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I have checked a lot of answers here but none gives a satisfying answer to the question. Doesn't different resistor dissipate different amount of energy in terms of heat? Now you might answer it's because one of the resistor will have larger electric field thus do same amount of work. How do electric field know how to be larger at some part? And there is another things called conservation of energy. I don't understand how you can apply conservation of energy to show there will be same voltage across parallel circuit. It is not necessary that terminal voltage should be same as voltage across circuit because different resistor dissipate different amount to energy think about friction. So why is voltage drop same in parallel Circuit?

(Please don't use hydraulic analogy I think it got a lot of issues in modeling accurate model. But one analogy you can use is a planet and asteroid scenario where one body represents electron and earth represents proton and asteroid belt as resistor which is fixed in place.)

banned
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  • If the voltage drop across two parallel elements is not the same you can run around them in a circle over and over again and attain any voltage to ground you want. – Jon Custer Apr 12 '22 at 13:45

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Let's imagine a simple parallel circuit like this one from hyperphysics

enter image description here

You are right that if $R_1 \neq R_2$, then the energy dissipated per unit time in $R_1$ will be different than the energy dissipated in $R_2$ per unit time. However, the energy dissipated per unit time (the power), doesn't just depend on the voltage, but also the current, through the equation \begin{equation} P = I V \end{equation} The fact that the voltage across $R_1$ is the same across $R_2$, does not contradict the fact that $P_1\neq P_2$, since the currents are different, $I_1\neq I_2$.

You likely learned earlier in your course that, in equilibrium, the electric field inside a conductor is zero) -- because if were nonzero, charges would rearrange themselves on the conductor to cancel the electric field. A circuit like the one shown above is in an equilibrium state (even though current is flowing, the current isn't changing). It is for this reason that if we model the wires connecting the battery connecting the resistor as perfect conductors, then there is no electric field in the conductors, and the voltage drop (which is the line integral of the electric field) is zero. Since a wire connects the top part of $R_1$ and the top part of $R_2$, the potential difference between these points is zero. The same goes for the bottom points on the wire. Therefore, the voltage drop across both resistors is the same.

In reality, there is some very small resistance in the wire, so the potential difference across the two resistors is very slightly different. However, for simplicity and because this effect is so small, we usually ignore it.

Andrew
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  • The fact that the voltage across $R_1$ is different across $R_2$, also does not contradict the fact that $P_1\neq P_2$. I understand wire at the bottom one will have 0 potential difference but how do you know top one will have same? – banned Apr 12 '22 at 03:41
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    @CuriousMind As I wrote, "Since a wire connects the top part of 1 and the top part of 2, the potential difference between these points is zero." – Andrew Apr 12 '22 at 03:43
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    @CuriousMind The potential difference between the top of $R_1$ and the top of $R_2$ is zero. The potential difference between the top and bottom of $R_1$ is $V_B$, and between the top and bottom of $R_2$ is also $V_B$. – Andrew Apr 12 '22 at 03:57
  • What confuses me is that when I consider a point charge as distance increases voltage drops. Now if voltage is same across a wire then there must be equipotential surface... To me it doesn't make sense if connecting to same wire makes them have equal potential. – banned Apr 12 '22 at 06:15
  • @CuriousMind That's a good question! The difference is that for a point charge, the field lines spread out around the point charge, and the field strength dilutes as you get far from the charge. For a battery in a circuit, the wires essentially focus the field in the circuit, so the field doesn't lose strength by spreading out. One way to look at it is that a circuit has many point charges. The electrons flowing in the wire cary some field with them. The battery pushes on electrons near the battery, which push on electrons further along the wire... on and on down the chain (...) – Andrew Apr 12 '22 at 11:16
  • (...) until an equilibrium is reached. In equilibrium, the charges from the battery to the top of the resistor flow without needing a net field applied. At the resistor, electrons build up on one end, and a net field across the resistor is created which "pushes" the electrons through the resistor. – Andrew Apr 12 '22 at 11:17
  • Some relevant PSE posts: https://physics.stackexchange.com/questions/143300/voltage-drop-more-electrons-on-one-side-of-resistor, https://physics.stackexchange.com/questions/186614/why-is-the-electric-field-highest-in-regions-of-highest-resistance – Andrew Apr 12 '22 at 11:19