☑️17 - Thevenin's Theorem: Circuits with Dependent Sources 1

In this lesson, we shall learn how to solve linear circuits involving dependent sources using thevenins theorem.
When solving a linear circuit with dependent sources using thevenin's theorem....
1. Deactivate all independent sources, mark the terminals and leave all dependent sources alone.
2. Introduce either a 1v source or 1A current source in between the two terminals
if you introduce Vo = 1, then we find io, but if you introduce io = 1, then you find Vo
3. The attempt is to find Rth, Rth = Vo/io
4. Find Vth from the original circuit, Vth is the open circuit voltage betwee the terminals ab
00:00 - Example 1
12:41 - Example 2
Playlists on various Course
1. Applied Electricity
• APPLIED ELECTRICITY
2. Linear Algebra / Math 151
• LINEAR ALGEBRA
3. Basic Mechanics
• BASIC MECHANICS / STATICS
4. Calculus with Analysis / Calculus 1 / Math 152
• CALCULUS WITH ANALYSIS...
5. Differential Equations / Math 251
• DIFFERENTIAL EQUATIONS
6. Electric Circuit Theory / Circuit Design
• ELECTRIC CIRCUIT THEOR...
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[@erdal.uzunlar]

Thank you very much for this awesome video!

[@SkanCityAcademy_SirJohn]

You are most welcome. Keep watching for more

[@nishantgupta5015]

Thank you
To help me in clearing my mid sems ❤❤

[@SkanCityAcademy_SirJohn]

You are most welcome ☺️

[@katzuev7786]

Thank you!💘

[@SkanCityAcademy_SirJohn]

You are most welcome

[@user-tj7zj8pj5e]

Thank you so much ❤

[@SkanCityAcademy_SirJohn]

You are most welcome

[@danthetutor6300]

What was the purpose of the polarity over the 5 ohm for vx .Shoudnt the voltage across ir be -5Ix.Since in the direction assumed there will be a negative drop?

[@SkanCityAcademy_SirJohn]

The polarity on the 5ohms just explains or indicates the direction of current through it, i.e. From positive to negative.

[@mohubemahlatsephaladi6246]

THANKS SIR WE BE GETTING 90'S BECAUSE OF YOU😭😍😍😍😍😍😍😍😍😍😍😍😍😍

[@SkanCityAcademy_SirJohn]

Awww that's great.
Which school do you attend?

[@mokhtarnasser2482]

is Vth always 0 if there are only dependant sources? like it was in example 2 since there are no independent sources to provide voltage.

[@SkanCityAcademy_SirJohn]

Yes, the vth will be zero if there are no independent sources.

[@yi_huimeng1936]

Can you explain why you set 1 V source on the a ,b？

[@nishjijiiiii1045]

Hi! May I know why are there two value of Ix? What is the difference between the Ix obtained in solving the resistance vs the Ix obtained in solving the voltage?

[@SkanCityAcademy_SirJohn]

The idea is.
To find the rth we need to solve for Ix, that's the first Ix.
For the second Ix helps you to find Vth.
The two are independent. So in another question it can be a different ball game altogether

[@irisianacani9316]

while finding vth at example one does no current flow from the 6v battery and we only consider 0.5 Ix?

[@SkanCityAcademy_SirJohn]

Current flows from the 6v battery. You can make 6v the reference source to do the current distribution, it is correct, but my reference was the 1.5ix current source that is why I have the 0.5ix moving towards 6v.
You can equally do a current distribution using the 6v as reference and get the same answer

[@fancybitch4159]

GOAT

[@SkanCityAcademy_SirJohn]

Thanks so much

[@zaidabdulkarim2125]

Hello sir, in example 2 when we found i1 it is in fraction if we change it to decimal it is 0.66A so in i0 when we replace i1 to find it it will be -1.32A hence, isn't Rth Vo/io = 1/-1.33 = -0.76 ohms. so only because i changed the fraction to a decimal it changed the answer totally?? if so then when should i just keep it as a fraction and when should i change it to decimal?

[@SkanCityAcademy_SirJohn]

Usually during the solution process, use fraction, and then you convert the final answer to decimal, but if you want to be changing to decimal every now and then, leave your decimal to 4dp.

