Resistors in Parallel - Why the Same Potential Drop ? (Electric Circuits, Physics)

[@PhysicsMadeEasy]

For more details on... What is an Electric Charge: kzbin.info/www/bejne/boexaYB6o5l2fKM What is an Electric Potential: kzbin.info/www/bejne/oGSqo4KBp8qSlZI

[@topjeeneethindi1064]

You are doing a great thing without even expecting something profitable out of it. No greed . Nothing . Consistent grinding. I am truly a fan of yours . Love you so much from India

[@PhysicsMadeEasy]

Wow, thank you so much for recognizing that! I deeply believe education should be for all, and totally free. My real-life students allow me to pay the bills, so the rest of my teaching time, I can dedicate to teaching others. Kind of a modest attempted contribution to make the world a more balanced place :-) ! Thanks again! Love from Europe.

[@topjeeneethindi1064]

@@PhysicsMadeEasy Please sir tell me this. I am a teenager now (17 years old) and after some time I'll be starting my company that is somewhat related to education. Can you please tell me how Education can essentially change the world. I mean taking you as an example why you want to spread you knowledge of physics to folks like me and others what do you see on the other side of the coin that our conventional methods of teaching in high school and even colleges cant see.Why education is important today as there are many ways to earn money rather than going to a college and stuff like that. What do you think is wrong with the education system now days. What is the single best change that must be introduced in our ways of learning and teachings that would revolutionalize the whole game for the learners.

[@dipuparihariya8157]

@@PhysicsMadeEasy pls upload video on why voltage gets distributed in series combination of resistor? I love watching your video 🤩🤩 you are doing great job.💖💖💖 lots of love.

[@avinavsony8159]

Really sir you have replied to the every questions in comments thats hats off to you. That shows how passionate you are towards your profession

[@PhysicsMadeEasy]

@@avinavsony8159 thank you very much Avina for recognizing this, yes, my passions are physics, music and teaching. And seeing students grow following my teachings fills me with joy.

[@nihalMaTh]

this is the video which serves the best explanation for this question, thanks a lot sir

[@PhysicsMadeEasy]

You are welcome Nihal, I am glad you enjoyed it!

[@MaaMeethecouple]

Omg, this concept gave me a headache when I tried to listen to the lecture of my professor. On another you explained this in an extremely easy way to understand. Thank you so much sir !!!!!

[@ulysses_grant]

It's amazing to hear a french physics teacher pronounce Coulomb's name. I finally can state that I know how to say the name of the man and that I had learned it from a 100% reliable source, lol! Thank you sir for your efforts and your work! Your teaching way is just staggering! God bless!

[@PhysicsMadeEasy]

Thank you Ulisses! I am glad you appreciate my work and that it helps you. As for the names of French Physicists, have you tried to pronounce "Louis De Broglie"? ;-)

[@ulysses_grant]

@@PhysicsMadeEasy Haha, no, I had never tried Broglie's name. Actually, I pronounce it with a Brazilian Portuguese accent, which results in a complete tragedy, lol 😂! But I'm quite sure you did say his name in one of your videos, I'll search for it, haha! Cheers!!

[@MelloCello7]

What is mechanically happening during a voltage drop? Theoretically, if voltage is a measure of potential energy solely reliant on position, then why does the amount of resistance in a circuit proportionally affect the amount of voltage drop? Theoretically any amount of resistance plugged in a 9 volt circuit should give you the exact same 9 V voltage drop. What does voltage drop really mean? Why and how does resistance affect voltage? Thank you for all of your hard work and dedication Dr. R

[@PhysicsMadeEasy]

Hi, The resistance does not impact the total potential drop of the circuit. This is set by the battery. The resistance impacts the current. Let me explain: Take a simple circuit with one battery (emf of 9V) and a resistor R. The battery provides 9 Joules to every charge that passes through it. Conservation of energy states that the energy that goes in the circuit, gets out. So for each Coulomb of charge that passes through the resistor, 9 Joules will be lost in the resistor. so there is a potential drop across the resistor of 9 Joules/Coulomb (9V) The key is to remember that whatever the resistance, the amount of energy lost in the resistance for each coulomb of charge remains the same. A high resistance for example, will limit the number of coulomb passing (current), but the energy lost per unit charge remains the same (9 Joules per coulomb). So, in the end, the resistance of a circuit modulates the amount of energy that transfers through that circuit by modulating the amount of charges that move through the circuit. If you have still trouble, think about: emf: amount of energy provided to the circuit by each Coulomb of charge Potential drop: amount of energy lost by the circuit by each Coulomb of charge Current: Amount of charge (in Coulombs) coulombs passing every second Resistance: limits the flow of charges I hope this helps!

[@jsivesh4323]

Thank you so much sir, Your explanations are very good and easily understandable

[@jsivesh4323]

@@PhysicsMadeEasy but i have one doubt in the above explanation sir. High resistance will limit the number of coulombs (current). 9 joules will be dissipated by every coloumb as heat. High resistance produces more heat but the number of charges losing 9J is less (as current is less) than a low resistance which will allow more current, so more coulombs will now dissipate 9J of energy as heat. But how can this be? Low resistance - low heat (energy ) right?

