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How could the potential difference be constant across all the resistors of parallel connecting resistors although each resistor has a specific resistance?

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  • $\begingroup$ Maybe because something else changes inversely, thus compensating the the differences between the resistors. $\endgroup$
    – babou
    Aug 20 '14 at 0:08
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Kirchoff's laws tell us that the potential drop across any closed loop in a circuit must be equal to the voltage sources in the loop, from which we conclude that the voltage drop across resistors in parallel must be equal.

Ohm's law states:

$$V=IR$$

From which we conclude that, since $V$ is fixed, if the different resistors have different $R$'s, then the current ($I$) through each must also be different (and obey Ohm's law).

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  • $\begingroup$ Thanks very much maan, your answer really made it clear to me, especially after using Ohm's law. $\endgroup$
    – Omar Ahmed
    Aug 20 '14 at 0:13
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This is because what changes in each resistor is the current passing through and not the voltage difference. One the other hand when resistors are in series they have the same current passing through, but different voltage through each one's nodes.

In essence when resistors are in parallel do not share same current path (i.e wire) but share same voltage. On the other hand when resistors are in series they do share the same current path. Since current $I$, voltage $V$ and resistance $R$ are related (macroscopicaly) by Ohm's law ($V = IR$). The rest follows.

Comment:

Once i asked my high-school physics teacher, how come resistance is defined as "difficulty" of current pasing through and at the same time, resistors in series have exactly same current passing through.

The answer was that although each individual electron has a "difficulty" passing through a resistor (and dissipates energy/heat etc.) The whole electron cloud (which defines macroscopic current) goes on at same average speed.

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  • $\begingroup$ Well, I am not sure I like much the explanation of your teacher. When in series, the resistors do have the same current (same number of charges passing trough), but to achieve that you have greater force pulling them in the more resistive resistors: the difference of potential is greater. $\endgroup$
    – babou
    Aug 20 '14 at 0:15
  • $\begingroup$ @babou, you misunderstand the question (to my physics teacher) :) $\endgroup$
    – Nikos M.
    Aug 20 '14 at 0:17
  • $\begingroup$ I misunderstand in what way? Why do you say current path rather than just current. $\endgroup$
    – babou
    Aug 20 '14 at 0:18
  • $\begingroup$ @babou The (my) question was not about increased resistance, but about current (defined as number of charges per unit time) and at the same time resistance was defined as "slowing down" the charges (electrons), so no force or increased resistance was mentioned. $\endgroup$
    – Nikos M.
    Aug 20 '14 at 0:21
  • $\begingroup$ That is not what you said. You said "difficulty" of current passing through. You have the same number of charges going through each resistor. I know it is a mass motion, rather than individual charges walking the whole way, but I do not see why it explains anything. $\endgroup$
    – babou
    Aug 20 '14 at 0:32

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