Circuit in series: $10\,\text{V}$ power supply, 2 resistors connected. Each resistor is $10\,\Omega$, total of $20\,\Omega$ resistance.

If I put an ammeter at the end of the circuit, will the current be calculated using 20 Ohms or 10? Electrons flow through the first 10 Ohm resistor and are slowed down, yes? Then they flow through the second 10 Ohm resistor. They have passed through a total of 20 Ohms BUT at 10 Ohms per resistor.

Will the current be the same as if it had passed through 1 20 Ohm resistor? That violates everything I understand.

  • $\begingroup$ Think about this: what is electric current? $\endgroup$ – Wouter May 10 '14 at 13:26
  • $\begingroup$ Flow of electrons? $\endgroup$ – Joel Aqu. May 10 '14 at 13:30
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    $\begingroup$ Indeed. More precisely: electric current is the rate at which charge passes through a certain surface (here that surface is some section of the conducting wire). Now, what is the effect of a resistor on the rate at which electrons pass through the wire? Think about this as follows: electrons go into the resistor at some rate $r$, how do they come out? $\endgroup$ – Wouter May 10 '14 at 13:33
  • $\begingroup$ In case you're still struggling, I'll rephrase my question: what has happened to the rate of electrons through any given section of the wire after the resistor? (as compared to before the resistor) $\endgroup$ – Wouter May 10 '14 at 13:55
  • $\begingroup$ Draw a circuit diagram. I'm a physicist who works with electrical circuits on a daily basis. If I can tell one single thing it is that whenever you do a problem or ask a question regarding circuits, you need to draw a circuit diagram. Whether or not you think it will help doesn't matter. Once you get in this habit you will start understanding things better. $\endgroup$ – DanielSank Aug 1 '15 at 4:44

The current $I$ has a value in one point of the circuit, in contrast to the voltage $U$ which is always measured between 2 points.

The definition of $I$ is the amount of charge $\Delta q$ that passes through a particular point in the circuit in the time $\Delta t$ (it's a quantity mathematically similar to the simple velocity in kinematics). So when you measure the current in one particular place of the circuit, you measure how much charge is passing through that point, divided by time.

So what do you think is the difference in the measured current $I$, if in between the power supply $U$ and the measuring point you have:

  • one resistor with $R = 20$ $\Omega$
  • 2 resistors, each with $R = 10$ $\Omega$
  • 4 resistors, each with $R = 5$ $\Omega$ etc.?
  • 1
    $\begingroup$ OK, I just noticed it's 3 months old... Damn you "Top Questions"! Well, at least I hope someone someday may benefit from this answer. $\endgroup$ – Lurco Jul 13 '14 at 20:56

If you rearrange Ohm's Law equation, you get $I=\frac {V}{R}$. When you have 2 equal resistors in series, the voltage drop across each resistor is $\frac {1}{2}$ the total voltage. Now calculate the current through both resistors, through one resistor and the equivalent resistor (20 ohms).

  • $\begingroup$ I don't think OP is looking for a way to calculate the current. They are trying to understand what's going on. $\endgroup$ – Wouter May 10 '14 at 13:35
  • $\begingroup$ @Wouter I didn't see his other question when I answered this one. So, it is apparent that he knows how to calculate the current, but he doesn't understand why the voltage drop is the way it is. As an example, there is a device called a rheostat which is a wire-wound resistor with a tap that can be moved along the wire. Since we know that the wire has a resistance, the amount of resistance depends on where the tap is positioned. With the two ends of the wire connected to the battery, you can see the voltage change at the tap when you move it. $\endgroup$ – LDC3 May 10 '14 at 13:44
  • $\begingroup$ This show that the voltage varies depending on the resistance in the circuit. Now if you connect the battery to one end of the wire and the tap, you can have a varying amount of resistance, but always the same voltage; which means your current will depend on the amount of resistance. $\endgroup$ – LDC3 May 10 '14 at 13:45
  • $\begingroup$ You're certainly right, but OP is trying to somehow understand what's going on microscopically - or at least get an intuitive picture of it. Observing that the voltage varies depending on the resistance in the circuit is great for phenomenology, but it's not really satisfactory if you want to know why that happens. And finding out why stuff happens is - ultimately - the entire goal of physics. $\endgroup$ – Wouter May 10 '14 at 13:52
  • $\begingroup$ @Wouter If that is the case, he needs to ask another question that is more specific in the answer he is looking for. $\endgroup$ – LDC3 May 10 '14 at 13:56

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