[@PhysicsMadeEasy]

@@jsivesh4323 “Low resistance - low heat (energy ) right?” Not necessarily… Still consider two resistors in parallel, thus having the same potential drop. The Power (Joules per second) dissipated is P = R x I^2. There will be more heat dissipated in the branch with the lower resistance because P increases as the square of I compared to P increasing linearly with R. Example: 9V across a 3 Ohm and a 6 Ohm resistance in parallel: In the 3 Ohms: I =V/R = 9/3 = 3A P = RI^2 = 3 * 9 = 27W In the 6 Ohms: I =V/R = 9/6 = 1.5A P = RI^2 = 6 * 2.25 = 13.5W (Note, you could also have done this using P = (V^2)/R)

[@jsivesh4323]

@@PhysicsMadeEasy thank you so much for the fast reply sir. I now understand that even low resistance can give more heat. I was thinking that if resistance increases the work done wil be more so more energy will dissipate but I forgot that with high resistance the current reduces. Thank you once again

[@Zivi-wl5sf]

I am from India and I literally had never studied the concept like this ever before

[@PhysicsMadeEasy]

Good Zivi, multiply your sources and approaches to an understanding, and that will allow you to truly master a concept;

[@riyasinha196]

sir you are really explaining well my teachers never thought me in this way this is really helping me in my exam thankyou so much

[@faisalzia481]

Thank you Dear Respected Sir..

[@Nikopaxton]

In last 2 months of my class 12 session I am extremely fortunate to discover this you tube channel

[@PhysicsMadeEasy]

Thank you Vishesh, I am glad my work helps you in your studies. You are passing your high school exams in two months ??? If so, have a look at the exam preparation playlist ont he channel. Good luck with your exams!

[@armypublic6386]

Very well explained☺

[@beatlebug2405]

Really nice explanations sir... Was finding it difficult toh understand this topic in detail.... Thank you

[@PhysicsMadeEasy]

Hi Beatle bug, I am glad my work was able to help you :-).

[@jessicafox4503]

This is the first time I actually understand this . TY!!!!! :)

[@PhysicsMadeEasy]

Hi Jessica, that's great! Now you should have an entire field of physics that opens to you: enjoy the discovery!

[@SaifAli-nj8fq]

Excellent explanation

[@asapstart5727]

Only video that explains that so effortlessly ❤

[@PhysicsMadeEasy]

Thanks Asap, many notions in physics are actually quite simple to get, if explained the right way!

[@murphy98969]

Wow what an explanation i understood this concept with as much certainity as i know 2+2=4

[@PhysicsMadeEasy]

Thank you for your feedback. I am glad my video crystallized your understanding :-)

[@theucovers4325]

Great Explanation Sir!

[@PhysicsMadeEasy]

Thank you Utkarsh

[@33ni]

Thank you.😊 You made the my doubt easy.

[@PhysicsMadeEasy]

Your doubt easy? or your doubt easily swept away? ;-)

[@laisandradesolar]

Perfect explanation!

[@isteakahamedimon]

Sir, when the voltage difference is 9V, 1C election will loss It's 9J energy. My question is : 1.Where the energy is transformed? Is it transformed into heat? 2. What does happen to the electrons after losing 9J. Do they loss their velocity? I know that the velocity remain same to keep the current constant across the circuit. So what is changed after 1C charge losing it’s 9J.

[@PhysicsMadeEasy]

Hello Insteak, 1/ Where the energy is transformed? Is it transformed into heat? It depends what you have as a load. If the load is made of resistors, yes, the energy "lost" by the circuit becomes heat. But if the load is made of light bulbs, it is converted mainly to light (and some heat), a loudspeaker, sound energy (waves), a motor, kinetic energy etc... 2. What does happen to the electrons after losing 9J. Do they loss their velocity? ... You are right in saying the KE of the electron remains the same even after passing through a resistor (the current remains constant across a series circuit). So where does this "lost" energy come from? It comes from the potential energy of the charge carried by the charge carriers. When the + terminal of a battery is at 9 Volts, it means that the energy needed to stack positive charges at that terminal is 9 Joule for each coulomb of charge stacked there. This is an energy of position, a potential energy. See it as the charges of the same sign stacked at this terminal repelling each other but they are forced to stay there until the circuit gets closed, and that the charges can flow away. This is the energy lost in the resistor When the charges come out of the resistor it is facing the negative side of the battery (say at a potential of 0V, meaning that there are no charges stacked there, and that the charge that arrive there are not repelled or repelling anything (in a battery you can buy in shops, these are actually neutralised in a chemical reaction.) I hope this helps your understanding! PS: In my answers, I am of course talking within the conventional model where we imagine that it is positive charges moving around carried by some kind of imaginary positively charged particles instead of electron. It is fully equivalent mathematically in terms of energy transfers.

[@KarmaDorje25]

Great explication 😎

[@yobabyyo]

Je viens de passer mon examen de professeur de physique hier et l'une de mes leçons portait sur le champ électrique. J'ai eu quelques problèmes avec mon approche didactique, en partie parce que je ne comprenais pas encore assez bien le sujet. Aujourd'hui, après avoir regardé quelques unes de vos vidéos, je commence à mieux le comprendre moi-même, ainsi que les concepts de potentiel électrique et de conservation de l'énergie dans un circuit - un autre point avec lequel je ne me suis jamais senti à l'aise à 100%. Votre vidéo m'a aidé à approfondir un peu plus ma propre compréhension ! J'aime vraiment la façon dont vous expliquez les concepts qui ne sont pas si intuitifs. Merci et continuez comme ça !

