Electrical Engineering: Basic Laws (12 of 31) Kirchhoff's Laws: A Harder
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- เผยแพร่เมื่อ 24 ส.ค. 2024
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In this video I will use Kirchhoff's law to find the currents in each branch of multiple-loop and voltage circuit.
Next video in this series can be seen at:
• Electrical Engineering...
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Best mathemati
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Q
I was stuck on this topic for such a long time and in only 9 minutes I understood it, you really saved me an all-nighter. Thank you so much. Your teaching is phenomenal.
Glad it helped!
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glad it helped.
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Thank you for the comment.
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Glad you found our videos! 🙂
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Bruh this is so beautifully smooth. I was literally melting my brain before but not I can answer these questions in 2 seconds.
Also, I’m not at university yet, but I can still understand this because of the really easy method of assuming the direction of current. Thank you so much sir
You out of all the videos I have seen have made me understand kirchoffs rule. Especially when it comes to the direction convention for voltages. Keep it up sir
Glad you found our videos and you found them helpful. 🙂
I really recommend this proffesor. Ive been struggling to anwer my homework for 3 hours but within just 9 minutes I manage to finish it with the help of this video. THANK YOU VERY MUCH SIR!!
Thank you. Glad you found our videos. We have covered all topics in physics on this channel.
I just wanted to say THANK YOU! I've watched like 10 videos trying to understand this Law and honestly, you've made it so easy to understand!! In very short steps which I love!
Only this guy in TH-cam clearly got an idea about kirchhoff's law.
Glad you like the videos on Kirchhoff's laws.
Not gonna lie! I missed some lectures of my Electrical Physics class. And I tried to make it up using Khan Academy and other videos and online resources unfortunately all pf them were crappy except this guy! He made Kricchoff's rule sound so easy now I can easily take nap knowing I'm prepared for my exams! Thanks Man!
Glad I could help!
After watching your videos I actually have confidence for my exams. You are truly a great teacher. Thank you.
Happy to hear that! Good luck on your exams.
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Thank you. Glad you found our videos!
Thank you very much..honestly..I went through several videos...but you gave me the enlightenment..even in school, this went terribly wrong! especially the direction of the flow, you mentioned it clearly! just fafded away my doubts! thank you, Sir!
You are very welcome
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The World need people like you profesor
Thank you, we appreciate your comment.
im back on your channel again for our last quiz and the hardest one yet (all about kirchhoff's law). i understood this very quickly thank you for ur tutorials!
Welcome back! Good luck on your exam!
Best mathematics and physics teacher in the world
3 unknown currents. Use Kirchoff's laws (1st law once, 2nd law twice) and Ohm's law to establish system of 3 algebraic equations. Put that in any math software or any TI, HP pocket calculator and it's done.
Yes indeed, solving the 3 equations 3 unknowns type of problem can be done with any app or program or calculator which has that programmed. But it is important that we learn how to do that.
Thank you so much sir. I will recommend you to all of my friends. Keep on making tutorial! By the way you look like Steve Jobs :)
THANK YOU! You are a gifted teacher and I learn so much from watching your videos.
Just left this open on my laptop for revision tomorrow. I bet it will help more than any of my dumb lecturers
Hope it was helpful.
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Finally i understand it,
After long time
Sir! You are the best professor I need.
Thank you. Glad you found our videos. 🙂
After watching multiple videos over a few months , i got even more confused but today i stumbled upon this video and i understand this topic. Thank you sir for the great work
Glad it helped!
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Thank god for this video. I was about to lose my mind.
Glad you found it, 🙂
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Happy to help
Thanks a lot, I am a senior high student and I have been facing problem but becus of you I now understand it well
Great! Glad you found our videos. 🙂
Thank you sir, you are really the best teacher ever.
Wow, thanks. We appreciate your comment. 🙂
Bravo! Mr. Michel !!! Brilliant and clean explanation ! Deep respect ! 🤓
Thank you. Glad you found it helpful. 🙂
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Thank you. Glad you found our videos. 🙂
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Sir thank you for always posting helpful videos. 🙏
Always welcome
I have a problem similar to this except with different numbers and have had trouble with the currents. Major Thank you for this
Glad it helped! 🙂
Thank you for this! I've learned so much from watching you videos. Kudos to you sir!
Awesome video, thanks
Glad you liked it!
Please dont stop teaching your a expert
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Thanks, this really helped me to understand it easier
Glad to hear that!
