Internal Resistance - A Level Physics

[@fidelsantos12]

Thank you soooo much! I was close to crying from not understanding the book that I have read a LOT of times!! You're my hero!

[@ItsRubenMorris]

fidel santos I’m past “close” at the moment. I hope this guy helps me

[@HannahCaeru]

My unit 1 exam's tomorrow...all I can do is say thank you for all these videos, they're what has taught me almost all of AS physics. Thanks!

[@DrPhysicsA]

Thanks. I hope the exam goes/went well.

[@ahmedeloufir]

I am a 38 year old software engineer who is supposed to know all this, yet I sleep every night watching your videos. I discovered your channel less a year ago and it has been my favourite one since then... yet, you haven't posted any new one for quite some years and still your demonstrations need more deepening. But if you ever decide to do so, please don't change the humanity of your pedagogy, talking and writing and filming in the same manner. Thank you thousands times and more !

[@yacinehannane2908]

"There is resistance in the battery itself" Batteryception DUUUN DUUUUUUUUUUN

[@lenakyony]

I feel like science would've been fun with you there

[@yacinehannane2908]

+Lena Kyony XD Thank you!!

[@swervo3065]

nerd

[@kamalthapa7860]

These classes are really useful. Thank you very much.

[@josefinej3921]

Thank you so much, this was the exact video I needed, you have no idea... I've been struggling to understand the idea of EMF and internal resistance for too long, and as "simple as it should be", I felt no one could give me a "simple" explanation. I believe my Physics book's description is far too inadequate, not to say confusing. This completely turned my level of understanding and I now finally think I understand this. This is primarily because you, in your explanations, leave few question marks to be left over as you more or less explain everything and in simple terms, I believe. So thank you :-)

[@simranjoharle4220]

This is one of the best explainations I have ever come across..... thank you Sir!... this really helped.

[@clementl6843]

Did you or can you make a video about the Norton & Thevenin models ? Thank you :)

[@denispoutchninecosta4886]

It's just what I was searching for Thanks

[@zawaideh61]

Thank you so much for those amazing videos I'm currently doing AS and I'm doing really well with your help.

[@legendfooty]

i don't get the graph for the variable resistor because as resistance increases current decreases so surely voltage stays constant due to v=ir not just v increases as r increases

[@DrPhysicsA]

What time on the video does this occur?

[@legendfooty]

17.5 mins

[@legendfooty]

DrPhysicsA please reply

[@ChristopheHerreman]

legendfooty The graph plots the voltage drop (over the resistor) against the current, not the battery voltage. When the resistor value increases, the voltage drop will be higher and the current will lower, and vice versa.

[@legendfooty]

oh so its the emf that stays constant not the terminal pd due to v=E-ir and I increases V decreases

[@Zipo214]

Its awesome because I have an exam over resistance tomorrow! Perfect timing. Thanks!

[@deeptavker3515]

Very helpful videos sir..:) nicely explained. Admire your work.

[@rahuldas-uv9eh]

the video was literally good and cleared all the doubt which i had in my mind regarding internal resistance..... thnxx alot☺☺

[@MrZubair708]

Thanks! This really helped me understand this section that i was having trouble with.

[@gertwallen]

0:48 Dr.Physics, someone told me that the classical concept of "electrons flowing in a circuit" is not correct but rather it's an electromagnetic wave (radiation) that travels in the circuit and not the electrons. Is this correct?

[@claraj6521]

How is that V= E - ( I * r ) ???? (15:46) Is it correct for Voltage across the battery =( E - Voltage across the external resistance or main resistance ). So it will be that V= E - (I * R ). There is no current, If Switch is open. So, Voltage across the main resistance is zero. Therefore, V = E - 0 => V = E.

[@titan8084]

small r = internal resistance

[@AhmadKhan-sp3qb]

at 18:30, if there is only a single resistor in a circuit, wont it take all of the voltage of the battery? so if we change the resistance, in any case it will use all of the emf of the battery thus the voltage remains constant? further, if we increase the resistance, the current will decrease so why wont the voltage remain constant? where am i going terribly wrong

[@tchtmm3112]

I have a question and I want from you to answer it please: what is the diference between the efficative voltage and the Emf

[@frango8352]

Fantastic video. Absolutely worth the 20 minutes.