[@PhysicsMadeEasy]

Merci Adrian. Quand j’ai lu ton commentaire il y a qq jours, cela m’a fait ultra plaisir! L’objectif de ma chaine, c’est justement de faire ça, de débloquer des lacunes dans les bases et permettre aux étudiants d’avancer, et aussi d’inspirer les profs sur des manières d’expliquer des concepts que leurs élèves peuvent trouver un peu trop abstrait. En fait la physique est souvent enseignée de manière élitiste, comme si l’on cherchait à sélectionner ceux qui ont le droit de comprendre comment marche l’univers et de s’extasier. Et ça, cette discrimination-là, on n’en parle jamais, et elle m’énerve profondément. Je m’en suis rendu compte en étudiant différents programmes de terminale internationaux, et voyant comment les élèves réagissent. Certains programmes sont pédagogiquement catastrophiques (AP Americain), d’autres juste passable (A-Level Anglais) , d’autres, il y a un effort dans ce sens (International Baccalaureat). Pour tenter de palier a cette carence, je fais l’effort de chercher un angle qui peut parler aux élèves, avec leurs mots et leurs façon de penser, et la, même le plus faible d’entre eux deviens capable de re-expliquer clairement ce qu‘est un potentiel électrique ou un champ gravitationnel. Pour le programme français (première + terminale), Je n’ai pas regardé le détail, juste les gros titres, la structure. Ma première impression est donc peut-être erronée. Pour les ados français, je l’espère en tout cas: j’ai trouvé ça ultra sur-complexifié, ce qui est pédagogiquement complètement idiot. Toi qui a du étudié le programme en détail, tu en penses quoi ? Vu que je rentre en France d’ici quelques mois (après plus de 20 ans à l’étranger !), je vais probablement m’y attaquer en 2022 et 2023. Et sur le côté, peut-être que je créerai une version française de Physics Made Easy! Félicitation pour ton exam, et bon courage avec ta première classe !

[@PhysicsMadeEasy]

Bonjour Adrien, je suis rentre en France. Comme vous le savez peut etre, je suis tuteur prive en Physique, et j' aimerai me familiariser avec le programme Francais (1ere et terminale S, prep au bac). Pourriez-vous me conseiller un bon livre de classe Physique-Chimie qui soit a jour avec le programme de l'education nationale? Merci!

[@ummenaznin5469]

Sir, in your electric potential video you said charge goes from high potential to low potential then why at 2:16 minute of this video you draw an arrow with the meaning charge going from low potential to high potential???

[@PhysicsMadeEasy]

Hello Umme, the arrows do not represent the motion of the charges, they represent the change of energy of the charges while they move (in the same direction as the current). When they pass through the battery, they gain 9 Joules for each Coulomb, above the resistor, they lose 9 Joules for each Coulomb. I hope this clarifies things!

[@umarmuzammil7332]

love you sir for dymistifying the problem i am struggling with

[@PhysicsMadeEasy]

I am glad I was able to help Umar!

[@udaanite]

2:09 sir for every +ve integer R it losses same potential

[@PhysicsMadeEasy]

Yes, when they are in parallel :-).

[@surendrakverma555]

Very good 🙏🙏🙏🙏

[@learningisfun889]

Sir, thank so much!!

[@diyhemantprojects2841]

Great explanation 👍

[@anustupbiswas9955]

two voltage sources with emf v1 and v2 (v1≠v2) are connected in parallel, what will be the voltage difference in the common terminal points?

[@PhysicsMadeEasy]

Well, at first, during a transitory period, the source of highest voltage would charge the second one... A large current would occur in the cables connecting the batteries (it's value would be related to the equivalent resistances of both batteries): The battery of smallest voltage would see its voltage rise, and the reverse for the other. While the difference between voltages decreases, the current would decrease too until it reaches zero. At this point, both voltage would be the same. I do not recommend you try with standard batteries... it would be like trying to charge a non rechargeable battery (risk of explosion)... and the max current at the beginning could lead the cables to melt, and the insulator to catch fire. So keep it a thought experiment :-)

[@riyasinha196]

sir plz make a video on how to solve numericals in physics

[@PhysicsMadeEasy]

Hello Riya, maybe you can check the exam preparation playlist. there are many examples of physics problem that I analyse and solve on the board: kzbin.info/aero/PLU0ETLdKNmc6jnVvy3vwcg4tg0bwcEBWA

[@riyasinha196]

@@PhysicsMadeEasy thank you so much sir

[@______4774]

Hlo sir Awesome explaination

[@durjoysen3866]

Thank you sir

[@DonMascarpon]

THANK YOU SO MUCH!!! 😁😁

[@hardpanchal2821]

Thank You So Much Really Helped Me a Lot TYSM

[@jsivesh4323]

Sir, according to V=IR, if R1 in 1st branch is greater than R2 in second branch. Say R1=5ohm and R2=2ohm , won't V=I1× R1 be greater than V= I2×R2 ? Then the potential difference won't be equal. Can you please clear my doubt sir

[@PhysicsMadeEasy]

If R1 > R2, that means that it will be more difficult for current to go through R1, therefore I1

[@piesmhh1833]

Could you please explain how the current remains the same in a series circuit?

[@PhysicsMadeEasy]

Hi Piesmmh, First, two things to know: Current is the amount of charge passing by per second. Conservation of charge: charges cannot be created nor destroyed. So if Q charges come out of the battery in a second, that means that Q charges need to go back in the battery every second… The current in needs to be the current out. If there are no branches in the circuit (series circuit), at the input and output of any component or cable cross-section, conservation of charge applies too. Whatever comes in comes out. That is why a current has to be the same anywhere in a series circuit.

