EMF, Internal Resistance, and Terminal Voltage of Batteries Explained | Doc Physics

[@alexander26bro]

If you were struggling with the find the voltage part. Add the resistances together, then divide the 13.2 V by that total resistance value. Then do 13.2 V subtract the value you got to get the voltage you are looking for

[@sciencenerd8879]

alexander26bro But why do you subtract? Once I got the current by dividing 13.2/R equivalent, I multiplied the current by the resistance which gave me wrong answers...

[@rimtoumi9302]

Sciencenerd You substract because the voltage you get without substracting is the voltage of both the battery and the resistance! So for you to get the voltage of the resistance alone you have to substract from it the voltage of the battery. I hope it was helpful.

[@youssefabdelhady4624]

+alexander26bro thanks that was helpful :) !

[@eeee8677]

We subtract because 13.2V/Rtotal is the voltage drop, we want the voltage on the resistor which is the total voltage minus the voltage drop. Kinda the opposite of what Rim said.

[@gretawilliams8799]

man but why should we find the voltage of resistor?? in the 2nd diagram the 1 ohm resistor takes 1.2 volt and the 10 ohm takes (13.2-1.2)=12 volts the voltage will be finished by the time it reaches the 10 ohm resistor so the PD across voltmeter is 1.2-0 -> 1.2 volt. i am confused about what and where voltmeter reads

[@M3lCHOR]

lmao "told you fairies" Thanks for the video

[@gh2110]

Ohhh my God! Ah Real Teacher” please’ hear me, man’ I didn’t get my license until I saw u. I didn’t understand it at first. I was going for the biggest thing rrrrrrr for years. Resisters were to small for me, kept fighting until I saw u, and let there be light 💡…. In the smallest thing. I fkn love u man! I had to say it… thank so much…

[@what_homework]

5:38 honestly jumpscared me so hard omg

[@murad1227]

It would be much more helpful to the audience if you showed the working for your calculations.

[@wise_fool]

Murad Cholak he's being a bitch

[@tayson2015]

I need your help. I did the math but my answers are: 6.6v (for circuit 1 wherein the total resistance is 2) 1.2v (for circuit 2 wherein the total resistance is 11) 0.13( for circuit 3 wherein the total resistance is 101). I think i did something wrong.Just can"t figure out what. I used the formula V= I*R by first taking out the current across the circuit and multiplying it with 1 which is the resistance of the resistor on my left. Why is my method wrong?

[@haajarnihal1579]

HOW DID YOU FIND THE VOLTAGE VALUES ???

[@michaeldollins6047]

Hi Dave, Great Vid.....you do a better demo than do in my classes. I will be showing this tomorrow....thanks again.....

[@DocSchuster]

+Michael Dollins Hey man! Happy to help!

[@prateekgurjar1651]

Nice, but who is Dave ?

[@prateekgurjar1651]

13:47 what a lovely pen, where'd you get that doc, what's the brand ?

[@harineeanandh6520]

sir, ur videos are really fun & interesting & it clears my concepts well. it looks like i don't have to got to my physics classes any more!!! thanks a tonne!!

[@jimc.329]

I wish I were younger, there is so much to learn.....thx so much!

[@thelastdragon3242]

Couple of comments at the bottom didn't get his voltages when they found the "I" of the whole circuit. Getting answers such as 6.6V, 1.2V and .131. That is the voltage drop on the whole circuit, it might make more sense if you did it like Kirchhof's rule. V-IR-IR=0. Easier to recognize that as a drop.You minus that from the original voltage( V0-Vdrop)=Docphysics Answers , get the answers he has at around 6:15 . ... BYE

[@sanchit1sanchit]

Just found this channel. I love your way of teaching, sir. Subscribed :D

[@secret_watcher5047]

im so astonished! we took this equation in our class but what I saw in this video is on another level and im glad I understood it and managed to solve for the voltages, this was entertaining thanks :)

[@simranjoharle4220]

wow those circuit problems really made it so clear for me👍

[@alejandrosiles5722]

"skip a bit brother"

[@Jnglfvr]