Nice explanation Profesor!
Thanks alot sir 😊
Excellence teaching
Most welcome
Sir I just want to say am very happy with the explanation it going to help alot and I also want say keep up
Glad you found our videos. 🙂
@@MichelvanBiezen yes sir and hope to see more
Nice job sir . Indeed i enjoyed the lecture . Keep on making life easy for us
Glad it was helpful! 🙂
I had an issue understanding how to do this..well.. not any more great teaching.. still not sure of the math procedure but very good.
Hi, is there any other way u could explain the general sign convention when applying KVL? And in the case of loop 2, why is the equation 8 i2+ 6 i3 -4=0 instead of 8 i2+ 6 i3 +4=0. Isn't the current flow opposite to direction of loop for 4V voltage source? Many thanks :)
Since you are going around loop 2 in a counterclockwise direction, you cross the battery from the positive end of the battery to the negative end, thus there is a 4V voltage drop. When picking current directions in each branch it doesn't matter if the directions are correct. (The assumed directions could be opposite to the real directions). Then work out the problem as is shown. If you picked the incorrect direction, your answer(s) will come out negative.
Thanks for the help sir ☺️.. great explaination 🤌🏼
Thank you. You are welcome.
Sir, In getting the I2 why 6(-0.308)-4 becomes 4+6(0.308). I'm confused. It should 4-6(0.308), right?
We moved the two terms to the other side and therefore the sign changes. A - B - C = 0 becomes A = B + C
THANK YOU SO MUCH !
You are welcome. Glad the videos are helpful.
very good and helpful lectures from you. Hope all the best for you.
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It's really impressive,,
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Thank you for your comment.
When you find the other currents, why do you plug in -.0308 Amps for I3. What if i assumed the correct direction for I3 at first, and got a positive value, then my I1 and I2 would be different values
+Shlomo No, you would get the same values for I1 and I2, no matter what direction you assume for any of the currents.
God bless you❤
Thank you. God bless you as well! 🙂
Thank you very much for the excellent video.
You are welcome!
I don't seem to get it.
Why are we multiplying it by 2 and 3? Is it because of the equation tag (Equation 2 & Equation 3).
It is an algebra technique to solve the two equations two unknowns type of problem.
Surely there's a better way to do these problems... Been stuck on these Loop and Junction rules all semester. How do you know which variable to solve for first for?
This is a very classic example of the use of Kirchhoff's rules. Hopefully after watching a few examples it will make sense. (We have similar examples in the physics playlists as well which is a prerequisite to this course).
When we draw the Loop 2 - its in the opposite direction to I3 - I didn't understand this part. Thank you.
The arrows in the center of the loops, simply show the direction of "travel" around the loop (which is independent of the current direction) in order to add up all the voltages. KVL
The loops must be in the same direction? Both clockwise, or both anti-clock. Or can I do ine clock and the other anti-clock?
They do not need to be in the same direction. You can go around each loop in any direction you like.
Thanks Man you rock
Glad you liked it.
Thank you
Please will changing the position of any one voltage source change any of the equation
Not if the voltage source remains oriented the same way and stays on the same branch. Once it is moved to a different branch, it will change the equations.
@@MichelvanBiezen you actually replied. I love you Siirrr.
Thank you, I understand
thank you Mr. Michel !
The video is indeed helpful☺👏
Glad you think so!
Thank you so much 😢
You are welcome. Glad you found our videos. 🙂
6.24 - why do you add the equations instead of subtracting? According to my notes we have been subtracting in class… I’m confused
You can do it either way. The algebraic method of elimination can be done either way.
Thank you sir
You are welcome. Glad it was helpful. 🙂
Why was equation one changed to negative ? It was I1 +I3 = I2 what happens when you solve it directly?
Excellent job
why did you multiply the 1st equation by 2, and 2nd equation by 3? is that just random?
That is an algebraic method of solving 2 linear equations. (method of elimination)
Can we apply Ohm's law for 4 ohm resistance and 10 v voltage? In that case current I1 will be equal to
10/4=2.5amp.which does not match with our I1.
No, you can't use the simple Ohm's law equation here because you have multiple voltage sources. Each source affects each branch in a way that is difficult to predict, unless you use Kirchhoff's rules.
@@MichelvanBiezen Thanks a million for the fast reply.
thanks
Welcome
you are literally my life saver! thank you so much!
