No please, the short circuit current will be 1A. So at the current source on the left, it will be 1.5ix divides to get 0.5ix and ix, and in 4ohms you have (ix + 1)
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?
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?
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
sir, for calculating the Vth in the example2, can I set the different positive and negative resistor terminals and write the equation as 4Vx+10Ix-5Ix+15Ix=0?
In general, you want to have positive voltage difference when the current flows from a low to a high potential point (voltage source in this example) and negative potential difference when the current flows from a high to a low potential point (any resistor). My point is that you have to use opposite signs for the source (4Vx) and the resistors (10Ix, 5Ix, 15iX). So the answer is no.
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
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?
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.
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 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_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
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.
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?
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
crystal clear explaining, thank you sir. greetings from istanbul !
Thanks so so much
Bro is dangerous 💀, I mean he is using loop equation rather than KCL or nodal. Damn bruh❤
Thank you
To help me in clearing my mid sems ❤❤
You are most welcome ☺️
Sir in example one ,we can also use short circuit current ,but short circuit current is equal to ix so the rth value is not matching
No please, the short circuit current will be 1A. So at the current source on the left, it will be 1.5ix divides to get 0.5ix and ix, and in 4ohms you have (ix + 1)
Thank you very much for this awesome video!
You are most welcome. Keep watching for more
Perfect explanation. Thank you
You are most welcome
THANKS SIR WE BE GETTING 90'S BECAUSE OF YOU😭😍😍😍😍😍😍😍😍😍😍😍😍😍
Awww that's great.
Which school do you attend?
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?
The polarity on the 5ohms just explains or indicates the direction of current through it, i.e. From positive to negative.
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?
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
In example1, why the current flowing through the 5ohn resistor is 0.5 Ix?
It's simple: 1.5ix is flowing towards the junction. And ix flows in 3ohms, hence
1.5ix - ix = 0.5ix will flow in 5 ohms
sir, for calculating the Vth in the example2, can I set the different positive and negative resistor terminals and write the equation as 4Vx+10Ix-5Ix+15Ix=0?
In general, you want to have positive voltage difference when the current flows from a low to a high potential point (voltage source in this example) and negative potential difference when the current flows from a high to a low potential point (any resistor). My point is that you have to use opposite signs for the source (4Vx) and the resistors (10Ix, 5Ix, 15iX). So the answer is no.
while finding vth at example one does no current flow from the 6v battery and we only consider 0.5 Ix?
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
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?
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.
Thank you so much ❤
You are most welcome
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.
Yes, the vth will be zero if there are no independent sources.
Can you explain why you set 1 V source on the a ,b？
1 question
when do we use test source
when do we use normal thevenin?
Normal Thevenin - Independent sources
Test sources - Dependent sources, like this.
i thought Vx in example 2, was supposed to be Vx= 15(Ix)
No please, it's rather -ix * 15. Because of the anticlockwise movement
Thank you!💘
You are most welcome
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
Kindly specify where you disagree with me.
@@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
@lugandaronald64 mmm are you sure?? Kindly double check
@@lugandaronald64 i agree
@@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
Rth= Vth/Isc and Isc across load is Ix so Rth=4ohms
What is wrong with this
Current in 4ohms is not Ix. Come again with your question
in example 1 why did u subtract 1.5 from 1
please state the time in the video
3.30
@@SkanCityAcademy_SirJohn
That's because 1ix leaves towards the right so the balance current is divided as 0.5ix. 1.5ix =ix+0.5ix
Sir
Does this tutorial video meant for senior high school level or university
University level
@@SkanCityAcademy_SirJohn okay no problem sir
Ok
Gate 2025 anyone
You had two different values for the same variable Ix... why are they different..
And does the Vth only affect that loop
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.
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?
Notice that in this case, there is a dependent source which also contribute to the circuit.
In my calculation vth is coming
16/3 v 😅😅
hehehe
Please 2:25
What if you use Io=1A. How one now find vth
Please sir
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
isnt the current in the 15ohm resistor ( i0-i1)-(i1 since i1 is also going through 15 ohm in the opposite direction ?
why did you use i1 only for the 2nd loop 15ohm resistor
No please
So that we can only solve for one unknown since we have only one equation around a loop.
If we use i1 for the other loop, we will struggle with the value of vx
Which country do you come from?
thank u so much
You are welcome
Where are you watching from?
@@SkanCityAcademy_SirJohn From Bangladesh
@sadiamubashira9354 that's great. Thanks so much for watching?