I have a question about the current flow in a series circuit.

In a series circuit the current flow is the same. It is also true that the current flow is inversely proportional to the amount of resistance in the circuit. I'm also aware that I can calculate the full amount of current flowing by applying the formula Itotal (A) = E total (V) / Rt (Ohms).

But if I consider the following case what will happen in reality?

I have a 9V battery capable of delivering 5A total. However I have three components in series: two of them pull 0.5A (according to the manufacturer) but one pulls me 2A. In this case because I don't know the components manufacturer internal resistances, I'm a little bit lost to apply the formula Itotal (A) = Vtotal (V) / R total (ohms). However it seems the total current flow is still inversely proportional to the amount of resistance I have on the circuit. Two components apparently have a high internal resistance whereas one has a low resistance.

My question is what final current will circulate on the circuit. 2A is just perfect for my third component but indeed too much for components 1 and 2 (0.5A). If the current is the same flowing on a series circuit what will be the final result to accommodate my three components (without damaging anything)?

Thanks for your attention, Antonio

  • 1
    \$\begingroup\$ What type of components? It seems there is some general misconception. \$\endgroup\$
    – Rev
    Commented Jun 21, 2013 at 7:56
  • \$\begingroup\$ You seem to be confusing a rating "two of them pull 0.5A" with the actual current. For 3 resistors in series the total resistance is R1+R2+R3 and the total current is V/(R1+R2+R3) It has nothing to do with their current rating which is probably stated for a single device with 9V across it. (see kirchoff's laws) \$\endgroup\$ Commented Jun 21, 2013 at 8:15
  • \$\begingroup\$ I think I might be confusing the total current on a series circuit with each component current rating (0.5A). I was not aware of that. So in that case the 0.5A component will only draw 0.5A even if the total current on the series circuit is higher (e.g. 4A)? Correct me if I'm wrong. As an example we could consider one stepper motor (2A) and then two kinect systems (we will say 0.5A just fictitiously per kinect) because in reality current rating is 1A each. For the theoretical example it will do. \$\endgroup\$
    – Antonio
    Commented Jun 21, 2013 at 9:33
  • \$\begingroup\$ The components that you are talking about are considerably more complicated than resistors. The equation that you're trying to apply is valid only for resistors. \$\endgroup\$
    – Skaevola
    Commented Jun 21, 2013 at 15:11

2 Answers 2


Perhaps a worked example would make things clearer.

Suppose I had three lamps - each has a different working voltage and takes a different current. What would happen if I connected all three in series and connected them to a 9V battery.

enter image description here

For the purposes of illustration only (keeping it simple) I am assuming the lamps work as pure resistances that don't change their values.

First we calculate the individual resistances of the lamps. In this case 6,24 and 15 ohms.

The total circuit resistance will be 6 + 24 + 15 = 45 ohms

The current that will flow around the circuit will be

                               9/45  = 0.2 Amps

This is much less than any of the rated values because there is MORE resistance in the circuit

Each lamp will 'drop' a different voltage

lamp1 will drop 0.2 x 6 = 1.2V

lamp2 will drop 0.2 x 24 = 4.8V

lamp3 will drop 0.2 x 15 = 3.0V

If we add up all the voltage drops we get 1.2 + 4.8 + 3.0 = 9V

In other words the sum of the voltage drops around the circuit is equal to the supply voltage (kirchoff's voltage rule)

Again note that these voltages are not related to the ratings of the devices.

enter image description here

  • \$\begingroup\$ So in this case the 9V are not sufficient to feed this circuit, I I will not have any lights ON? The minimum voltage to get it up an running would be 48V? So the total current of the series circuit has to cover at least the maximum current rating of each individual component (in this case the lamps 1;0.6;0.5A)? If you any links where I can learn more about total current, individual current and voltage rating, that would be great. \$\endgroup\$
    – Antonio
    Commented Jun 21, 2013 at 11:21
  • \$\begingroup\$ @Antonio No - because each of the lamps is rated at a different voltage and current. You must apply Ohm's law (V = IR) to work out the current that would flow for a given voltage. You cannot use the rating. Putting 48V across all of them in series would put 48/45 amp (about 1.07 Amps) through EACH of them. The voltages would be distributed so that lamp1 would drop 6.42 Volts (no problem, just about rating), lamp2 would try to drop 25.68V (and probably blow - exceeds rating) and lamp3 would try to drop 15.9V (and probably blow - exceeds rating). \$\endgroup\$ Commented Jun 21, 2013 at 15:24
  • \$\begingroup\$ I see. I have to be careful with drop voltage (for each component) which is different from rated voltage. It seems series circuit doesn't give me enough flexibility for what I want (parallel is the way to go forward, same voltage, different currents in each branch). The problem on series is that the current flowing is always the same. So having different components with different current ratings is problematic (in order to calibrate one I might damage the others). Revising on your initial example (3 lamps) there's no series solution that would turn the lights ON with the current configuration? \$\endgroup\$
    – Antonio
    Commented Jun 21, 2013 at 19:08
  • \$\begingroup\$ @Antonio Yes - The rating is there to give you a guide for each component. Provided they have the same working voltage you can connect them in parallel. Series solutions only work when all the components take the same current. \$\endgroup\$ Commented Jun 22, 2013 at 6:22
  • \$\begingroup\$ Thanks very much for this explanation. I think I got the main idea. I'm software engineer however I do my own bits of electronics. I'm constantly learning thanks to you guys. \$\endgroup\$
    – Antonio
    Commented Jun 22, 2013 at 15:31

It seems you're getting confused by thinking of current as "pulled" or "drawn." While that's a convenient way to put it linguistically, what really happens is the voltage source "pushes" a current, and the components resist that current to make it smaller.

When you have components in series, their effective resistances add up. The same amount of current flows through all of them, because there's nowhere else for the current to go.

However, keep in mind that the type of component is very important, because they don't all add up linearly. An ideal LED, for example, will allow as much current as necessary in order to keep its voltage at a certain level, called its forward voltage. If you have an LED and a resistor in series, the current through the LED will change depending on the resistor value, even though the voltage across the LED is relatively constant.

For a more complex component like a stepper motor, the manufacturer will specify a maximum current, but the actual current is constantly changing, depending on how fast the motor is turning, what part of the cycle it's on, what the load on the motor is, etc. That means if you put components like that in series, there's really no way to predict what the total current would be at any given time.

That's why except for very basic components, you generally power them in parallel rather than series. That lets each component use as much current as it needs without affecting the others.


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