[@piesmhh1833]

@@PhysicsMadeEasy ohhhhhh thank you so much sir i finally understood this concept!

[@ashleyhemanth9997]

Electric potential depends on position right? So why is it that if a wire has low resistance, then it conserves electric potential? The more closer you are to the negative terminal, the more the potential decreases right? How does electric potential depend on resistance?

[@PhysicsMadeEasy]

Hello Ashley, Connect the extremities of a wire to the two poles of a battery (in other words, build a short circuit). The wire has some resistance, so the potential will decrease progressively from the + to the -. If the wire is uniform, the electric potential will decrease proportionally to the fraction of the wire already covered. So you see, the electric potential is only dependent on the position. The resistance of the wire is very small, so you will get a huge current, and consequently the power dissipated may be enough to melt the wire... Plug in a resistor in the circuit (in series). The resistance of the resistor is huge in comparison with that of the wire. The resistance of the wire can be neglected. This approximation allows to consider the wire as being an equipotential. (but in reality, it is not...it is just that the decrease in potential in the wire is small enough to be considered as negligible)

[@ashleyhemanth9997]

@@PhysicsMadeEasy "The wire has some resistance, so the potential will decrease progressively from the + to the -" Even if there is no resistance, the potential will decrease right? Why is resistance a necessary condition?

[@PhysicsMadeEasy]

​@@ashleyhemanth9997 Hi Ashley, When you say ‘no resistance’, and mean ‘infinite resistance’, then yes, the potential will decrease progressively with position. If you mean zero resistance, then you deal with a transitory period where the two potentials of the battery are going to tend to equate. However, zero resistance doesn’t make sense in that context: The huge current you will get will heat up the wire, which will lead to an increase in resistance… Besides, there is no zero resistance conductor that exist (in standard température conditions above a critical temperature Tc under which electron coupling processes like BCS take place). Then, there is superconductivity (for T

[@iankelly49]

Thank you for this video, here's my question, total novice I'm afraid. I have a 9v battery, with a resistor to stop my LED's from burning out. I have 2 LEDs in parallel. If I use 2 LED's with similar voltage requirements (around 2.8v) they both light up, if I exchange one LED for an LED with a lower voltage requirement, around 1.5v, only the lower one lights up. Why? Many thanks

[@PhysicsMadeEasy]

Hi Ian, At the moment you switch on the circuit, the voltage across the LEDs increases until it reaches a 'stable' point where current can flow. The first stable point is at 1.5V across the LEDS (the other 7.5V being for the resistor). Now current flows and the LED with lower voltage requirement lights up: the circuit is in a stationary state. The 2.8V LED, being in parallel with the 1.5V one, is thus submitted to 1.5V which is not enough to make current go through it (Check the VI diagram for that LED). It remains turned off. When both LEDS require 2.8V, the first stable state for the circuits is when there is a voltage of 2.8V across the LEDs, so both lit up.

[@ayushkhanal36]

Hey sir , can you tell me why does we feel cold , if we turn on the fan although fan increases the speed of molecules around us ??

[@PhysicsMadeEasy]

because the wind from the fan increases the rate of evaporation of sweat. When water passes from liquid to Gas, it sucks energy from the surroundings. The human body uses this phenomena: when sweat evaporates, it cools the skin. That's why we sweat in hot weather and need to drink a lot to keep sweating! As for the increase of speed of the fan: the effect is negligible compared to the speed of a molecule has already at ambiant temperature. For example a molecule of oxygen at 25C is moving already at speeds around 500 m/s (1800 km/h).

[@ayushkhanal36]

@@PhysicsMadeEasy But , my teacher told me when fan is turned on , the air molecule strikes our body and take some heat with it , as a result we feel cold . Is he correct ??

[@PhysicsMadeEasy]

@@ayushkhanal36 Hello Ayush, I don't think so... Wind (or air from a fan) allows to increase the rate of evaporation, so our sweat evaporates faster, which uses heat from the skin/body to do so (The latent heat of vaporisation of water is large, that's why it is so effective). And that's why we feel colder in presence of an air flow. In a way, one can consider that the 'air molecules' take the heat away by transporting with them the vapor produced by the evaporation of the sweat. That's maybe what your teacher meant?

[@shiekh366]

Sir why and how does electron gain energy as they move across the battery

[@PhysicsMadeEasy]

Actually, it is not rigorously the same electron that enters the battery and that comes out of it. But due to conservation of charge, any quantity of charge coming in, comes out. So it is a useful analogy to see the battery like a petrol station and the electron as a car, where it is the same electron (car) that gains energy (petrol).

[@ummenaznin5469]

Sir, which country are you from??Your teaching method is eye popping😍😍

[@PhysicsMadeEasy]

Eye Popping? haha, first time I hear this about my teaching approach lol! I am French.

[@TechToppers]

Not really great with electronics,but if the resistance literally takes all the energy of the charges, how do charges move after passing through the resistance?

[@PhysicsMadeEasy]

High Techtoppers. Your question shows a confusion: Charges are carried by charge carriers. The charge carriers move, thus posses kinetic energy energy of motion. The charges themselves carry electrical potential energy... The energy dissipated by the resistor is provided by the charges, not the charge carrier. This is why the drift velocity of electrons in a closed loop is constant (the current is constant all around a series circuit), but the potential energy of the charges decreases when energy is dissipated. As a very approximative image you can imagine a car with a passenger that has a credit card in his pocket with money on it... The car is the electron, the passenger is the charge, and what's on the passenger's credit card, the money, is the electrical potential energy. When passing through an automated motorway paying boot, one that just detects the car when you pass by it (= the electron passing through a resistance), the money on the card (electrical potential energy ) of the passenger (charge) is automatically removed from the card to pay the motorway fee (The electrical potential energy is dissipated). Yet the car is still moving.

