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We know that the electric field inside resistor produced by battery is the actual thing that pushes the electrons in it to move and thus producing some current. We also know that strength of electric field can be described by the electric flux which is the total amount of fields passing through an area. That flux when combined(multiplied) with charge of electron gives the amount of force on it. Now comes our two resistors R1 and R2. Assuming a number of field lines coming out of a terminal going towards another terminal through these resistors, the field lines should get divided into the resistances R1 and R2. Therefore giving a reduced flux through each resistor than in a series. Which will then give a lower voltage for both resistors.

Wasn't the electric fields strength divided across resistors in parallel?

What is the main cause of voltage? How electrons get a force to produce current even when the wire is very long in? I think it is better to understand with reality than analogy. Please don't give misleading pictures

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Why is voltage same for the two resistors connected parallel in circuit?

If there are no time-varying magnetic fields involved, voltage difference between two points is path independent. It is like altitude on a mountain. Suppose you are on top of a mountain, and there are different paths from the top to your car in a parking lot. The distance you travel will be different depending on your path. But the total change in your altitude will be the same regardless of which path you take. [There is a mathematical formalism to explain this. If you are interested you can research "conservative fields". One of the basic laws of electromagnetism states that when there are no time-varying magnetic fields, the electric field is conservative.]

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  • $\begingroup$ @Maths Keeps Me Busy, #adding voltage in parallel# you are right in saying the height remains same that's potential difference is same. But in electricity V = E×l . Then changing the source charge can change E. So, when I connect two positive terminals and two negative terminals. The total amount of charge should increase. On terminal the charge density is same as it was when we used only one battery. But the electric field from both both terminals in our case is passing through same resistor. Therefore, field density must increase and so the voltage. $\endgroup$ May 26, 2021 at 8:12
  • $\begingroup$ You write $V = E \times I$. I'm not sure what you mean. Ohm's law says $V = R \times I$ where V is voltage drop, R is Resistance, and I is current. Watts law says $P = V \times l$, where P is Power. When E is the electric field, and when it is conservative, $E = \nabla V$. But I don't know what you mean by $V = E \times I$ $\endgroup$ May 26, 2021 at 10:58
  • $\begingroup$ It was " L" length of conductor! Sorry 'I' current and 'I' lentgh looks the same $\endgroup$ May 27, 2021 at 7:38
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Voltage in a circuit is defined in terms of energy per charge (J/C). So for simplicity, imagine the simplest parallel circuit with one battery and two branches with a resistor each. Now imagine electrons flowing around this circuit. The battery gives each electron the same amount of energy. So when the electrons leave the battery and reach the first fork in the road, some electrons go down one path, while the rest go the second route. But each electron still has the same amount of energy attributed to it. So the energy per charge in each branch is the same as it was at the battery. But that’s what voltage is, and so voltage must get copied into each branch at a junction.

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    $\begingroup$ Means that the energy required to put a unit charge in battery is the voltage of it. When we double the terminal, the density of charge remains same and we the same use amount of energy as it was needed for putting the charge in battery. But still no one is talking about electric fields which is the main thing here! $\endgroup$ May 26, 2021 at 8:17
  • $\begingroup$ @amazonpime but when we add batteries we are actually doubling the terminal and therefore the total charge it(assuming charge density is same). The double amount of charge says that the potential at terminal will be double as V is proportional to E and E is proportional to Q(charge on source). This should double the potential difference too. Am I not right in saying this. $\endgroup$ May 31, 2021 at 8:05
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enter image description here

The image is from the website https://www.allaboutcircuits.com/technical-articles/resistors-in-parallel-understanding-current-and-voltage-in-parallel-networks/. This image shows two resistors connected in parallel. There are a few things to note about this circuit configuration:

  1. Assuming that the current flow is from "A" to "B", note that the "upstream" sides of both resistors are at the same voltage because they are connected together.

  2. Note that the "downstream" sides of both resistors are at the same voltage (but different than the upstream sides of the resistors) because they are connected together.

  3. Voltage for resistors actually means voltage drop across the resistors, so "the voltage is the same for both resistors" means that the voltage drop across both resistors is the same.

Since the upstream voltage is the same for both resistors and the downstream voltage is the same for both resistors, the voltage drop across these resistors must be equal, regardless of the values of R1 and R2.

