Circuit example with multiple batteries

[@watermel]

around 18:09 you made a mistake, its not 200 ohms its 800 ohms, it took me a while to realize

[@bramschmiddy]

thank you so much. my teacher's lecture made it seem like brand new math that I never even imagined. you made it so much simpler to understand just by breaking it down the way you did. its not new math at all! its just a different (and necessary) way of using using ohms law! thank you!!! i feel more prepped for this exam than i ever did for a physics test. lol

[@EMlNEM2020]

Of all the videos I watched, this was the most helpful, thank you

[@lanceneidhardt4283]

ik i'm here 5 years after this was posted but great video. really helped me a lot

[@ccerdamusic5467]

There's a mistake at 18:03 if anyones wondering. When he multiplied that loop equation by a factor of 4 he forgot to distribute that 4 factor to the 200 ohm resistor.

[@generalzheng9453]

I was wondering that too !

[@anirvinvaddiyar7671]

I watched your video 5 minutes before a test knowing nothing and aced the test.

[@clairmatialeto5106]

These videos have helped me a lot in my studies

[@abdallahabdelsalam6080]

so beneficial till now, nice work

[@Football-zt4vy]

Good content 💯

[@jonathanblake2656]

If he would have used [32V-(I2)(800ohms)=(I1)(400ohms)] the final answer for I2 would be 0.027Amps correct?

[@Tim-Jaeger]

Yes it should be 0.027A, ended up with the same result. If you ever want to check a circuit you could build it at www.falstad.com/circuit/ and see if the measurement is the same, if not check what you built and then your math

[@umihikari5199]

@@Tim-Jaeger 0.027A is also my results. I guess he might have some sign flipped. The general method is good though; and that is what matters. Thanks Andrew Bennett much for the knowledge!

[@SilverscreenSurgeon]

trent is correct. you forgot to multiply the one coefficient by 4. aside from that error great video

[@AndrewBennettScience]

+Ben Hunter Thanks for the feedback! I've added annotation to point that out, and will correct the video when I get a chance.

[@xAbergAx]

Great vid! Thanks for the help, needed this before my physics exam!

[@miguelfh21]

This was very helpful, thank you!

[@ChangedNames]

Can someone recommend me a video that explains the same circuit but how to find one of the emf’s? so like they give you all the resistance, the middle current, and one of the emfs

[@manveejaiswal8419]

You can do it by using POTENTIAL EQUATION

[@ChangedNames]

@@manveejaiswal8419 Too late, graduated, but thanks.

[@cyclohxn]

What if one of the batteries' voltages is unknown?

[@trentdeines1090]

I think that you may have forgotten to multiply I2(200ohms)*4... which should give you I2*800

[@AndrewBennettScience]

+Trent Deines Thanks for the correction. I added an annotation to the video to point that out until I have a chance to correct the video itself.

[@manveejaiswal8419]

Here I have doubt When current flow , it decreases with time then how can we use the equation I1=I2+I3

[@theregalproletariat]

It doesn't decrease with time, current stays constant between junctions. You thinking of PD?

[@xxtremmewizard1001]

Very helpful, thanks

[@zachbryant2124]

Do you have videos with batteries facing the same direction?

[@AndrewBennettScience]

I don't think I do, but the only thing that changes in that case is the sign of the delta V associated with the battery when you write a loop rule equation.

[@mohammadmoawad9934]

can you please post solving triple batteries problems ?

[@AndrewBennettScience]

I'm not sure when I'll have a chance to get to this, so for now I'll just address this here. Solving a problem with three batteries (or four, or twenty) involves the same reasoning as having two batteries. Each new battery will show up as a constant potential difference in any loop rule equation for which the loop passes through the battery. Since (often) the values for the potential difference across the battery is given, this doesn't usually even complicate the work, as there aren't any new unknown quantities.

[@yasinalam2951]

Love from Bangladesh ❤️

[@jisupark9606]

nice video

[@spoder22]

Why is the sum of the voltage drop equal to zero as opposed to being equal to the voltage of the battery

[@AndrewBennettScience]

It's not the sum of just the voltage DROPS that equal to zero, but the sum of all of the voltage changes (including any batteries included in the loop). In a simple, one-battery problem, you could just subtract the delta V for the battery from both sides and end up with the equation you suggest. The idea is that electric potential at some chosen position on the circuit has a set value. If you track along any path that takes you back to the point, you will end up back at that same electric potential, too. All of the change in potential due to batteries and resistors and anything else, then must add up to zero. Any other result would mean that a charge going around that loop ends up going to higher and higher energy levels, which would approach infinite electric potential energy as time approaches infinity.

[@veronicanoordzee6440]

Why do people in electrical engineering courses care so much for the effects of RESISTORS in series or parallel and care much less for the effects of VOLTAGESOURCES in series or parallel?

[@edwinnajera7454]

so I_2 = 3/110 will be the answer Right?

[@AndrewBennettScience]

Yes, that's correct.

[@esorampapote7483]

E1 + E2 / R1 + R2 = simplified

[@asiya3004]

If you do not know the circuit, leave the circuit and do not bother others

[@kaitlynnmotley]

I believe that the way you have your current flow drawn for the battery on the right is incorrect. Typically the longer line on the battery notation signifies the positive end of the battery, which means that the current should be flowing in the opposite direction than what you have drawn. It kind of makes the way you solved the rest of the problem obsolete.

[@AndrewBennettScience]

With the presence of two opposing batteries, it is possible that one battery will have current flowing the "wrong" direction through it. It turns out in these problems that guessing incorrectly on the direction of the current isn't a big deal; you'll know you had it wrong because the current will turn out negative when you solve for it.

[@kreepercrewLewis]

It makes no difference which way current is flowing as long as it is used consistently. The direction of current flow does not affect what the current does.

[@bigchicharito4813]

pretty good video , just a word of advice , dont do what everyone else is doing do something different.Do something like solving interesting concepts , (like a circuit with two batteries)

[@theregalproletariat]

That's...what this is?