[@drspinz955]

You cant plot that graph, because in order to get that ..SOMETHING should be a constant !!! which exactly is the constant?? R?? V?? I?? I don't get this help!!!

[@silverchain5525]

You are autistic. Go to 20:50. I don't care ur comment is 3 yrs old. Ur still autistic

[@davidbellfilm]

At 15:00 you say that the current will be 0 and hence the emf will equal the voltage displayed on the voltmeter. Surely there will still be a current through the voltmeter?

[@jeanpierredaviau7478]

Is not the resistor kind of diluted in the mass of the battery? The électrolitic or chimical? So the internal resistor is in between the two poles. That would put the variable resistor in PARALLEL with the internal resistor intead of in series.

[@ModestMaker]

If I understand this correctly, this is why the measured bus voltage on my battery-powered circuit decreases as the current demand from adding loads increases. It's a function of that negative sloped V to I curve which is set by the internal resistance of the battery. So, if I have a battery with a large internal resistance, the voltage on the circuit decreases dramatically as I draw more current. Is that correct?

[@TheFrazerboy]

great video your better then my physics teacher

[@jdnandi6617]

Than* You may want some help from your english teacher.

[@JamDaaMan]

pikachu sucks I could say the same for you, your grammar is defective.

[@JamDaaMan]

pikachu sucks Except that you put words straight after commas. Flawless is one word.

[@JamDaaMan]

pikachu sucks At least I can retain some maturity as opposed to resorting to vulgar language. I guess it reflects your intelligence in a way - your limited vocabulary. Your grammar is terrible. Its is incorrect to leave spacing after commas. So don't lecture everybody else when you yourself can barley string a sentence together.

[@arocks1234567]

pikachu sucks u wrote "you didn't noticed a simple mistake", that's a grammar mistake not spacing -_-but u are right, people should look at themselves first! :)

[@duaaaq4193]

this video is just great (: Thank you so much Dr PhysicsA

[@AmineChM21]

this is just beautiful

[@cooler3603]

At 13:31 when i calculate i find voltage drop was 0.5. But when i measure it, will the voltmeter reading be 4.5 volt ? And also will the reading of voltmeter in the resistor be 4.5? Thanks!

[@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

[@jesuschal]

EXCELLENT PEDAGOGIC!!! I would wish all my teachers would have had your teaching skills. After watching this video I did my own test and I came up with a question. Let's say the my circuit has an intrinsic impedance. I would have three resistances in my circuit: the R (resistor on your explanation), r (internal battery resistor on your explanation) and Rc (impedance of the circuit without battery and without resistor R). How would I add Rc to the calculation?

[@DrPhysicsA]

Since we are using a battery it will be DC. So any AC impedance effect would not arise and you would simply look at the resistive element of the reactance. you would simply treat that as another resistor.

[@fatimazohra2224]

i had a hard time with internal resistance, but now i finally get it ! thank you !

[@MdSajir-bn6vb]

but shouldn't the gradient of the voltage current graph give the resistance of the object and not the internal resistance...?

[@Vulpes_Lagopus115]

Hi Dr. do you think you could do a short series of videos as an introduction to lie groups and lie algebras, at least SU(3) SU(2) U(1) because I think they are really important in physics and there isn't a lot of accessible material in an appropriate level, everything is very rigorous. They would also be complements of your video on rotation operators. I think it would be a great idea and would help a lot of people that in the future, like me, would like to become a physicist. Thanks :D

[@johnlawrence5672]

Yeah, I think that is a good idea because although I downloaded some books they are too specific. I just want to get some idea of it not to much but at least something.

[@DrPhysicsA]

There is some basic introduction to these unitary matrices in my series on particle physics.