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  • $\begingroup$ I mean if I divide the electric fields in two resistors, the drop of voltage will also become half not the same. I am trying to understand the phenomena with what is actually going on their not the height. Can make a explanation by using Electric fields. E×l = V which we know if we change E , V should also change. What are the possible ways to change E without changing l(length). $\endgroup$ May 26, 2021 at 8:28
  • $\begingroup$ @PredakingAskboss, there is no length or height inferred in the picture, and the electric field is not "divided" between two resistors. There appears to be a hidden assumption in your question, but I can't determine what that assumption is at this point. $\endgroup$ May 26, 2021 at 12:28
  • $\begingroup$ The voltage always remains the same no matter the resistance. The only thing that changes is the current. Think of 2 light bulbs tied to a battery. Both light bulbs feel the same voltage. There is nothing inside, outside or dependent with the resistors. They are simply a load on the voltage. $\endgroup$
    – Rick
    May 27, 2021 at 1:04
  • $\begingroup$ @Rick Why connecting terminals of battries to resistor creates voltage/potential difference across resistor? $\endgroup$ May 27, 2021 at 7:24
  • $\begingroup$ @David White Yes! That was my assumption because I know that i am not getting the solution with the way I've tried but I need to tell you how I get to this question, simply by making an assumption that electric field are coming out of terminals of battries and passes through resistor and because there is fixed number of field we can't increase the field lines. Is assuming that electric field comes out from terminals and passes through circuit to give push on electrons, completely wrong? $\endgroup$ May 27, 2021 at 7:35
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The simplistic answer is they share the voltage.

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  • $\begingroup$ Why do they share? And if the share voltage then why the voltage doesn't decrease fro each. Don't say that they both are in contact with the same thing. I mean can you please explain it with electric fields inside resistor which is the actual thing going on. I have read a lot of analogies but now i want to know the correct thing in correct way . Use electric fields to explain them. $\endgroup$ May 26, 2021 at 8:24
  • $\begingroup$ they share the voltage simply because there is only a single voltage drop is a parallel circuit. The 2 resistors have a resistance of Rt = 1/(R1 + R2) giving you a current of I=E/Rt. However the voltage is simply the voltage $\endgroup$
    – Rick
    May 27, 2021 at 1:09
  • $\begingroup$ So what do think why electrons even move? Is it just because they have a potential difference? But we need a force to accelerate these electrons. Saying Potential difference doesn't explain what is actually applying the force and how much it is. In the case of water analogy the gravitational field of earth provides force on water but in our case it seems to electric field of battery(maybe). We also know that its strength is dependent on amount of charge then, why increasing charge(here: doubling (+) and (-)terminal by adding battries in parallel) doesn't increase field strength. Error! $\endgroup$ May 27, 2021 at 7:21
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This question is more frequent than imagined. It is assumed that resistors or electronic components connected to the same two nodes have the same potential difference. It may seem obvious. But why does this happen? Let's prove the obvious! :)

The question then changes to: why connecting components (for example R1, R2 and R3) to two nodes A and B through a perfect and ideal conductor means bringing the potential of point A and point B to these components?

These doubts are due to the way in which certain quantities are explained (the electrons that "have" a potential energy) compared to the fact that we lose focus on what these quantities are.

Let's tkink crazy! We reason by absurdity.

If the potentials were perhaps different, for example in the electrons "outgoing" from the resistors, we would each have them with a certain "electric potential energy" still to be "spent" and in a different quantity for each branch converging towards B in which they are found. We also know that on the conductors carried by current there are surface charges that generate the electric field useful inside them to push the electrons against the resistivity: in perfect conductors this resistivity is zero so the electric field will not be generated; in fact, the surface charges will naturally arrange themselves so as not to have an electric field (and potential difference) along the path. However, this also means allowing the surface charges placed on the three branches to arrange themselves so as not to have a net electric field in any part of the (perfect) conductor: the currents will pass but there will be no energy expenditure and the potential (energy to be spent) will thus be the same for all the electrons contained in each of the branches. Therefore, connecting the components in parallel to two nodes A and B does not mean "bringing" the potential of A and B to the components, but "making the potential of A and B equal" for all components thus connected.

In addition to the microscopic form of Ohm's Law it may be useful to examine the matter with Kirchhoff's laws, especially the law of voltages based on the principle of conservation of energy within a closed path.

A circuit is a system, it adjusts itself.

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