☑️19 - Superposition Theorem: Circuits with Dependent Sources 1

Summary

This 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

00:00

🔌 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.

05:05

🔧 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.

10:09

🔍 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.

15:09

🛠 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.

20:11

🔗 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.

Keywords

💡Linear Circuits

Linear circuits are electrical circuits that follow Ohm's Law, where the current through a resistor is directly proportional to the voltage across it. In the video, the theme revolves around analyzing linear circuits involving dependent sources, which are a type of non-linear element. The script discusses using the superposition theorem to solve for voltages in such circuits, indicating the complexity and importance of understanding linearity in circuit analysis.

💡Dependent Sources

Dependent sources are circuit elements whose output is determined by the state of another circuit element. They are not independent of the circuit's conditions. In the script, a dependent source is described as '0.1 VX carrying Source,' which means the current it provides is 0.1 times the voltage across it (VX). This concept is crucial for solving the circuit using the superposition theorem, as it requires considering the interaction between the dependent source and other circuit components.

💡Superposition Theorem

The superposition theorem is a fundamental principle in circuit analysis that allows for the simplification of complex circuits with multiple independent sources. It states that the voltage across (or current through) an element in a linear circuit is the sum of the voltages (or currents) produced by each independent source acting alone, with all other sources turned off. The script uses this theorem to find the voltage VX by considering the effects of the 20-volt source and the 4-ampere current source separately.

💡Independent Sources

Independent sources are power supplies in a circuit that are not affected by the circuit's conditions, such as voltage or current. In the script, there are two independent sources mentioned: a 20-volt source and a 4-ampere current source. These are deactivated one at a time during the application of the superposition theorem to isolate their effects on the circuit.

💡Current Distribution

Current distribution refers to the division of current in a circuit when it encounters multiple paths. In the script, when the 20-volt source is acting alone, the current produced by this source is distributed through different branches of the circuit, creating different currents (I and I1) that are analyzed to solve for VX.

💡Ohm's Law

Ohm's Law is a basic principle in electrical engineering that states the relationship between voltage (V), current (I), and resistance (R) in an electrical circuit: V = I * R. In the script, Ohm's Law is used to calculate the current produced by the 10-volt source when it is acting alone, as well as to find the voltage across resistors in the circuit.

💡Source Transformation

Source transformation is a technique used in circuit analysis to convert between voltage and current sources while maintaining the same circuit behavior. The script mentions transforming a voltage source in series with a resistor into an equivalent current source in parallel with the same resistor, and vice versa, to simplify the circuit for analysis.

💡Parallel Resistors

When resistors are connected in parallel, they provide multiple paths for current to flow. The script explains how to calculate the equivalent resistance of parallel resistors, such as combining 20 ohms and 4 ohms into an equivalent resistance of 10/3 ohms, which is crucial for simplifying the circuit and applying the superposition theorem.

💡Circuit Deactivation

Circuit deactivation refers to the process of temporarily removing or short-circuiting a source in a circuit to simplify the analysis. In the script, the 20-volt source and the 4-ampere current source are deactivated one at a time to isolate their effects using the superposition theorem.

💡Voltage Division

Voltage division is a principle stating that when resistors are connected in series, the total voltage across them is the sum of the individual voltage drops. In the script, the voltage VX is found by considering the voltage drops across series resistors when the 20-volt source is acting alone.

💡Circuit Analysis

Circuit analysis is the process of determining the behavior of electrical circuits, such as calculating voltages, currents, and power. The script is entirely focused on circuit analysis, specifically using the superposition theorem to solve for unknown voltages in a complex linear circuit with dependent sources.

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