[@Vulpes_Lagopus115]

DrPhysicsA Yes I know thats why i tell you, I saw your video and I thought it was a good idea for a sort of miniseries :D Also I thought you had finished your projects and you didn't know what to do next. By the way, do you have something in mind?

[@jackt4274]

So does it mean that for an internal resistance of 0, the graph will plot a horizontal line, meaning no matter the current values you will always get the full voltage?

[@mahmoudm451]

I don't understand why when the circuit is close and you measure the voltage across the battery you get 4.5 V? shouldn't the battery and resistor work as potential dividers? it's like putting 1 ohm resistor beside a 5 ohm resistor, with a 5V battery, does that mean that when you measure the voltage of the 1 ohm resistor you will get 4.5? I'm puzzled.

[@vishaljain4915]

A voltmeter measures the difference in potential between the two terminals. If u look at the diagram, when the voltmeter is put across the battery, the right hand side terminal measures 0 V as all the energy the charges are carrying has been dropped across the resistor, as when charges come back to the battery they have no energy or in other words, all the energy charges carry is dropped across the circuit. So the right hand side terminal measures 0V. The left hand side terminal measures 4.5V as the charges gain 5v from the battery but lose 0.5V from the batteries internal resistance, so they leave the battery with a P.D of 4.5, this means the voltmeter measures 0 on the RHS and 4.5 on the LHS so the voltmeter reading is 4.5V

[@anonymousrhino5121]

But what does the area under the line give us?

[@serpentineflame]

The last time I heard conundrum it was because the whole saw I was using for cutting in the coach light outlets had a bent pilot bit, and my project manager lost his damn mind.

[@davidsweeney111]

Hi, Im getting stuck on Karnaugh maps in my digital electronics, do you have any videos that can help me please? thanks.

[@DrPhysicsA]

Sadly no. Sorry.

[@tomrp-yt2409]

Hello DrPhysicsA, I am just curious whether this lecture still applies to yhe current ALevel Specification, because if so, it would be a great help to me! Thanks

[@whiteguyplays6229]

I don't get graph of E/V against 1/R to find r (internal resistance)???

[@infuze6647]

18:34 as the voltage increases the current decreases? Wait I’m confused, shouldn’t it be “as the voltage increases the current increases?”

[@babysuyash]

which potential drop is great is it of variable resistor or internal resistor ?

[@mrpeanutguy4719]

Thanks a ton for these videos, they're really helpful. Would you ever consider doing an A-level maths series?

[@DrPhysicsA]

Thanks your very kind comments. I have no plans to do the maths series.

[@ZrxGR3Gx]

examsolutions does a level maths.

[@jimdogma1537]

Dr. PhysicsA, can you do a video on the new BICEPT experimental results with the gravitational waves, what these results are, mean, and their implications? Thanks

[@DrPhysicsA]

I'm not sure there's very much to say at the moment beyond what has already been promulgated.

[@jimdogma1537]

DrPhysicsA Thanks for replying. Actually PhysicistMichael's channel did a great report in the BICEP experiment I recommend everyone check out. I even gave your channel a plug there cause both your channels are cooool ;)

[@viknesavetrivelan3352]

why do they still maintain the fiction that current flow from postive terminal to negative terminal ?

[@MathraMania]

Probably for the hassle of correcting every single book that's already been published. :/ The new books nowadays mentions clearly about this problem.

[@julienferte3795]

Hello, I would like to know: what camera are you using? Are you using an external microphone?

[@DrPhysicsA]

I use a Panasonic SDR SDR S26 and I switch off the autofocus. I use the internal microphone of the camera.

[@julienferte3795]

Thank you for these details, I am interested in creating similar teaching videos, in other fields of knowledge. How do you make the camera stand above the papersheet? A peculiar device or a handmade thing?

[@DrPhysicsA]

Julien Ferté It's just a standard camera stand. I arranged for the Camera to point vertically down onto an A3 sheet of paper.

[@kurdman12345678]

Is this AS or A2 because im doing AS AQA unit 1

[@williamjarrett97]

Very good video, well explained. Thank you!

[@angelachu8359]

thank you so much. your video is so awesome, and I finally understand the idea of internal resistance.

