When using loop analysis to find Vth, wouldn't you need to account for I times the 4-ohm resistor as well? Vth equation should equal 4I + 20(I+3) right?
For the first loop it was supose to be instead 42=-8i-6i-2(i+3) bcs when current enter a resistor the is a current drop accross that resistor which makes it negative sir!
No please, we can equally used the anti-clockwise direction, and that is also correct, the only way we will get a wrong answer is if we use the loop containing the current source.
I have an inquiry on the second problem,in calculating the thevenin voltage,why wasn't the 4 ohm resistor used ? We only used the 20 ohm resistor yet the 4 ohm resistor is also in the loop in consideration.
It depends on how you assign current in the circuit. You can either distribute current from the voltage source or the current source. I used the voltage source: to say that current in both 8 and 6 is I and current coming from the right is 3A. Then current in 2 will be (I + 3). You can use the current source to say that: 3 = I + I2 that is also correct.
Not only that, other elements in the circuit contribute to that. And it's not strange if the thevenin's voltage is larger than the voltage source in the circuit.
When solving using Thevenins theorem, you remove the resistor you want to find the current flowing through, thats how 4 ohms was ignored when finding vth
i think your explanation for that answer is wrong because we are dealing with the 8 ohm's resistor without the 4 ohm's resistor@@SkanCityAcademy_SirJohn
@@godswayzottor5555 yes I was confused too. when finding Vth use loop anyalsis and make sure to include the 4 ohm resistor. that equation should be Vth = 4I + 20(I+3)
Assume that there is a voltage source in between the terminals ab. Now you realize that there is the left side of the circuit and the right side. 6 and 2 are connected end to end so they are in series and the combination is connected to the same two nodes with the 8 which means that (6+2)//8
@@SkanCityAcademy_SirJohn i think the person is referring to the 4 ohm resistor in problem 2. I'm also confused as to why it wasn't included in finding Vth
@@yammy_yams the 8 ohm was on open circuit therefore no current was flowin through the 4 ohm.that why it was used According to my understanding hope i make sense..
why is it that when finding the thevenin's voltage across the circuit from the right hand side the current will not flow the 4 ohms resistor? please i will be happy with somebody can help me with the understanding.
Please are you aware of the steps to take when solving using thevenin's theorem? You know you are supposed to eliminate the resistor you want to find the current flowing in. When that happens you have an open circuit and current cannot flow in an open circuit.
man I'm watching different parts to your videos it makes sense but the steps always confuse me its always the beginning as well, where you'd divide the voltage and then at other times you are doing a different step where you just say a number parallel to another number I'm just confused i know its me. its makes sense but if i were to do it without watching's the vid I'm just back to square one i just wish there was a specific formula to keep it the same its always different every time i watch not saying its you.
I understand, with applied electricity, you just need to understand the fundamentals. You wouldn't have a general formula to help you solve all types of questions, because each question presents how you approach it due to the positions of the circuit element. The fundamental things, how to determine how current circulated in a circuit, relating to which resistors are in series, and parallel, which one has been shorted or open circuited and all that counts a lot
@@SkanCityAcademy_SirJohn that's absolutely correct i sir its just im confused at times when i see it im not able to write down and understand it untill you start wrtiting. question when you have 2 voltages sources you are solving for both I and I1? and when you have a single voltage source and a current you are solving for I including the brackets for the current? for example ( I + 3) my question is if its clockwise its I plus a number and the opposite for anti clockwise ?
Yes, if you take any arbitrary direction, if a current moves in the same direction, then it's positive, if it moves in the opposite direction, then negative
Yes, so if you want to solve any network with Thevenin's theorem, you first remove the resistor you want to find the current flowing, that's why the 4 ohm was removed
Really helpfull with my final exam, From Malaysia
thank you
You are most welcome. Thanks so much for watching
Ideal explanation. ❤
Thanks so much
Watching from South Africa, your videos are unbelievably helpful, please keep it up man 🫡🫡
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How I was struggling to understand this,,, but you've made it very simple for me
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Thanks so so much Rose
@@SkanCityAcademy_SirJohn welcome ♥️
Thank you so much! This video was very helpful. Thank you for not skipping any steps :)
You are most welcome. Where do you watch me from?
Also, should loop analysis always be the first approach to find Vth?
This is very helpful
Thank you
You are most welcome
When using loop analysis to find Vth, wouldn't you need to account for I times the 4-ohm resistor as well? Vth equation should equal 4I + 20(I+3) right?
i find this piece very helpful, thank you very much.
You are most welcome and keep watching
For the first loop it was supose to be instead 42=-8i-6i-2(i+3) bcs when current enter a resistor the is a current drop accross that resistor which makes it negative sir!
They are rather supposed to be +s. Because we are using the conventional flow of current here.
@@SkanCityAcademy_SirJohn so will it always be positive no matter the direction of the loop sir??
@Jambenoire-kn4lk yes, unless there is another current opposing your direction
Why we use the clockwise direction in 4:41 ? It would be a mistake if we have used the anticlockwise?
No please, we can equally used the anti-clockwise direction, and that is also correct, the only way we will get a wrong answer is if we use the loop containing the current source.
in second question vth should be equal to vth=4I+20(I+3) ? right? im confused
No please because of the open circuit, there will be no current in 4 ohms, so it's just vth = 20(I + 3)
I have an inquiry on the second problem,in calculating the thevenin voltage,why wasn't the 4 ohm resistor used ? We only used the 20 ohm resistor yet the 4 ohm resistor is also in the loop in consideration.