At 2:56 you have the same "terminal" voltage for both 3 and 4 bulb loads. This is obviously a measurement error I suspect because you are using an analog voltage meter or have simply interpolated the scale incorrectly. What you refer to as the emf voltage is more accurately described as the Thevenin voltage of the source (Vth) and what you refer to as the "circuit" voltage is usually known as the load voltage (VL). Finally the "internal" battery resistance is the Thevenin resistance of the source (Rth). The formula expressed at 10:48 is an equation expressing the load voltage as a function of the current. If you were to plot this (load voltage versus current) for your data at 2:56 would would get a (approx) straight line whose slope is the negative of the "internal" (Thevenin) resistance. The "y-intercept" (at zero current) would be the Thevenin (open circuit) voltage and the "x-intercept" (zero voltage) would be the short circuit current (which is the ratio of the Thevenin voltage to the Thevenin resistance). I.e., V (load) = Vth - I*Rth. If you do this you would find that the straight line is best approximated ignoring the "3-bulb" voltage (which is incorrect). This line is very nearly straight and a least squares approximation yields an "internal" (Thevenin) resistance of 0.3 ohm in this case. Finally, a battery is not an "ideal" voltage source as an ideal voltage source has no internal resistance and thus will supply ANY current necessary to maintain a constant voltage. I.e., its voltage versus current graph is a horizonal line.

[@chilekotarsepai]

I love u and ur channel. You're a great teacher! Thanks man.

[@ellenhodder9842]

Are you secretly Ryan Reynolds helping me to pass physics? And us budding physicists never see your face.. Coincidence? I think not!

[@YEUHNBD]

great video and explanations to everything! one personal opinion and you may or may not agree is i dont like that formula you proposed at the very end. because if I was to just look at that on a formula sheet, quite frankly i would have no idea where that came from or how it was derived, therefore i would not understand any theory behind it. with just getting the same answer like i did in the first couple of minutes of the video i actually remember it and it makes complete sense and makes me think. with this formula its just plugging in numbers. regardless though, awesome job and keep it up 10/10

[@ibrahimalotaibi2399]

is it possible to share the program that was running the battery ! thanks .

[@mrcastor9093]

Aha! so when we measure a branch with the voltmeter, we consider only the characteristics of that branch, like the resistance of the battery!. The only variable that depends on the entire circuit, and affects the branch measurement, is the current, because the current goes through the entire circuits and comes into the branch you're measuring as being modified. The internal resistance doesn't sum with the exterior resistance when measuring the branch with the voltmeter, that was my mistake.

[@nshaiman]

how can you get 6.6v,12v and 13.1 volt against Rinternal as you pointed towards Rint and ask viewers to calculate voltage?Vx=Rx*Vt/Rt

[@rpphysics1369]

Good explanation .simple to understand.thanq

[@TheCzm123]

+rama prasad kota I read ur cmnt in an indian accent

[@Joel-oe7ud]

Hey the voltage reading of 3 bulbs and 4 bulbs are the same??

[@mahadevankm2621]

=New subscribers even in 2018. wonderful video Doc.

[@nanak3363]

U nailed it :) Thank u so much!

[@davidontiveroz8295]

Thank you so much for making this video sir. Helped me with physics 2 and also entertaining. Subscribed

[@MrPickycat]

My PHYS teacher in HS was named Schuster Great guy.

[@mtalhashahzad899]

when we open the switch then current should flow through battery and thus internal resistance will be offered and then the terminal voltage us not equal to emf

[@deeppatel8276]

you taught really nice..

[@PedroLucas-fs8mc]

Hey. How did you come across the last value? intuitively it makes sense one could measure electric potential regardless of a current but how can I achieve the result?

[@fayokanmifashanu1333]

is it because of the number of bulbs

[@dans8285]

how i wish you were my physics teacher

[@abhishektomar_5046]

what is the potential diff between two battries of different emf E1>E2 joined parallel to each other with internal resistance is zero.. these battries are joined is such a way that positive terminal of one is joined to the positive terminal of other battery and negative to negative ..

[@sunainamadgula8664]

why can't the last circuit diagram resistance be zero instead of taking it as infinity? practically zero is possible right? not infinity.

[@frankbucciantini388]

"fairies who live in batteries carry electrons and get sweaty" lol

[@coopers9522]

You are Hilarious! Fun to watch even if I don't need to learn this exact topic, I wan't to learn it because its so fun to watch haha! :)

[@mybikeknowsnospeedlimits3169]

why do you substract the voltage with the ampere?why not ohm's law?

[@AA-nh5wo]

Looks like a voltage divider? Doesn't it?

[@michelleybelly-l9q]

How would you find the voltage drop?

[@sahilkhan_cs50]

Is 1 ohm the internal resistance of the cell

[@Hydrojen_]

Lol this made me laugh and then do the problem! Thanks!

[@ashrithabs6205]

very well explained. ...👌👌

[@0_0RuRu]

Is it doesn't matter if the resistance are parallel or iin series

[@piaakasyapa]

Wbt potenstial difference and external resistance?

[@kwazikhuzwayo1060]

Callin the light bulbs suckers!! LOL

[@-AnweshaDas-]

that was great! thank you💙

[@amandanorazman401]

Hey Doc, I know, old video.. but I can't get the calculations right.. Isn't calculating through a voltmeter uses volts but the results you give are used with VR=I, i.e. 13.2 x 1/2 = 6.6 A not 6.6 V? Or am I missing something here..