[@irreproachableincorrigable1378]

Hi , At the beginning of the minute 2 you mentioned that there is a potential drop across the resistor which means resistor absorbs energy of (9 V) that is carried by the flow. At the minute 2.40 you draw 2 resistors in parallel . Hence there is a flow across both parallel paths, each resistor is supposed to absorb equal amount of energy (9V +9V=18v ). in this case energy emission by battery is 9V and energy absorption by resistors is 18 V in total. This is violation of energy conservation in this circuit. how to overcome this discrepancy? I haven't got a proper explanation and not comprehended a solution so far. Thank You

[@PhysicsMadeEasy]

Hi, The answer to your question lies in an accurate understanding of what is an electric potential (and Voltage, which is the difference of the electric potentials between two points). 1 Volt is NOT an energy, but at energy per unit charge… That is a crucial point to understand… When there is a potential drop of V Volts across a resistor, it means that each coulomb of charge that passes through the resistor loses V Joules. of a energy. When Q coulombs pass through the resitor, it results in a dissipation of Q * V Joules of energy. Illustration: A single resistor with a potential drop V across it will lose, V (Volts) * Q (Coulombs) in energy, when a charge of Q coulomb passes through it. If you replace it by 2 resistors in parallel, thus with the same potential drop V, the amount of charge will be split in 2 so that Q = Q1 + Q2 (Conservation of charge). The first resistor will see Q1 coulombs passing through it, and the second will see Q2 coulombs. The energy dissipated in the first one will be V * Q1, and in the second one, V * Q2 The total energy dissipated will therefore be V*Q1 + V*Q2 = V*(Q1 + Q2) = V*Q. This is the same amount of energy dissipated than when there was a single resistor… I recommend you check this video, to clarify your understanding of an electric potential and a voltage. It should be useful to you. kzbin.info/www/bejne/oGSqo4KBp8qSlZI Be well,

[@irreproachableincorrigable1378]

@@PhysicsMadeEasy Thank you a lot for your quick and fruitful response. Now it is time to scrutinize the details

[@ankitthakurankit4764]

Why is total resistance in a parrallel circuit less than any individual resistance of the same??

[@PhysicsMadeEasy]

because by putting two resistance in parallel, you actually increase the number of possible paths an electron can take. Remember that the R = resistivity of material x length of resistor / cross section of resistor. That means that by replacing a resistor R by two resistors R in parallel you double the cross section... It's like increasing the diameter of a canalisation where water passes through.

[@ankitthakurankit4764]

@@PhysicsMadeEasy one more question . By H=I^2rt resistance is directly proportional to heat but using H=Vt/r resistance is inversely proportional to heat can you clarify these two contradictory statements.

[@PhysicsMadeEasy]

@@ankitthakurankit4764 heat delivered by a resistor r where a current I flows during a time t is H =I^2rt. The potential difference of this resistor is V = rI, thus I = V/r. Substitute I for V. H = (V/r)^2rt = V^2rt/r^2=V^2t/r. (don't forget the square on V...). You must make sure before you use the word proportional that the proportionality 'constant' remains independent or is set to be independent . Here I depended on r...

[@shiekh366]

Sir if the battery provides 10 joules of energy per coloumb how can there be loss of 10 joules of energy in all resistors doesn't it violate the law of conservation of energy

[@PhysicsMadeEasy]

In both resistors in parallel, the loss is not 10 joules, but 10 Joules per coulomb passing through it. Practical example: Suppose the battery lets 1 Coulomb of charge flow in the circuit, and provides via these charges 10 Joules of energy to the circuit (The battery as a voltage of 10/1 = 10 volts). When the 1 Coulomb arrives to the junction before the two identical resistor, one half flows in the first resistor, and the other half in the second resistor, each half delivering half of the energy to the resistor (5 Joules). In each resistor , a half coulomb lost 5 Joules, that is 5/0.5 = 10 Joules/Coulomb (10 Volts).

[@jackbidds7082]

Say you’ve got a 9V battery, how can both resistors drop 9V. That’s 18v in total with a 9V supply isn’t that breaking the conservation of energy.

[@PhysicsMadeEasy]

Hello Jack, A Volt is not an unit of energy, but of energy per unit charge... 9V means 9 Joules per Coulomb... When two resistors are in parallel, the current splits in two: there is less amount of charge that passes through each resistor, but there is still a loss of 9 Joules per Coulomb across each resistor. Reflect on what a volt is (1V = 1 joule per coulomb) and everything will make sense :-) I know the concept of potential and voltage confuses many students, so I made a video about this. You should check it out: kzbin.info/www/bejne/oGSqo4KBp8qSlZI Be well

[@pankajkumarpradhan6333]

Sir why there's no induced current generated in a coil when light rays falls on it ? As we know that light is an electromagnetic wave consisting oscillating magnetic and electric fields.. It should induce current in a coil due to oscillating magnetic field in accordance with faraday's law of electromagnetic induction.. but it doesn't happen so . Why ??