[@MacyBoy11]

Fantastic summary! Thank you very much

[@w1zzard418]

Very well explained sir!

[@TheEngraver100]

So basically, Electromotive force can be analogous to potential energy? If so, is it a force?

[@DrPhysicsA]

No it is not a force. Even though it is called the force. As you say, it essentially is the potential energy stored in the battery. It is measured involves which is work done (or potential energy) per coulomb of charge moved.

[@sahilthakur1322]

sir what is internal resistance

[@deepakchandravanshi6274]

powerful explanation...

[@augustsobolewski64]

Probably a stupid question, but when "Voltage drop" is said is "Voltage output" meant.

[@so.meYeah]

best video on this subject ever!!!

[@harshitkhanna1918]

Won't the voltmeter provide potential drop

[@PeterWalkerHP16c]

yes. Modern volt meters have a very high input impedance to reduce this effect. Older meters had a 'table' on the dial to allow you to calculate the effect of the meter on the circuit being measured.

[@CudiTV.305]

The voltmeter doesn't have resistance ?

[@mbikov]

+Rony Polisaintvil It does I recon. But very insignificant.

[@titan8084]

Serkan Demirel what the actual fuck is wrong with you " Ideally, an ammeter should have a zero resistance value, but materials with zero resistivity are not present" CAN YOU FUCKING READ???? i wasnt even talking to you

[@imagination7710]

Actually voltmeters have a massive resistance, so they don't take any current from the circuit

[@mohammadaniyal]

17:46 one question. can the voltage go zero if the value of the current reaches some particular value because the graph you made indicates so as the gradient is touching the axis

[@DrPhysicsA]

No. The voltmeter operates by having a very high resistance and so a very tiny amount of the current flows thro the voltmeter. Most of the current flows through the resistance. But if the resistance value becomes too high (and higher than that of the voltmeter), then most of the current will flow thro the voltmeter and the experiment ceases to be valid.

[@mohammadaniyal]

DrPhysicsA i understand thank you

[@faizmalik9210]

It would be less confusing if the practical battery is drawn as box containing an emf in series with an internal resistance.

[@ReMoBoY97]

Thanks very much ! very useful video.

[@skyleonidas9270]

You are genious mannn how u know all diss?

[@m9u7shuriken]

12:45. It's called a 'parallel' voltmeter. Quantum Electricity guys!

[@faizasikandar6050]

What do yoy mean by internal resistance. What thing is resistance

[@PeterWalkerHP16c]

A battery is not an ideal current source. If you have a perfect conducting wire and short the battery out, the actual current will be less than the ideal current because the battery has some resistance internally.

[@krisnegi2626]

drPhysicsA to the rescue ......... great video !!!

[@fasanayorinde1194]

I really enjoy the lesson

[@ahmedal-ebrashy3691]

Why I has a large value when V is 0? I thought no voltage no current!

[@ali-bb7so]

Very good video indeed

[@alimoazzam11]

Sir, I have the same question as someone in the comments. When you are trying to measure the Internal resistance at 16:15 in the video, you put a variable resistor in the circuit and plot the voltage drop across it against the current. My question is, shouldn't the voltage remain constant? Since you increase the value of the resistance, current decreases correspondingly, so that V remains constant. Where am I muddling it up?

[@kinshukdua]

are you talking about the Voltage in the Voltage Current graph around 17:13??? Then he is talking about the voltage *through the resistor after the voltage drop* so when you change the resistance R the the voltage *through the resistor after the voltage drop also changes, it's NOT the total voltage* and because the total resistance has changed the current also changed (because V = IR) I had the same confusion when I had first learned about it.