[@prithividharamdass9191]

It's V=IR not VR=I. If you were to make I the subject of the formula then it would be I=V/R and the units would be in A. To get his answer I think he most likely did his calculations like this: He found the total current in the circuit by using I=V/R which would be I= 13.2/ 2, your answer here would be 6.6A. To find the reading on the voltmeter he would have used the voltage passing through the 1 ohm resistor. Which would be V=IR or V=6.6 x 1. The answer would then be 6.6 V.Of course it is possible that I could be wrong and he used some other method but I think this is what he did. I hope it helps :)

[@larrybanks1961]

why is it 6.6V, 12V 13.1V and 13.2V yet it should reduce?

[@tarunpahadiya3803]

If one day I will be the teacher I would try to be like you

[@andreatorres8676]

You saved me! Thank you

[@nurzulaikha6089]

umphh, the current supposed to move from terminal positive to terminal negative but why for this topic the current move inverted?

[@DocSchuster]

+Nur Zulaikha How did you get that impression? I always treat current as coming from positive terminal.

[@texury2714]

In my school, we learn in physics that electrons flows from negative to positive but in electronics (different subject) we learn that current flows from positive to negative. This is because we originally thought the latter, but later is was discovered the former was true. The formulas still work with both concepts however, so electrical engineers kept with the idea that current flows from positive to negative for simplicity.

[@MsTommyknocker]

But only the inner voltage decreases with increasing number of bulbs.

[@rambhar9949]

I didn't get the first circuit diagram. Shouldn't the effective resistance be 1/(1/1 + 1/1) = 1/2? Then the voltage reading of the voltmeter should become 26.4, should it not?!! Can someone help me understand if there's any calculation error, because right now, I,m not able to!

[@2206arnab]

Ram Bhar 1. The resistances are in series not in parallel..they'll be 1+1(Req)=2 and the voltage drop would be 13.2/2= 6.6 2. The voltage..can it ever be 26 when the battery can produce max as 13.2?

[@Jayremayah]

Doc how am I supposed to calc the battery emf if given only current flowing as 20mA

[@texury2714]

are you not given the voltage of the battery too ?

[@texury2714]

Also what other components are in the circuit, if any?

[@booesha2]

this video so boss man

[@TheDoritos777]

So the emf is the maximum potential difference between to electrodes? Excuse my english, it's not my native language

[@DocSchuster]

TheDoritos777 That's right. Under no load.

[@clashofkings3139]

I got 6.6V,1.2V,0.131V cause currents had the same magnitude,since they multiplied with 1ohm,we got these voltage values but this decimal place is confusing me

[@clashofkings3139]

We learned that the bigger resistor got greater share of voltage so the resistor right next to our battery had a smaller voltage for itself so I felt like it's a bit reasonable

[@christopherwong998]

What you got is the current throughout the circuit, not the voltage. To find the voltage drop multiply what you got by the internal resistance, which is 1 ohm (because V=IR), and then subtract from the 13.2V. Idk if you still need this cause it was a year ago but hope this helps

[@ayn4671]

thank you

[@hirshitharajee9322]

thanks a lot

[@dahazz]

Yo yo yo.. I accidentally took the resistors in parallel (Silly mistake, the paralle voltmeter threw me off.). So I got Rtot using the parallel resistors equation. And then followrd by another silly mistake,(V=VR.) Getting the exact same answers as shown in the video. Any reasons behind this coincidental error? Help Doc!

[@DocSchuster]

+Hamza Ali I don't know how you use V = V/R. Sounds like an inverse mistake or something. Two wrongs made a right.

[@dahazz]

+Doc Schuster Doc doc, The last equation to this video. Quite similar to V=VR, don't ya think?

[@MrOoblek]

Fam, it's V = IR Check your units.

[@texury2714]

+Hamza Ali Voltage = Current * Resistance

[@surya8891]

dude , you're cool

[@amersheikh11]

i got 13.7 for the third one, can someone explain third and fourth?

[@billwilliams6338]

DOC SCHUSTER,,,, ,,,, Eric Johnson is saying that overdrive & fuzz guitar pedals that take batteries will sound different when using different types of 9volt batteries NON-Rechargeable 9 volt batteries like Alkaline, Carbon zinc, Lithium, Mercury because of the DC resistance of the 9 volt battery. //// The sound will be different if you use an Alkaline, Carbon zinc, Lithium, Mercury battery each will sound differently because of the batteries DC resistance --->>> 1.) Why does each 9 volt battery type Alkaline, Carbon zinc, Lithium, Mercury have a different battery DC resistance? Why different battery types will have a different DC resistance compared to another battery type, any reasons why? Yes I know that the batteries DC resistance will drop the 9volts when the guitar pedal is drawing more current but what Eric Johnson is saying is that 9 volt Alkaline, Carbon zinc, Lithium, Mercury will have a different DC resistance which will drop the 9 volt down differently when the overdrive or fuzz pedal is drawing more or less current.