[@PhysicsMadeEasy]

Hi Pankaj, light does induce a current when hitting a material: this current is just every small. Charges in the coil will experience an oscillating force under the influence of the oscillating EM field (oscillating dipoles form). Because of the very high frequency of the light, the amplitude of the oscillation of the charges will be very small: the AC current will just be of small amplitude, thus tiny… (note that this answer is given under the perspective of classical physics...). That is the classical view. For a quantum view: in a material, there are bands of energy that correspond to all possible energies that an electron can have. When an electron receives energy (heat), it will jump to a higher level of energy in the allowed energy bands. When it relaxes, the electron emits a photon…

[@pankajkumarpradhan6333]

@@PhysicsMadeEasy Thank you sir . You have cleared my doubts... but can you suggest me some books for physics from where you came to know about this thing . And sir please suggest some books to make our concepts crystal clear in physics..

[@PhysicsMadeEasy]

@@pankajkumarpradhan6333 Hello Pankaj, hummm, To be honest, I do not know of such a resource. All the text books I know are often quite elitist, (and this is something I fight: science is for all!), and when they are not, they simplify the wrong things, making things actually incorrect which brings much of confusion in students minds. I have witnessed this by myself so many time, with my students, with having to correct the damage that was done by these books. I fight this, this is why maybe I am quite successful as a teacher: in my career, I encountered some students which had light mental handicaps, and I helped them pass some prestigious high school exams, like IB HL Physics. For me, that proves that anyone can succeed! So I can’t really propose any books for you, maybe the best compromise between accuracy and simplicity would be the Tsokos, IB Physics, (Ed: Cambridge), that I kind of like. But I do because I am already familiar with the concepts: some of my students do not like it. The best I can suggest is that you go to a library, read on various topics you already know, and think to yourself: “Would I have understood this if this was the first time I read it…”. The book for which you answer yes the most often, is the one you need to buy.

[@pankajkumarpradhan6333]

@@PhysicsMadeEasy Thank You Sir for your guidance.. Sir I'm From India and you are from which country?? SIR NOW I'M IN CLASS 12TH( i.e in India we call the 2nd year of intermediate studies which is afterpassing class 10th as class 12th.).. AND SIR I WANT TO DO RESEARCH IN PHYSICS.. SO SIR CAN YOU TELL ME THE LIST OF BOOKS YOU STUDIED IN YOUR UNDERGRADUATE & GRADUATE YEARS.. IT WILL BE VERY HELPFUL FOR ME IF YOU GIVE ME THE LIST.. AND IN FUTURE I WILL TRY TO MEET YOU ONCE.

[@PhysicsMadeEasy]

​ @Pankaj Kumar Pradhan Hello again Pankaj, During my studies, I didn’t use books outside of what was given to me: no money to buy them, no guidance, no Internet… I struggled a lot in high school but still managed to pass with very average grades (50% was needed, I got 50,2%!). I went to a low class University, but I was very attentive in class, worked a lot on the material the teachers gave me, and when I didn’t understand something would work with my friends, and go to the University library to try and figure out what I didn’t understand from class. Because of this approach, my grades rose, and I was able to choose a Master’s studies. I took one which was not very popular, still of low prestige , but it had topics I enjoyed, and it was a small class (24 students): I chose this format because I could be in contact with the professors more easily. During that phase, I approached myself from the Research community by trying and getting summer placements / work in the local laboratories. That was also a good move. I developed my network, my marks skyrocketed, and I was offered to go up to a PhD with a great topic , then a post doc and published a lot. I made myself an international name in my field. For that, I didn’t hesitate to leave France (I’m French). What I am trying to say by telling you my experience, is that you have to make your path for yourself. So mot always follow what people expect from you. We all have a path that fits us. It’s up to each of us to find it. There is a place for you in scientific research if you really want to become a researcher, no doubt, but it’s up to you to find this road, unique to you, that will bring you there. Just be pro-active, do not follow the crowd systematically (if you have average marks, don’t aim for the most prestigious universities, it would just hurt you). Be honest with yourself, always make your mid-term goals achievable, and you will reach your final one. I wish you the best of luck.

[@equinoxxx5453]

My question is If electrons are flowing because of voltage . Then won't electrons flow after the point of passing resistor ? There is 0 voltage

[@PhysicsMadeEasy]

Imagine you have a system that releases heavy balls from a certain height ( = a potential difference). The balls arrive in a large bucket. At the base of the bucket, there is a hole that can allow a ball to pass through. At first the balls would fall in the bucket and stay there. But after some time, once the base of the buckets is filled up with balls, the next ball falling will push one of the balls already in the bucket through the hole (the negative side of the battery). If you want a better understanding of that analogy and how it answers your question, check my video "What is an electric potential).

[@godisgood2482]

I have a very much confusion how does battery provide potential difference i am not able to understand current electricity suppose there is a 12 volt battery so i mean potential difference is 12v so does inside the battery something happens and it gives a magnitude of charge q??? And that q amount of charge flow through a conductor?? Is it always that if there is a battery of 12volt will it give same amount of charge?? I am not talking about current i am talking about just charge magnitude of q And same if there is 2volt battery then of how much magnitude of charge(meanshow many electrons it will derive) ?????

[@PhysicsMadeEasy]

Hi, If you understand what an electric potential is, all your questions will find an answer. Check this video: kzbin.info/www/bejne/oGSqo4KBp8qSlZI The “12 Volts” of a battery does not tell you how much charge you get until it is depleted. It tells you how much energy one unit of charge will have. The amount of charge that can be provided by a battery simply depends on how big it is. For example for chemical batteries, that is how much reactant there is inside. You can have two batteries, one of 1.5V and another of 48V, of the same size. In a way, you can see the potential difference indicates how much each charge will ‘feel the need’ to move from one terminal to the other. Both will provide as many charges, but the 48V will provide more energy. I hoep this helps!

