Circuits I: Example with Power Calculations

This video works through an example problem where we calculate power absorbed or delivered by individual elements in a circuit with resistors, current sources, and voltage sources.
University of California, Davis
ENG17: Circuits I, Spring Quarter 2015
Instructor: Dr. Kyle Montgomery
Course Website: www.kmontgomery.net/teaching/e...

[@WyattC100]

That's some pretty skillful backwards writing you have there

[@Kinoable]

can we talk on how he literally wrote things in backwards?!!!>?! that's talent right there

[@Kinoable]

and also the glass wall he wrote on are super clean and the marker pens they used are so nice! i thought it was fake until he wipe off the screen :0

[@billyray1172]

lol Left handed aswell! That's a clue btw.

[@ThePhDEngineer]

Magic!

[@faysalaslam1]

amazing way to teach.....thank you Dear Sir

[@danikokodrile2003]

Thank you a lot, this video was pretty helpful! I have a question, what would be the power supplied by the 4 A and the 3 A current sources? 0 W? Is it possible for a source to not consume or supply power?

[@vonfiller5166]

THIS IS A GOOD PROBLEM THANK YOU!

[@SirishaBhumi918]

super ...marvellous. thank u sooo much sir.

[@glassofwater3767]

Awesome video!

[@covfefe2936]

For the current of the 1 Ohm resistor between the 3A and 5A, can't you just do i = (5A-3A)? Like the case of mesh currents where mesh 1 is the 3A and mesh 2 is the 5A?

[@movearound6037]

Sir can you tell me how to calculate the power when two power suply is given with three current. Here iam unable to provide you my circuit diagram

[@indytube4343]

Great explanation

[@ThePhDEngineer]

Thanks!

[@dawoodaltai6113]

thank you so much i hope you will go on

[@ThePhDEngineer]

No plans to add to these right now...but we'll see...

[@mjnicolas8135]

thank you sir! :D

[@tedvanmatje]

the backward writing threw me off a little bit...had to re-watch the first couple of minutes because my brain was throwing a proper 'wobbler' ;) very well presented Dr. Kyle! I'll be showing this to my eldest son (saves me explaining it to him) and will be sharing this to get you some more subs (speaking of which - add me to that list too) thanks for posting - live long and prosper!

[@Risingnshouting]

Thumbs up for backwards writing or fancy editing of the video, either way, nicely done!

[@christiancoffey7563]

Quick question; for I1 and I2 at around 7:10, how did you know their directions? Or is it arbitrary? Thanks!

[@raghavendragupta8931]

Based on the 3V Voltage source , the I1 current direction can be known and by using KCL at node 1 we can knoe the I2 direction....If i am wrong correct me....Thanks

[@lulufish07070707]

But he said it doesn't have any direct voltage source?

[@electrikul9060]

While studying source transformation, we study that when ever voltage and current sources are connected in series, voltage source is considered dummy/deactivated....how you even calculated power across it?????????

[@joshuadeleon7943]

Sir Why didn't you use the formula P=V^2/R in Determining the power of the 5A current source?

[@horcrux8329]

if we had to use the formula V^2/R for determining the power of 5A Current Source , we would need to know the internal resistance of the Current Source. You see , the R in V^2/R is unknown to us . So , instead , we used the formula P=V X I , here both the Current I = 5A ( from the question) and the voltage across the Current Source V was already determined by us. Hope this helps you understand .

[@Risingnshouting]

Yes, I'm easily amused.

[@chiprock804]

Great, but why do Americans always want to use fractions?

[@altuber99_athlete]

*For the power of the 6 Ω resistor:* _P_ {6 Ω} = _I_ {2}^2 * 6 Ω = (1/2 A)^2*(6 Ω ) = +3/2 W. *For the power of 3 A independent current source:* _V_ {1 Ω} = _I_ {1}*1 Ω = (2 A)*(1 Ω) = 2 V. (1 Ω) || (3 A) ⇒ _V_ {1 Ω} = _V_ {3 A} = 2 V. _P_ {3 A} = (_V_ {3 A})*(−3 A) = (2 V)*(−3 A) = −6 W. The polarity of _V_ {1 Ω} = _V_ {3 A} is the positive (+) terminal up and negative (−) terminal down, because _I_ {1} = +2 A is a positive current, therefore in order to satisfy the passive sign convention the current must enter through the positive terminal of the resistor; and since _V_ {1 Ω} = _V_ {3 A}, the polarity of the voltage _V_ {3 A} is the same of the voltage _V_ {1 Ω} of such resistor. *For the power of the 4 A independent current source:* The voltage from the upper node to the lower node is _V_ {1 Ω} + 3 V = 2 V + 3 V = 5 V. If we assume that the voltage _V_ {4 A} of the 4 A independent current source has the positive terminal on its upper terminal and has the negative terminal on this lower terminal, we get: _V_ {4 A} = 5 V − 10 V= −5 V. _P_ {4 A} = (4 A)*(−_V_ {4 A}) = (4 A)*(5 V) = 20 W. *Applying the law of conservation of energy to verify the results:* Σ_P_ {absorbed} = −Σ_P_ {generated} ⇒ (_P_ {3 Ω} + _P_ {1 Ω} + _P_ {6 Ω} + _P_ {3 V} + _P_ {5 A} + _P_ {4 A}) = −(_P_ {10 V} + _P_ {3 A} + _P_ {4 A}) ⇒ (3 W + 4 W + 3/2 W + 15/2 W + 10 W + 20 W) = −(−40 W − 6 W) ⇒ 46 W = 46 W, which is true. Therefore, the results are correct.

[@wesleykizer79]

What the HELL this guy is writing backwards!!!!!!

[@GalibFida]

I've seen other youtubers do that. Always wondered how they do that though!

[@GalibFida]

I think I've figured out how he did it. He's probably writing normally, in which case the writings should appear horizontally inverted. But the video was horizontally inverted with some video editing software which inverts it back to normal.

[@elijahvivio6199]

just flip the video in post editing, what are ya, a flat earther?

[@lesspoint4413]

bad, the pen sound is bad...

[@lesspoint4413]

still pretty good, but the pen sound is bad

[@lesspoint4413]

180th like btw

[@ThePhDEngineer]

Agreed...