# Potential difference of resistors in parallel linked systems

Before the question I want to explain somethings to see if I understand anything wrong please tell me if im wrong: When there is a Potential difference on a system then the electrical charge will move from high voltage to low voltage and the amount of charge flows at one point is called the Current. In parallel linked resistors total current of each resistors is equal to current of the generator's but the voltage of the resistors are same and equal to voltage of the generator's,

So my question is how is the voltage of the different resistor's with different currents are equal to each other? in my logic if there are more charge flowing from a resistor then the voltage must be bigger but how is this working and what is the defination of the potential difference of a resistor, isn't it same with difference between two generator's potentials? and please don't make the classical explanation(potential difference is the difference between two point bla bla) I've read this 1000 times and it just doesn't make sense to me because if I gave x voltage to a system then I have to get x voltage out I can't get 2x because with this logic if parallel resistors have equal voltage then x+x=2x voltage but my generator has x voltage and how did I get 2x voltage with x voltage? Anyway hope you get it and can help me to end my complexity right now xd

Start with one resistor resistance $$R$$ and connect it to a power supply so that the potential difference across the resistor is $$V$$.
The current in the resistor is $$I= \frac VR$$.

Now set up the same circuit with a resistor with the same resistance $$R$$ as before with the same potential difference across it $$V$$ generated by a different power supply.
The current through this resistor is $$I= \frac VR$$, the same as before.

Now connect up the positive terminals of the two power supplies together and also the negative terminals.

The potential difference across each of the resistors is still $$V$$ and the current through each of the resistors is still $$I$$.
The total current is $$2I$$.

Keeping the resistors connected to one power supply remove the other power supply.
The potential difference across each of the resistors is still $$V$$ and the current through each of the resistors is still $$I$$.
The total current is still $$2I$$.

You can do a similar analysis with any number of resistors and varying sizes of resistance.
In each case the potential difference across the parallel resistors will still be $$V$$ but the total current will vary.

• Very nice. Reminds me of Galileo's (linked masses) argument for why a heavy body and a light body must fall with the same acceleration. Commented Dec 26, 2018 at 14:36

At some stage you will need to get to grips with "the classical explanation [that you've read] a thousand times". I'll just tackle one misconception that you have: "if parallel resistors have equal voltage then x+x=2x voltage". This is confused because there is only one voltage… If you have two resistors, R1 and R2, in parallel and connect a voltmeter first across R1 and then across R2, you're reading the same voltage twice, because the ends of R1 are connected to the ends of R2. So there is no sense in adding the two voltage readings together.

It's just like this… Two staircases go between the same two floors of a building. Each staircase takes you through a vertical distance of 2.5 m. I can't think of any circumstances when it is useful to add the two 2.5 m together.

Now suppose that there is a burst water pipe on the upper of the two floors, and water is cascading down both staircases. If the rates of flow are 2 litre per second down one staircase and 3 litre per second down the other, then it does make sense to add the two flow rates together to deduce that water is descending at a rate of 5 litres per second. In the same way, it makes sense to add together the electric currents through two resistors in parallel.

Electrical potential difference ('voltage'), then, is analogous to height difference. Electric current (rate of flow of charge) is analogous to (volumetric) rate of flow of water. These analogies mustn't be taken too far, but many people find them useful.