[@aliekamaraa445]

Sir
Does this tutorial video meant for senior high school level or university

[@SkanCityAcademy_SirJohn]

University level

[@aliekamaraa445]

@@SkanCityAcademy_SirJohn okay no problem sir

[@SkanCityAcademy_SirJohn]

Ok

[@hleketanicarl2984]

How is it possible that the voltage drop across 4 ohm resistor is 5.33V? That means 5 ohm and 3 ohm resistor only dropped 0.67V from the 6V battery. Can you please explain that?

[@SkanCityAcademy_SirJohn]

Notice that in this case, there is a dependent source which also contribute to the circuit.

[@KinguinMalfoy]

You had two different values for the same variable Ix... why are they different..
And does the Vth only affect that loop

[@SkanCityAcademy_SirJohn]

The values for Ix are different because you are doing two different things:
1. You want to find RTh hence you need to excite the circuit because mind you, the independent sources are removed and if independent sources are removed, dependent sources have no purpose because they are controlled by these independent sources. So ones you excite the circuit with a 1v or 1A source you get a value for Ix. Also when finding Ix or any current value, do not consider the loop with a dependent current source.
2. After finding RTh, you need to find Vth, and this time you leave the terminal open, hence the value for Ix will surely be different.
For the last part. You can use the other loop from vth to the far left. Ignore the loop with the current source.

[@sadiamubashira9354]

thank u so much

[@SkanCityAcademy_SirJohn]

You are welcome

[@SkanCityAcademy_SirJohn]

Where are you watching from?

[@sadiamubashira9354]

@@SkanCityAcademy_SirJohn From Bangladesh

[@SkanCityAcademy_SirJohn]

@sadiamubashira9354 that's great. Thanks so much for watching?

[@21khalid8]

in example 1 why did u subtract 1.5 from 1

[@SkanCityAcademy_SirJohn]

please state the time in the video

[@caesarlesikiba6417]

3.30
@@SkanCityAcademy_SirJohn

[@onthewayofficial5889]

That's because 1ix leaves towards the right so the balance current is divided as 0.5ix. 1.5ix =ix+0.5ix

[@viwemkhothi6349]

isnt the current in the 15ohm resistor ( i0-i1)-(i1 since i1 is also going through 15 ohm in the opposite direction ?

[@viwemkhothi6349]

why did you use i1 only for the 2nd loop 15ohm resistor

[@SkanCityAcademy_SirJohn]

No please

[@SkanCityAcademy_SirJohn]

So that we can only solve for one unknown since we have only one equation around a loop.

[@SkanCityAcademy_SirJohn]

If we use i1 for the other loop, we will struggle with the value of vx

[@SkanCityAcademy_SirJohn]

Which country do you come from?

[@dcmar-ava5150]

Please 2:25
What if you use Io=1A. How one now find vth
Please sir

[@SkanCityAcademy_SirJohn]

Okay,
Do current distribution. That is
Current in 4ohms is (ix + 1), in 3 ohms is ix and in 5ohms is 0.5ix.
Apply kvl to the loop containing resistors, 3, 4, 5.
Ix = -8/9
Vo = 4(-8/9 + 1)v
Rth = vo/io = 4/9/1 = 0.444 ohms

[@lugandaronald64]

This is contradicting to content I read from Basic engineering circuit analysis by David Irwin 11th edition page 192, I suggest u check that page and see where u go wrong if at all u are wrong

[@SkanCityAcademy_SirJohn]

Kindly specify where you disagree with me.

[@lugandaronald64]

​@@SkanCityAcademy_SirJohn
The part where you apply an independent current source, the book says you only do this if the entire circuit is made up of only dependent circuits, yet here yo dealing with one that has both independent and dependent

[@SkanCityAcademy_SirJohn]

@lugandaronald64 mmm are you sure?? Kindly double check

[@volerdenuit1582]

@@lugandaronald64 i agree

[@lugandaronald64]

​@@SkanCityAcademy_SirJohnI rechecked and both methods work, the first time I replied I had not yet tried both, but now am sure they both work, thanks for the great content

[@DoggyD.electro]

In my calculation vth is coming
16/3 v 😅😅

[@SkanCityAcademy_SirJohn]

hehehe