[@fayokanmifashanu1333]

if 4 bulbs = 1.325 and 3bulbs= 1.325 how come the resistance is different

[@DocSchuster]

That's what I'd like to declare "experimental error." I would graph the data and see that one of those two is a bit of an outlier (inconsistent with the pattern). Perhaps the contact resistance was too high in the three-bulb case.

[@hablahabla6653]

yay new marker

[@prateekgurjar1651]

Uhm, what actually is an "active" circuit.

[@meenakshibm3897]

tq...tq..tq..verymuch sir....

[@alphonsesynrem28]

More of mathematics than electricity or internal resistance of a battery. Simple yet making it so complex with all the math work. Teach algebra.

[@thatguy1306]

Lolz what a legend!!!!!

[@byronanderson3291]

awsome

[@TheDoritos777]

Two*

[@willyrivero470]

Why don't you get those fairies out. They are messing it all up.

[@MysticMD]

Umm thank you :)

[@TheIzugec]

Okay, the more I go through your videos, more I have urge to comment. I don't know if You ever watched the show named Big bang theory but in analogy to that show I will argue for some future sitcom that you become heroic professor that everyone remembers!! :) Now seriously, if I ever make even slightest contribution in physics or not, EVEN if I become what I want, that is physicist, ergo accomplish my dream :D. I'll make sure I'll visit USA to personally thank you. I know I can thank you this way as well, but I want to make a photo of that proud smile you'll have :)) Also, I would be really happy if for some amount of years I would be able to make lectures on my language and even remotely help people like you do :D Oooh, also, if you'll have time I would really like to see your thought on modern physics, by that I mean physics today (Problems with standard model) PS.I have big question in my head connected to view of looking on world, I would tell you background theory, but post is long enough as it is, but do you think that we can know every's particle that exist, position and velocity at any time. But actually what i'm asking is do you think that heisenberg's principle and quantum mechanics are just our phase of not knowing the physics enough, or do you think that nature doesn't let us know everything at any place at a given time. A whole lots of greetings from sunny Croatia :)

[@TheIzugec]

I Zugec Aaaaaaaaaand you found it rather creepy hahahah, I get it what it's like from your perspective. It's just that i find it so cool and had this moment when I had to express :D , sorry doc, thought the best. Cheers

[@dogeeatsveggies]

my savior lol

[@victordolman2646]

I don't like that R_int. I think all our lives would be better if it were zero. I do like puzzles though. Suppose I have a box, and I paint it black. (This is essential, because after we are done I want to seal the box and keep its contents secret: a true 'black box'.) The idea is that we put a battery in the box, and no other sources of electrical energy. We can put in any electrical components that we want. But no other electrical energy source! Just the battery. Can we then get rid of R_int? We can put in transistors, opamps, IC's of whatever nature, capacitors, resistors, bulbs, coils, wires, enitire voltage regulator chips, a computer, whatever. But no other electric energy source (did I mention that?). Can we then do it? Can we eliminate R_int and have the box produce a voltage, possibly a little less than the original V_emf, that does not depend on the external load? Or is this impossible in principle? To get puzzlers going, imagine that we put in an entire current meter and a small embedded computer, powered by the battery. Then when someone hooks up an external load, the computer 'sees' the current go up, and might take action to compensate for the voltage loss (by switching some transistors or something). This is a straightforward exercise in voltage regulation, that is done all the time in power supplies. When you begin to feel that this might be possible, then realize the starkness of the question: the box is an ideal voltage source, without R_int. Seriously? In reality? Electrons travel through the box, they are pushed only so much. Whichever complicated path they follow inside the box, there must surely be some resistance along that path. No way this can work. Do you feel the two approaches fighting inside your brain? Good. :-)

[@phasorsystems6873]

Are your teachers busy? Track down androidcircuitsolver on google

[@Vunami]

lmfao fairies

[@alexwestwood2352]

didn't solve the problems lol

[@mtalhashahzad899]

when we open the switch then current should flow through battery and thus internal resistance will be offered and then the terminal voltage us not equal to emf

[@MsTommyknocker]

But only the inner voltage decreases with increasing number of bulbs.

[@kamarajdevendran1952]

Thank you

[@balendrabalendra9095]

thanks a lot