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I'm following a course about electrical engineering at university and I have a question about an example the professor made: he said that if we were to take an ideal voltage generator and connect it in series with a resistor, the voltage measured between the free end of the resistor and the free end of the generator would be V = E + RI where E is the voltage imposed by the generator, R is the resistance of the resistor and I the current that flows trough the resistor.

What I don't understand is the reason why when I try this at home with a 220ohm resistor and a 9V battery, the voltage V is equal to E with the resistor having no effect on the voltage but only on the current the battery manages to output. Why does this happen?

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    \$\begingroup\$ The trick here is I=0. \$\endgroup\$
    – Matt Young
    Commented May 28, 2014 at 15:36
  • \$\begingroup\$ If I'd know you were going to accept a answer within a few minutes (while I was typing), I wouldn't have bothered to write a detailed answer. \$\endgroup\$ Commented May 28, 2014 at 15:57
  • \$\begingroup\$ i'm sorry, i'm new here so i don't really know how everything works. I really am sorry. @OlinLathrop \$\endgroup\$ Commented May 28, 2014 at 16:00

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The answer should be easy to see if you look at the complete circuit you created:

The battery is modeled by the 9 V voltage source in series with R1. R1 represents the internal resistance of the batter, and V1 is its open circuit voltage. V1 and R1 together form a Thevenin source, which is a better approximation of a battery than just being a perfect voltage source.

R2 is the resistor you put in series with the battery, and R3 is the resistance of your voltmeter. The meter will show whatever voltage is accross R3.

Now put some plausible values on R1 and R3 (you already said what R2 is). R1 for a 9 V battery is probably in the 10-20 Ω range. Let's say 20 Ω, because as you will see shortly, R1 doesn't really matter in this case. A typical modern voltmeter will have about 10 MΩ resistance, so set R3 to that.

Now you have a voltage divider with R1+R2 forming the top resistor and R3 the bottom. What is the attenutation ratio of a divider with a top resistor of 240 Ω and a bottom resistor of 10 MΩ? From the fact that the bottom resistor is much much larger than the top, you should be able to see this is very nearly 1. In reality it is so close to 1 that you don't notice the difference on the voltmeter between the top resistor being 20 Ω (just the battery) and 240 Ω (the battery with your extra resistor in series).

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Your voltage meter has a very high resistance, so there is virtually no current flowing, which means that the voltage across the resistor is (almost) zero.

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take an ideal voltage generator ... i try this at home with ... a 9V battery

A 9 volt battery is not an ideal voltage generator. It has an internal resistance r.

the voltage measured between the free end of the resistor and the free end of the generator

If you connect the meter in series ("free end"), then you are measuring current, not voltage. If the circuit is open, then the current is \$I = 0 A\$.

And you got that formula wrong. It is:

\$E = V+Ir\$ or \$E = I (R + r)\$, where E is the electromotive force of the voltage generator, r is the internal resistance of it, I is the current through the circuit, and V is the available voltage at the voltage source output.

So, if you get \$E = V\$, then either your battery is an ideal voltage source, or you have an open circuit (\$I = 0 A\$). If you connected the voltmeter in series, then I tends to 0 A.

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