Because the terminal ab is now opened, there will be no current in that branch (4ohms)
@@SkanCityAcademy_SirJohn Thank you so much for good explanation.
You are most welcome
where do you watch me from?
@@SkanCityAcademy_SirJohn I still dont get it ? can we use the loop containing the 24V source ? watching from uct SA
In question 1 Why is the 3A current not being divided between the 2ohm and the 6 ohm resistor
It depends on how you assign current in the circuit. You can either distribute current from the voltage source or the current source. I used the voltage source: to say that current in both 8 and 6 is I and current coming from the right is 3A. Then current in 2 will be (I + 3). You can use the current source to say that: 3 = I + I2 that is also correct.
I used Nodal analysis in finding the Vth, and my answer is 10.25V. Is my answer wrong?
Thanks for trying with a different theorem. Which of the questions did you attempt and had 10.25volts for vth
Why was the 3A not used to find the vth
When finding the vth, we do not consider a loop with a current source
Please explain in details the computation of Rth.
It's simply the resistance that is seen between the two terminals when all sources (voltage and current sources) are deactivated
@@SkanCityAcademy_SirJohn thanks . I see.
You are most welcome
You missed the voltage across the 5 ohm
No please
How can the thevenin voltage be greater than the voltage source in problem 2. Is it because of the 3A current source?
Not only that, other elements in the circuit contribute to that. And it's not strange if the thevenin's voltage is larger than the voltage source in the circuit.
why have you ignored th 4ohm resistor when finding VTh
When solving using Thevenins theorem, you remove the resistor you want to find the current flowing through, thats how 4 ohms was ignored when finding vth
i think your explanation for that answer is wrong because we are dealing with the 8 ohm's resistor without the 4 ohm's resistor@@SkanCityAcademy_SirJohn
@@godswayzottor5555 yes I was confused too. when finding Vth use loop anyalsis and make sure to include the 4 ohm resistor. that equation should be Vth = 4I + 20(I+3)
😊
Thanks so much for watching
But please why is the (6+2)//8. I thought they were all on series?
Assume that there is a voltage source in between the terminals ab. Now you realize that there is the left side of the circuit and the right side. 6 and 2 are connected end to end so they are in series and the combination is connected to the same two nodes with the 8 which means that (6+2)//8
The 4 ohm resistor wasn't used why??? In calculating the Vth
According to the rules, you need to remove the resistor you want to find the current flowing through
That's how come, the 4ohm was removed
@@SkanCityAcademy_SirJohn i think the person is referring to the 4 ohm resistor in problem 2. I'm also confused as to why it wasn't included in finding Vth
@@yammy_yams the 8 ohm was on open circuit therefore no current was flowin through the 4 ohm.that why it was used
According to my understanding hope i make sense..
That's correct
why is it that when finding the thevenin's voltage across the circuit from the right hand side the current will not flow the 4 ohms resistor? please i will be happy with somebody can help me with the understanding.
Please are you aware of the steps to take when solving using thevenin's theorem?
You know you are supposed to eliminate the resistor you want to find the current flowing in. When that happens you have an open circuit and current cannot flow in an open circuit.
Where from the rth
What do you mean, by where from rth?
man I'm watching different parts to your videos it makes sense but the steps always confuse me its always the beginning as well, where you'd divide the voltage and then at other times you are doing a different step where you just say a number parallel to another number I'm just confused i know its me. its makes sense but if i were to do it without watching's the vid I'm just back to square one i just wish there was a specific formula to keep it the same its always different every time i watch not saying its you.
I understand, with applied electricity, you just need to understand the fundamentals. You wouldn't have a general formula to help you solve all types of questions, because each question presents how you approach it due to the positions of the circuit element. The fundamental things, how to determine how current circulated in a circuit, relating to which resistors are in series, and parallel, which one has been shorted or open circuited and all that counts a lot
@@SkanCityAcademy_SirJohn that's absolutely correct i sir its just im confused at times when i see it im not able to write down and understand it untill you start wrtiting. question when you have 2 voltages sources you are solving for both I and I1? and when you have a single voltage source and a current you are solving for I including the brackets for the current? for example ( I + 3) my question is if its clockwise its I plus a number and the opposite for anti clockwise ?
Yes, if you take any arbitrary direction, if a current moves in the same direction, then it's positive, if it moves in the opposite direction, then negative
@@SkanCityAcademy_SirJohn Thank you sir john you are filling in these lost puzzle pieces for me and its starting to make so much more sense
Thanks so much
why 8 ohms resistor wasn't used when calculating 4th?
can you please indicate the time stamp so i can help you out?
Where the 4 pass
can you please come again with your question?
Yes the 4 ohm resistor wasn’t used
Yes, so if you want to solve any network with Thevenin's theorem, you first remove the resistor you want to find the current flowing, that's why the 4 ohm was removed
Can you please come again. I didn't get it.
Oh okay. I get you know. That's a great point. Thanks so much Sarah.
poor explanation
Thanks so much for your input
@skancity can I get your personal contact
Ideal explanation. ❤