[@mahmoudm451]

I get where you're confused. Think of the variable resistor as one resistor split into 2 parts by a sliding contact, one part is the one you use and you take an output voltage from it, and the other part you don't use it, but it acts as a potential divider with the other part. The voltage of the output you're using can be increased by increasing the length of the part of the resistor that you're using, by simply moving the sliding contact. Remember that the sum of the voltages of the 2 parts of the resistor is always equal to the voltage of the battery, so by increasing the voltage of one part, you're in turn decreasing the voltage on the other part. Now when you increase the resistance, you also increase the voltage, but the current drops...you may say that when the voltage increase, the current must also increase and that's true, but think of it in this way: when the voltage increase, you can't get extra electrons from outside to compensate for this increase in voltage and so the number of electrons in the circuit is limited, and thus the current is constant. Think of it as 2 horses pulling a cart across a rocky path (lets call it x), if the path is short (small resistance) then they will spend little energy crossing it (low voltage) and they will pass it very fast (high current) becuz although it's rocky, but it's very short. Now when you increase the length of the road( increase the resistance) you're inturn making it MORE ROCKY, and so the poor 2 horses will have to spend more energy (more voltage) to cross it, and since the path is more rocky, they will take longer time to cross the distance x they crossed that they crossed before and thus the current falls, since there are no other horses to come and help. hope that makes sense? I tried to make it as simple as possible.

[@alimoazzam11]

Mahmoud Matar thank you so much and also the previous commenter :) I had forgotten about my comment but maybe others can benefit from your kindness

[@LifeOnHoth]

These videos are very nice. My question is no doubt gonna show that I have never had a physics class in my life, and I never did any good in math classes either :). However, I was (I like to say) born with an "unhealthy" interest in electronics. So my interest in that greatly helps me understand physics and math I never really got at school - and you teach it in a VERY good way - thanks :). Back to the question. When I watch the video, it is obvious to me that for the math to work out, assumptions need to be made, and based on that you get the basis for calculating this stuff. I'm just philosophizing here - couldn't there be an internal resistance in the voltmeter and ammeter to take into account too? And also other factors that would affect the results, like ambient temperature etc.? Hehe - I know it will seem obvious to those of you who are used to understanding math, physics and all that - that I might blend stuff together which doesn't belong - but if there is something to my philosophizing here - is it so that the math in the video above is really a simplified picture of what is really going on in the circuit, taking out certain elements to make it less complicated to understand...? Or am I just totally off? :) (it probably is so :p)

[@Ovulationman97]

There is a small amount in the voltmeter and ammeter but it is such a small value therefore we don't really need to take it an account. Same again with the temperature... I don't know if I'm right or not :)

[@naimza1]

Mourmour Man voltmeter has a vert very VERY high resistance rating. this is to prevent current flow in the voltmeter. this is also why it is connected in parallel. whereas ammeters have negligible resistance that is why it is connected in series.

[@grïnj966e]

Thank God for DrPhysicsA!

[@larrybanks1961]

"you should be able to draw this in your sleep" -Physics Don

[@pervezjan3077]

Hi! I am a fan of your vidoes and I use them in my classroom very often. Thanks for such a nice work Man! I do ask for some more help specially for Practical papers of Cambridge for AS and A level. (Physisc 33 and 52) If possible , up-load some examples papers which I can show to my students. Jan

[@MrPowellfactor]

I thought electrons exchange photons this was electrical charge , why do I keep hearing people talking about the flow of electrons?

[@roseroseey6381]

Thank you so much, you really helped me

[@mStephanLouis]

You just saved my life

[@mtj1999]

Still can't understand why voltage drop across the battery equals to E.M.F minus voltage drop across "battery". If anyone has evidence then plzz do share

[@IvanKalamazoo55]

very good video before sleep... you've got such a relaxing voice!! thx for posting

[@sahilpatel6138]

Very Excellent video, thank you :) and do you use marijuana recreationally sir or have ever have experienced with it ? (Legitimate question)

[@SorryCrane16]

And where did that come from 😲🔫

[@rithvikshindihatti1215]

I think the circuit is not in series.:(

[@naimza1]

thank you this was very helpful

[@kimmidoll9717]

This saved my life.

[@RoyAlexander]

brilliant - thanks for this.

[@physicsSriLanka]

Thanks sir, we are from Sri Lanka, can you do this videos on 1080p full HD quality. they're really helpful. your voice is very clear to listening us than other teachers. Good luck

[@MrAlbashiri]

Thank you very very very very very very very very much.