[@himalaysharma8]

hi, I have one Question - , When I am connecting a 1000Ohm resistor to single LED in series connection with 5V input supply. the voltage drop across the resistor is 2.14v and for single LED is 2.68v , but when I add one more same type of LED in series connection , the voltage drop reading are for resistor .10v for LED 1. 2.48v for LED 2 .2.48v please explain.?

[@PhysicsMadeEasy]

LEDs are not ohmic components. They have to respect a certain VI diagram to function. The current passing through the circuit will adjust itself to their functionning mode, that is why the potential drop at your resistor changes so drastically.

[@tanujsharma2812]

I have a doubt In case of series connection ε = v1+v2 Why in case of parallel connections ε is not equal to sum of all the potential drops across all the resistors?

[@PhysicsMadeEasy]

Hi Tanuj, In a circuit where two resistors R1 and R2 are in parallel to a cell of emf Epsilon, Epsilon = V1 = V2. You can prove this by two ways: 1/ By using the loop law you find, Epsilon = V1 and Epsilon = V2 2/ By considering the electric potential at each point of the circuit. Remember a voltage (= potential drop or emf) is a potential DIFFERENCE. Because cables are equipotentials (the potential is the same along the cable), you can realise that before the resistors the potential is Epsilon for both resistors, and after it is 0. So both potential drops across the resistors are Epsilon - 0 = Epsilon, thus they are the same for both resistors. I hope this lifts your doubt!

[@AmanSingh-fl4lh]

@@PhysicsMadeEasy does emf of battery always equal to voltage,in parallel connection.?

[@anuragkaushish5992]

Good morning sir It's a humble request to please make a video on derivation of Kirchhoff voltage law by energy conservation method

[@PhysicsMadeEasy]

Hello Anurag, great suggestion, Duly noted!

[@faisal7650]

Thank u sir

[@furqankhan3741]

Thanks Sir

[@andrew-br6et]

Lose energy in form of what? When it passes through a resistor at 2:24 please explain properly????

[@PhysicsMadeEasy]

The Energy of the charge is transferred to the structure of the resistance material which then starts to shake. In other words the Electrical PE of the charge decrease in favor of the kinetic energy of the particles that form the resistor. Temperature is a measure of the average KE of the particles of a system. So the T of the resistor increases above that of the environment it is in. That leads to heat going from the resistor to the surroundings. We say that the energy is dissipated (it heats up the surroundings).

[@andrew-br6et]

If at the negative terminal side of the battery we say it has lost all energy and it is 0 joule then why does the charges move then shouldn't they stop there 0 joules energy means charge should not move please explain I am confused 😢

[@PhysicsMadeEasy]

Hi, The 0 joules refer to the electrical potential energy of the charge. Do not confuse this with the kinetic energy of the charge carrier. Please note though, to be precise, that when we refer as a voltage of say 9V, it means that the electrical PE of a charge that goes from one pole to the other changes by 9 Joules for each Coulomb of charge. So you could have 9J/C at the + terminal and 0J/C at the - terminal , but you could also have , for example, 11J/C at the + terminal and +2J/C at the - terminal... I hope this clears things up

[@ummenaznin5469]

@@PhysicsMadeEasy then what happens to that kinetic energy of charge when it finally arrive to its destination.Sir,please give reply.

[@pragyanjitozahphukon7201]

Please explain the concept of how force acts in a forced oscillator as well as how the graph of force and time will look🙏🏼

[@PhysicsMadeEasy]

If you draw the amplitude of the oscillation as function of the frequency of the applied force, You will get a resonance graph... This might deserve a video one day... but not right now... But I do not think it answers your question. If you could formulate it in less general terms, I maybe could answer...

[@pragyanjitozahphukon7201]

@@PhysicsMadeEasy Thank you so much Sir for replying , I just want to see a animation of how applied force works, and how this force is periodic 🙏

[@naveenkumarbasiri4850]

What happens when I introduce a battery in the middle of wire, in the video 1:42

[@PhysicsMadeEasy]

You mean where normally you would place a resistance? If the positive side of the battery you place there faces the positive side of the first battery. then one of them would be charging the other. Otherwise, this would be equivalent to one battery of emf equal to the sum of the emfs of the two batteries, and the circuit would be a short circuit.

[@naveenkumarbasiri4850]

@@PhysicsMadeEasy thank you sir for your response....

[@saravanasiva1804]

Sir,please explain alternating current

[@PhysicsMadeEasy]

Hi Saravana: when a DC current flows through a resistor (and energy is released out of the circuit), it is due to + charges pushed by the + side of the battery (or if you prefer to think with electrons, the - charges they carry pushed by the - side). Imagine that every second you flipped the battery in the circuit (replacing the + terminal by the - terminal and vice versa). The charges would be pushed one way , and then after the flip, pushed the other way… they would end up oscillating, and some would pass through the resistor continuously. That is an AC current. Charges are oscillating around the same position and continuously gaining and losing energy, Does this answer your question?

[@kacgibibiter6640]

thx veryu much

[@PhysicsMadeEasy]

You are welcome Umutcan?

[@faisal7650]

Sir, we say that emf is the maximum potential difference between two terminals of a cell when no current is drawn from it, the answer is that potential drop occurs but how. What is potential drop, explicit it sir.

