Electrical Engineering: Ch 4: Circuit Theorems (28 of 35) Maximum Power Transfer Ex. 1
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- เผยแพร่เมื่อ 18 ต.ค. 2024
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In this video I will find P(max)=? the maximum power transfer of the linear circuit to a load resistor.
Next video in this series can be seen at:
• Electrical Engineering...
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I wouldn't call myself a "hero" but I am glad these videos are helping students around the world.
He's a real hero, I'm angry with myself right now because I didn't get to know him earlier and exams are really close by. It's sad because had it been I knew him earlier I would have had this problems I electricity 😪
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Thanks to you, I learned a lot about circuits. Thank you so much. Greetings from Turkey.
Glad these are helpful. Welcome to the channel
Once again, thank you for the videos , I learned a lot off you than in college.
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Hi. I have a question. What if the circuit is in parallel? example is a battery and two resistors. all of them are connected in parallel. I want to determine the value/setting for the other resistor (rheostat). resistor 1 is a fixed resistor having a resistance of 120 ohms and resistor 2 is a rheostat (variable). voltage source is set to 12V. what setting should be the value of the rheostat to be able to achieve maximum power transfer to the rheostat?
If I will be solving the thevenin resistance, it will be zero, Rth = 0. It means, I need to short circuit the terminals of the voltage source?
Thanks for those who will answer my question.
Hi, thank you for the video.
I have a quick question. Why is the Thevenin Voltage = 24V? Isn't the Voltage measured at the load resistor = 24V - the voltage observed at the 25 ohm resistor?
In retrospect, I understand that resistors placed in parallel will experience the same voltage but I always thought of voltage as something that drops as it goes across a resistor. In that case it is hard to see that the voltage drop across the load resistor is the same as the voltage across the 40ohm resistor because the 25ohm resistor should take some of the available 24V away before it goes across the load resistor. Could you clarify this for me?
To find the Thevenin voltage, you remove the load resistor and then measure the voltage between A and B. Not that no current flows through the 25 ohm resistor and thus there is no voltage drop across the 25 ohm resistor. Thus Thevenin voltage is the same as the voltage drop across the 40 ohm resistor which is 24 V.
nice. thank you. i had the same doubt
Good day Sir, does this mean that the 25 ohm resistor can be ignored? i also tried calculating V1 using the node-voltage method. it did not work? why is this method not valid?
@Aidan Higgs Nodal analysis cannot be used because that 25Ω resistor doesn't see any current, so there is no voltage drop there. The 30V just goes through the 10Ω and 40Ω resistors on its way from A to B :)
@@TheChosenOne66501 Thank you very much, i understand now
Thank you for helping me pass Electrical Engineering
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Thank you sir. Your tutorial really helps.
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Excuse sir , i would like to ask some questions
I know no current flow through from node 1 to A because of short circuit. current flow from one place to another place because of the voltage difference, place with high voltage will flow to place with low low voltage. If no voltage difference, then no current flow, so we conclude v1= va, is my concept right?
Yes indeed. V1 = Va as long as nothing is connected between A and B
Hello sir , is there no current run through the 25 Ohm resistor? (when A and B is open)
That is correct. If there is not a closed loop current cannot flow.
Professor in this example can we use nodal analasys to find the thevenin voltage ?
First current enter the nodal and 2 other current come out so I1 = I2 + I3
typically, you can use any method on any circutuit. Eventually you will personally prefer one method over another, and you'll choose what you prefer. (In algebra there are 7 different methods to solve linear equations and just like in algebra we learn all the techniques in an introductory circuits course).
@@MichelvanBiezen right . However voltage devidor fits more in this example , and its a short way
Thank you teacher,from vietnam
Indeed, the best
Hello sir, if there was no 25 ohm resistor, would the thevenin's voltage be equal to the voltage across the 40 ohm resistor?
Since there is no current through the 25 ohm resistor (without the load attached), the Thevenin voltage will be the same, which means it equals the voltage across the 40 ohm resistor. Once a load is attached, the 25 ohm resistor will make a difference.
@@MichelvanBiezen you are amazing sir, to still reply 7 years later is insane, thank you so much for your work.
Thank you from Ontario.
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Good job! It was your hard work that go you there.
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Happy to help
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Thank you from Thailand
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very good, explanation, the actions i wouldve taken are the same as urs.
God Bless You Sir
Wow nice, still little unclear with 40/10+40? Thanks
That is the calculation for the voltage divider. The voltage drop across the 40 ohm resistor is (30V) x (40/(40+10)) = 24 V (the other 6 volt drop is across the 10 ohm resistor
What shd I do when there is a dependent voltage source?
thx
very well explained.!
Glad it was helpful!
I thought it P_max = 1/2(v^2/r) not 1/4(v^2/r)
I know both equations are used, please help
I think one of those equations are for input power - P=IV rather than Ploss=(I^2)(V)
(30-v1)/40 = 30/50 , is this equation correct , cause m getting v1 as 6 , ie vth =v1 = 6.
between thank you sir for giving such great knowledge :-)
No, Vth is as shown in the video. What principle were you basing your equation on? (Always start with something you know to be correct, like Kirchhoff's rules, etc.)
ok sir , btw i figured out the mistake, thankyou for replying and saving our time by making such awesome videos.
Great, that is how we learn. :)
the best
I don't think the thevenin voltage is correct, the 25 ohm resistor does matter...Pretty sure
Actually, was wrong it didn't matter
current doesn't flow through the 25-ohm resistor when R_L is removed - thus no voltage drop can occur over than resistor.
great
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Thank you. 🙂