When we have resistors in series, the current through all the resistors is same and the voltage drop (or simply voltage) at each resistor is different.

Question 1: It is fine that voltage drop (potential drop) across each resistor is different because each resistor offers different resistance (suppose). but how is the current through each resistor same? If we have resistors of different resistance, shouldn't the current be different through each resistor?

Similarly, when we have resistors in parallel, the current through each resistor is different but the voltage drop at each resistor is same.

Question 2: Current through each resistor is different because resistance of each resistor is different (suppose). but how is the voltage drop across each resistor same here? Shouldn't the voltage drop at each resistor be different because each resistor offers different resistance?

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    $\begingroup$ Hey Muhammad, you may have some misconceptions about resistors in general. Current is constant since charge is conserved. You may look up Kirchhoffs law and what a resistor actually is. All the best. $\endgroup$ – Robert Filter Jan 9 '13 at 10:43

I'll start with current first...

1) "Current flows in a circuit" is the simple answer. In other words - It's the rate of flow of electric charges. Other than $i=\frac{dq}{dt}$, Current is also given by $I=nAEv_d$ which says something that it depends upon the drift velocity of electrons. The drift velocity is the average velocity between two successive collisions. This velocity prevents the electrons from accelerating continuously. Ok. Let's consider a circuit with three resistors with resistances in an increasing order $R_1>R_2>R_3$.

First, current enters $R_1$. After some collisions (causes heat generation), it exits the resistor. Now, the same current enters and exits $R_2$ & $R_3$ in the same manner. One point is to notice that, the rate of flow of charges is always the same (the current entered and exited the resistors with same magnitude). Only the drift velocities vary in different resistors.

If the same are connected in parallel (Now, we look into the resistors), current flows through $R_3$ easily. Because, $R_3$ requires a lesser $v_d$ (i.e) Electrons entering $R_3$ would exit within a small period of time relative to the other two (thereby increasing the rate)... As a result, larger current would be observed.

2) Voltage is simply the "Energy per unit charge" in both electrostatics & current electricity. Let's assume that the electrons have some energy and they pass through the series of resistors.

When the electrons encounter collisions within the resistor $R_1$, they lose their potential (which is referred to as "voltage drop") accordingly with their drift velocity (i.e) number of collisions which evolves as heat (depends upon the resistance of each). The remaining potential is still present in those electrons which drops across other resistors. Thus, the sum of each potential drop gives the total potential in that circuit.

As in case of parallel circuit, current has divided. Now, though the charges are taking different paths, they have the same potential (energy). Hence, the potential drop across each resistor would be the same... Man, It's all the consequence of Ohm's law $V=IR$.

Note: And, sorry for that confusion. Electrons gain energy only as they pass through the battery. But, they're always continuously accelerated by the electric field. Also, resistance is also given by $R=\frac{mL}{nAe^2\tau}$ which Simon Ohm replaced it as a constant (for a given temperature)...



The diagram shows first four resistors in series then four resistors in parallel.

For the resistors in series the current flowing into the wire, $I_{in}$ must be the same as the current flowing out, $I_{out}$ because the current can't escape from the wire. There is only one route for the current to flow through the wire so the current has to pass through all the resistors in turn. That's why the current passing through every resistor must be the same.

Now look at the resistors in parallel. The point here is that the top ends of the resistors are all connected together so they must all be at the same voltage $V_{in}$. Likewise the bottom ends are all connected together so they must be at the same voltage $V_{out}$. That means all the resistors have the same voltage drop across them of $V_{in} - V_{out}$.

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    $\begingroup$ @john: (in series resistors) current is the charge passing through a certain area per unit time. let this certain area be the resistance R1, R2, R3 and R4. now if we see theoretically, all the resistors should offer some resistance so, shouldn't the current through a certain area (R1, r2, r3 or r4) be different as all of them offer different resistance. im veryyyyyyyyy confused in this. $\endgroup$ – Rafique Nov 9 '12 at 12:55
  • $\begingroup$ Remember that the current is just electrons. Suppose the number of electrons flowing through $R_1$ per second was different to the number flowing through $R_2$. In that case the wire between $R_1$ and $R_2$ would either be depleted of electrons or the electrons would pile up there. This would change the potential at that point until the electron flows became equal again. $\endgroup$ – John Rennie Nov 9 '12 at 15:27
  • $\begingroup$ I've just spotted that you said current is the charge passing through a certain area per unit time. The area of the conductor is not counted in the current. The current is coulombs per second not coulombs per second per square metre. $\endgroup$ – John Rennie Nov 9 '12 at 15:29
  • $\begingroup$ I'm still confused by all this. If a 100Ω resistor reduces the voltage by some amount in series, why should that same resistor do absolutely nothing to the voltage in parallel? Is a resistor's effect on voltage not at all related to its resistance? $\endgroup$ – user3932000 Jul 11 '16 at 4:38
  • $\begingroup$ Maybe this article can help. physicsclassroom.com/class/circuits/Lesson-4/… $\endgroup$ – Delali Sep 12 '17 at 8:25

When I was studying about electric circuits, I had a similar problem too. But when I discovered where I was going wrong, it made things pretty clear for me. I will share with you what problem I faced, as it may be true that you are facing the same.

Current is the quantity of charge passing through the circuit in unit time, by definition. Observe the definition, its 'quantity', not 'drift velocity'. Yes, the drift velocity will be different, but think for yourself, the quantity will remain same. Consider a system of drainage. In the system, let there be a big sized pipe, analogous to the wire which provides almost no resistance. Then there is an extremely small sized pipe, analogous to the resistor providing max resistance, and let there be a normal sized pipe, analogous to the resistor with less resistance. Now, water starts flowing through the system. It flows freely through the big sized pipe, as there is less hindrance. The speed of the water is slow. Now, the water enters the smallest pipe. There is a huge pressure, and the speed becomes high. But note that in unit time, same volume flows. Now, as for the normal pipe, the speed will be high, but slightly slower than the smallest pipe, but again, note that same 'volume' flows. Similar is the case with resistors. So, its about that 'volume'(charge), not the 'speed'(drift velocity).

I hope my answer helped you, even a little bit.

  • $\begingroup$ In this model, does the speed of the water represent voltage? If so, are you implying that increasing the resistance will also increase the voltage? Which is demonstrably false, since a 100Ω resistor and a 100,000Ω resistor in parallel will both have the exact same voltage. $\endgroup$ – user3932000 Jul 11 '16 at 4:46
  • $\begingroup$ @user3932000, read it carefully. I never said that the speed was the voltage, instead I specified the OP about current, which is in this case the quantity of water. $\endgroup$ – codetalker Jul 11 '16 at 9:30

To be clear, current is charge passing through a certain area per unit time. This does not imply a second parameter in the denominator of the formula for current $\frac{dq}{dt}$; just a guideline for how to measure $dq$. The larger the cross sectional area, the larger the perceived current will be. This is why resistance is seen to decrease for larger cross sectional areas when selecting wire gauges.

Secondly, it is true that different drift velocities along a circuit would be impossible as the electric field would vary in the areas wherever charge density is different and force an equalization of the rate of flow.


Look friend. The number of charge carriers in a circuit is constant. So current means flow of charge carriers. It is constant. Voltage source provides energy to these charge carriers to flow through the ckt and some energy is droped in moving through resistance. Understood???

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    $\begingroup$ The question asks about both parallel and series circuit. You don't specify which your answer is referring to. What is ckt? Also the tone is a bit confrontational. $\endgroup$ – Rick Oct 14 '15 at 12:38

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