Why I1 and I2 goes down together instead of them splitting some of their currents to the both resistors i did not get that.
Oh i get it now, divided the problem into two sections and calculated the currents of each voltages then combined them together and got the same results.
Great. Glad you figured it out.
At 7:50 when you moved 6(-0.308) to the other side you changed te negative sign to a positive which i get but why didn't the positive 6 change to a negative. I'm still learning algebra so I'm still learning all the tricks of moving everything around. Thanks David.
- 6 (0.308) = 6 (- 0.308) but that is NOT equal to -6 (- 0.308) = 6 (0.308)
@@MichelvanBiezen looking at your answer I couldn't figure out but after doing the calculations now it makes sense. doing it your way gets a positive1.848 which is right. doing it my way would have resulted in a -1.848 which is wrong. As a beginner I would not have known that and would have changed the 6 to a negative. Is that allways the case, if a number is multiplied by a number in brackets you change the sign of the numbers in the brackets but the number outside the bracket you don't change. Isthere some kind of albebra rule that states that. Thanks for your help David
Yes, you have to be VERY careful with the signs doing these types of problems.
thank you
Welcome!
Sir , I have a question
What will happen if we take one loop clock wise and the other one anti clockwise
Or one of them anti clockwise and the other clockwise.
It doesn't matter. You will get the same result. (but you have to keep track of your voltage rises and drops according to the direction you choose. Note by reversing direction, you'll get the negative result of the result you got from your original direction.
Could you correct me? I put i3 as clockwise which created i1 = i2+ i3. I moved the terms to get i1-i2 = i3. I plugged this into resistor 6 creating 6(i1-i2). Doing the system of equations I get an i1 of -0.18 which is different from your answer.
The answers in the video are correct. 6(i1 - i2) is not correct as a voltage drop
Thanks alot
Happy to help
Excellent
Thanks
Thanks so much sir for this 💯it made me understand better as beginner,
But where my problems lies now is at your final answer in I2 i couldn't understand how you got 0.731A
Pls sir can you explain it to me
I'm just a beginner 🙏🏾
Use equation 3 which only contains I2 and I3 and I3 is known. So substitute for I3 and now you have an equation with only 1 variable I2.
@@MichelvanBiezen thank you so much sir I've gotten it
I really appreciate, you're one in a million teachers 💯
sir, why did you assume loop2 in a clockwise direction?
It doesn't matter. You can assume either direction and it will work.
sir you are god in disguise... thank you
Not in the least. Just another person trying to do his little part in the world.
is it correct to say that the voltage in that bottom node that I2 passes into via the 8 ohm resistor is 0?
Since the node is not connected to ground we don't know. Only if that section of the circtuit is connected to ground do we know that the voltage is zero.
Is it possible to obtain the currents in a situation exactly like this but with another resistor at the second loop? I reckon it wouldn't be since there'd be 4 unknowns but only 3 equations...
You can have any number of resistors in any of the loops and this method will still work. We have examples like that.
should I get the same result if I used the mesh analysis method?
Yes, you should obtain the same result no matter what method you use.
Thankyou sir! Now I already understand it.
Great!
I love you Sr. Thanks so much.
Sir, do we have any short trick to solve these types of problems?
I am preparing for NEET ( medical entrance exam in India ), so we have very little time to solve such problems...
It would be great if you can tell a short trick. please
For this type of problem there is really no short cut. But that said, it does depend a lot on how the questions are asked on the exam. We are currently making a number of videos on the JEE advanced tests and often you don't have to work out the problem in detail to find the correct answers. We illustrate that in those videos. We haven't looked at the NEET exam yet. But we will and we'll try to determine how to best prepare for that. Our time is very limited, so we don't know when we can get to it.
@@MichelvanBiezen oh ok thank you, sir
@@vedangisekra9399 There is a shortcut. It's the method of superposition.
To give an overview, you "turn on" one source at a time, and "turn off" the other sources.. Voltage sources become shorts when "turned off", and current sources become open gaps when "turned off".
You then assign a point, or several points at which it is of interest to find the voltage, and would help you solve for everything else you'd like to know. Then you find the voltage at that point, exclusively due to each individual source. Then you add up the partial voltages, to find the total voltage at the waypoint you assigned. You then have the voltage at that point, which you can use to solve for everything else.
This turns algebra into plug-and-chug arithmetic.