[@JackSteadlan]

very useful

[@anshumanmishra2555]

Good

[@magn8195]

18:49-18:50

[@aryagupta3649]

Electricity flows because the - side of the battery has MORE NUMBER OF ELECTRONS than the + side of the battery which has LESS ELECTRONS. It is totally wrong to say that charges attract, so electrons move from - to +. Here, + & - are not charges, they simply denote that - means more number of electrons = higher potential while + means less number of electrons meaning lower potential. Thus electrons move from higher potential to lower potential i.e, from - to +. By convention, we say that current flows opposite to the direction of flow of electrons, i.e from + to -. Guys this is a very important concept. Although f*cking lazy, seeing such crap being posted in a video and it being on the internet where thousands of unknowing people will learn from it. And if you feel like I'm wrong, think logically. If unlike attract, if there even WERE unlike charges in place of the + & -, wouldn't the attraction have been equal for BOTH resulting in an attraction from both the sides resulting in the +ve & charges meeting ? Instead of a one-sided flow of CURRENT ? ;))

[@realname712]

Really? 22 minutes for internal resistance

[@itstwinklerkate]

i am too distracted to listen because his accent makes me feel like im in Hogwarts lol

[@tatevarevikyan6852]

God bless you

[@kagogoveya6213]

V=IR is not ohm's law....

[@ThatBigBlackClock]

Ohm's Law deals with the relationship between voltage and current in an ideal conductor. This relationship states that: The potential difference (voltage) across an ideal conductor is proportional to the current through it. The constant of proportionality is called the "resistance", R. So yes it is.

[@kagogoveya6213]

NO!! V=IR is not ohm's law.... ----------------------------------------------------------------------- "We stress the relationship V=I*R is NOT a statement of Ohm's law. A conductor obeys Ohm's law only if its V vs. I curve is linear, that is, if R is independent of V and I. The relationship R=V/I remains as the general definition of the resistance of a conductor whether or not the conductor obeys Ohm's law. . . . . . . . . Ohm's law is a specific property of certain materials and is NOT a general law of electromagnetism, for example, like Gauss's law." -

[@JohnSmith-cy8hq]

Hello, I did my exam yesterday and I'm pretty happy with how it went and I've got you to thank for that. Thanks, I really appreciate your videos. There was one question in the exam that kind of bothers me now. Besides the strong nuclear force, which force acts between protons in a nucleus and what is the exchange particle of this force? I wrote electrostatic and exchange particle virtual photon. Virtual photon is definitely correct but I'm not sure if I should've written electromagnetic force instead of electrostatic. What do you think? Thanks again

[@DrPhysicsA]

Personally, I would give you the mark.

[@mohammadaniyal]

awesome thanks :D

[@TheCsePower]

Something is bugging me If im correct you said, V=Emf-Ir then at I=0, V = Emf but didnt you say earlier that V=IR? If that is true then why not Emf = 0R =0?

[@DrPhysicsA]

emf is total voltage capability in battery/cell. V is the "external" voltage (i.e. the amount available after driving through the internal resistance)

[@raykay126]

DrPhysicsA Isn't there current flowing around the circuit connected to the volt meter though?

[@DrPhysicsA]

Ray Kay Yes but the voltmeter has a very high resistance so the current is extremely small.

[@maviosmachethe144]

i get it now thanks

[@harrisonrossiter5608]

you a goat

[@amaanihamid528]

NOT ALL HEROES WEAR CAPES

[@gretawilliams8799]

+drphysics A. Pls clarify this solution for me. physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-187.html question no 913. ... iv) 2 and 3 I thought this in two ways, and it would contradict the other answer through either one of them. ... 1) Pd is the amount of voltage a component can take, with this... T pd across A = 4.4+ Ir (which is wrong) and T pd across B= 2.3 +Ir (which is correct) 2) terminal pd across a component is the (total emf - pd of that component) which gets me the answer to 2 but 3 becomes wrong.. aagghhhh