[@faisal7650]

Is there any relation between charge separation and potential difference

[@PhysicsMadeEasy]

"Sir, we say that emf is the maximum potential difference between two terminals of a cell when no current is drawn from it, the answer is that potential drop occurs but how. What is potential drop, explicit it sir." Because you have to consider internal resistance in a battery. A battery is made of a cell of constant emf, E, and internal resistance that can be modelled by a little resistor, r, in series with the cell. The terminal potential difference (across the full battery) is thus: V = E - RI. When the battery is not connected to a circuit, it doesn't deliver current, you see that V = E - 0 = E. If it is connected to a circuit, some current will be delivered by the battery, so you see rI is not zero anymore. therefore V = E - rI < E.

[@PhysicsMadeEasy]

" Is there any relation between charge separation and potential difference" Absolutily. You have to understand first what is an electric potential. I encourage you to check this video (What is an electric potential) kzbin.info/www/bejne/oGSqo4KBp8qSlZI. When charges are packed at a position, the electric potential around that position is high. so when you compare with a position that has no charges (where the potential is zero) the potential difference between these two points is high. Does this answer your question? or did you mean something else?

[@ultimatewaybigx7685]

For voltage drop to be the same across all the resistors wouldn't that be assuming that they are all the same amount of ohms meaning the all take the same amount of joules from a charge. And also voltage drop doesn't have to mean that the end voltage will be 0 because the voltage the battery will supply lets say is 9v and then the resistor takes 7v. This means voltage drop will be 2v.

[@PhysicsMadeEasy]

Hi, “For voltage drop to be the same across all the resistors wouldn't that be assuming that they are all the same amount of ohms meaning the all take the same amount of joules from a charge. “ You are forgetting current. The current represents the amount of charges passing per second… The voltage is the number of joules lost per Coulomb passing across the voltage drop. So there is no problem with having the same voltage drop across parallel resistors: a small current and high resistance or a high current and low resistance can provide the same amount of joules lost per coulomb. “And also voltage drop doesn't have to mean that the end voltage will be 0 because the voltage the battery will supply lets say is 9v and then the resistor takes 7v. This means voltage drop will be 2v.” Assuming there is not other resistors in series or an internal resistance to the battery, that means the battery would have an emf of 7V (+9V on the positive terminal, and +2V on the negative one). Which is totally possible. A comment: be careful when you say “the end voltage will be 0”. A voltage is apotential difference, you should say “the potential at the other side is 0V”… I hope this helps!

[@25cmx23m]

shoutout to you gang, I hope your wife cooked up a hot plate for you, you deserve it bro

[@PhysicsMadeEasy]

Hummm it was nice :-)

[@TH-kn9lw]

In parallel the charges loose 9 V of potential so technically the total potential drop should be 18 should it not?

[@PhysicsMadeEasy]

9V means 9 Joules per unit charge (per Coulomb)... When two resistors are in parallel, the current splits in two: there is less amount of charge that passes through each resistor, but there is still a loss of 9 Joules per Coulomb across each resistor. Reflect on what a volt is (1V = 1 joule per coulomb) and everything will make sense :-)

[@TH-kn9lw]

@@PhysicsMadeEasy no but when voltmeters are connected both show 9V so that technically means a total of 18 V drop

[@PhysicsMadeEasy]

@@TH-kn9lw If you connect the voltmeter before and after where the cables join, you will also see 9V... You will also find 9V if you connect your voltmeter before and after a single resistor. (Again, I encourage you to reflect on the meaning of voltage: 9V = 9 J /C)

[@mahalakshmibalaji1842]

why does current decreases when voltage decreases but sometimes it increases when voltage decreases? thank you for this video!!

[@PhysicsMadeEasy]

Hi Mahalakshmi, V= RI, so V across the resistor and I flowing through it evolve the same way... Can you be more specific by giving me an example (or referring to somewhere in the video?

[@SaifAli-nj8fq]

But you gave me an intuition

[@PhysicsMadeEasy]

Glad I could help :-)

[@AmanSingh-fl4lh]

Owesome

[@puskarkumarray4694]

When current(electrons) gets divided in a parallel circuit, then energy held by electrons should also divide, which means potential drop across each resistor should also be different 🤔🤔 But this doesn't happen, why?

[@PhysicsMadeEasy]

If the current is split in two, yes the number of electrons passing through will split also, therefore the energy dissipated in the resistor too... But a potential drop across a resistor is not the total energy lost in the resistor but the energy lost FOR ONE UNIT OF CHARGE. Roughly you can see the potential drop as a value proportional to the energy lost by ONE electron... That value will not change even if you have less electrons passing through. What changes when you split the current is that you have less electrons passing through per resistor., that's all. I recommend you check this video explaining what is an electric potential (thus a potential drop): kzbin.info/www/bejne/oGSqo4KBp8qSlZI I hope this helps

[@sriramsriram9144]

@@PhysicsMadeEasy We can also say that voltage (joule per C) is the pressure / the energy each coulomb of electron holds and so, how much ever coulomb of electron (when it comes to count of electrons which means the current (q/t)) flows, it will possess the same energy irrespective of the no. of electrons, which is nothing but the current!. Hence voltage remains the same and current is different in parallel. When series, it's Vice Versa....

[@SaifAli-nj8fq]

All others gave me a vague meaning

[@rajivkumarsharma2201]

👍

[@domainmojo2162]

I still do not understand. Now I'm even more confused.

[@PhysicsMadeEasy]

Can you tell me what confuses you?