is the independent this is
dependent
that is 0.1 VX and then we have this to
be
X
now we have four amperes
0.1 VX combining at this node and then
moving in this direction so that is four
plus zero point one of the X
now
the voltage across
this resistor
is VX the voltage across this resistor
is VX so we have
VX
to be equal to
the current
4 plus 0.1
the x times the value of the resistor
which is 10 over 3.
so at this point you can cross multiply
so that's going to be
3 v x equals
here we have
4 plus 0.1
VX times 10. so we multiply 10 across we
have three v x equals
40 plus
v x we transpose vx to the left hand
side we have 2 VX that is equal to 40.
we divide 2 by we divide 2 by
2 by 2 and then we have a x to be
20 volts now because this is as a result
of 4 amps acting alone let's call that
VX prime prime therefore
the value for VX is giving us v x
prime plus VX prime prime
and notice that we had VX Prime to be 5
volts
so 5 plus v x prime prime 20 and that is
equal to
25 volts so the value of VX
is equal to
25 volts now let's move on to the next
example
so in our next example we are going to
use superposition theorem to find V
naught in the circuits below
so as usual we have more than one
independent sources so we are going to
consider one and then deactivate the
other and then later we consider the
latter one and then deactivate the
initial one so first let 10 votes act
alone
okay
so we deactivate two amps
so at this point we've been able to
redraw the circuit deactivating the two
amps current source now we are supposed
to assign current and then find the
value of venotes
now sometimes it's very difficult to
easily identify whether these two ohms
resistor is in series with this one ohm
resistor or better so the two are in
parallel now what we are going to do
here is to borrow the principle of
Source transformation now what does this
principle say he says that assuming you
have a voltage source let's say you have
a 10 boot source
which is in series with
let's say a two ohms resistor
okay then you can transform
this circuit such that
you have a current source
which is this time in parallel
with
the resistor
so a voltage source in series with a
resistor can be transformed into a
current Source in parallel with that
same resistor and vice versa so if you
want to find the value of the current
Source basically we are going to use
Ohm's law so from Ohm's law we have V is
equal to I times R now if you want to
find the value of I then it's nothing
but V divided by R you have V to be 10
you have R to be two so this is equal to
5 amperes so you have this to be 5
amperes in parallel with
two ohms so this is how the principle of
Source transformation works so we are
going to deploy that here so instead of
having one ohm in parallel with 0.5 V
naught we are going to
find the value of the voltage notice
that this is a carrying source so to
find the voltage is basically the value
of the carrying source
0.5
Vin knots times one
which is still 0.5 V notes notice that
this time the unit is in volts so we are
going to have 0.5 Vin notes which is in
votes in series with this one ohm
resistor
so let's redraw that circuit so that's
going to be
now again notice that the polarity of
the voltage source should be in line
with the direction of the current so you
realize that we have the positive
pointing upwards so the direction should
be apples the same way applies to this
we have the direction of the carrying
Source moving to the right we have the
positive also at the right now we assign
currents let's say I and then we want to
find the value of current produced by
the 10 volts later we can find the value
of V naught
so let's try to find the value of I
so we have I
moving throughout the circuits and then
we are taking the clockwise direction we
have the source voltage to be 10 plus
0.5 a knot notice that we have
the polarity
in the same direction so it's going to
be
10
plus 0.5 V notes that is equal to we
have
2i Plus
I
Plus
4i
now this combines to
7i so we have 10 plus 0.5 V naught
equals
7 I now again
the voltage Vin notes across the four
ohms resistor is equal to the value of
the current flowing in the branch times
the value of the resistor so
we have V naught to be equal to current
flowing in this branches I
so that is 4 times I this is the value
of V naught or the expression for V
naught we are going to put that in this
equation and then you realize that we
have 10 plus 0.5
in place of V naught we have four I
equals 7 I so this becomes 10 plus 0.5
times 4 is 2 so 2i equals 7 I you
transpose this to the right hand side we
have 10 equals 7 I minus 2i which is
equal to 5i so we divide 3 by 5
by 5 we have I to be equal to 2 amperes
so this is the value of I now to find
the value of V naught that is equal to 4
x i and then we have 4 times 2 which is
equal to eight therefore V naught
is equal to 8 volt now because V naught
is the value of the voltage
across four ohms resistor as a result of
10 volts acting alone we call this V not
Prime
Windows Prime so next we are going to
consider two ampers acting alone
so considering the original circuits we
are going to let
two amperes
act alone
two amps acting alone this becomes a
short circuit
we have this to be
the carrying source to amperes
and then again deploying the principle
of source transformation we are going to
have
the resistor in series width
the dependent voltage source
so that's going to be 0.5 V notes from
the previous section
and then we have the resistor
4 Ohms
venotes
now let's try to find the current
flowing through
V naught
so again we have an issue here
it's very difficult to determine how the
current to be distributed in this
circuit again we are going to use or we
are going to deploy the principle of
source transformation the second time
so this is what we are going to do
we are going to transform
this current source which is in parallel
with these two ohms resistor to be a
voltage source in the series with these
two obvious resistor now if you want to
think of this as in parallel you can
let's say clean this and then come and
put that here
and so you realize that it is in
parallel okay so we are going to
transform this so that we have
a voltage source so using
Ohm's law from Ohm's law we have V
equals 2 times 2 which is equal to four
two amperes times two ohms that is four
so you have four votes in series with
two ohms
and then you have
this one oh
and then dependence voltage source 0.5 V
knots
four ohms
so
we assign current I and then we take the
clockwise direction we have the sum of
the source voltages to be
for
Plus
0.5
notes that is equal to
going around the loop we have two plus
one
plus four which is seven
so seven
I seven I
seven I and then we have way notes to be
equal to
I times four so four I we're going to
put this back into this equation
so we are going to have
4 plus 0.5 V notes and then we have V
naught to be
for I so four I equals seven I so this
becomes four plus
2 I equals 7 I and then we have 4 equals
7 I minus 2 I that is equal to 5i we
divide through by 5 by 5 we have I to be
equal to
0.8 amps now since we are interested in
the value of a naught we have V naught
to be equal to
4 I so 4 times 0.8
which is equal to
3.2 volts
so since this value of V naught is as a
result of 2 amperes acting alone we call
that V naught prime prime
therefore the original value of V naught
is equal to V naught prime plus we have
Vin naught Prime
plus v not prime prime
now in the previous section we had Vin
notes Prime to be eight foot
so we have 8 plus
re nodes prime prime three points two
and that is equal to
11.2 volts so this is the value of V
naught so that's it for today's video
thanks for watching and see you in my
next video bye bye
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