Unexpected Results for the Simplest of Circuits

[naragon1]

Hello,

I'm trying to get a solid understanding of basic electronics. I got a hold of a used copy of the second edition of the The Art of Electronics and the Student Manual for the Art of Electronics,  and I'm planning to work my way through all the labs. As I set up the first lab, which is basically verifying Ohm's law, I got some strange results.

The lab is basically to measure voltage vs. current for a 20k resistor. Not having a 20k resistor, I put two 10k resistors in series and constructed the circuit as shown in the first image - easy enough. I was also using an adjustable power supply and taking measurements at 0.5V intervals. As I noted the results,  I got twice the value of the current I expected each time.  For example, for 5V I expected to get 250 uA, but instead I got close to 50 uA. I've posted a plot of the values I was getting, and sure enough I confirmed the linear result for Ohm's law, but the values of current appeared to be always 2x the expected value. I measured the resistance of each of the resistors and I got ~9.9k.

What gives? Maybe someone can point out something obvious I may be missing.

[pmbrunelle]

What kind of power supply do you have?

Consider that your ammeter will have a certain resistance; it is not zero Ohms.

[Simon]

are you sure they are 10K resistors ?, something does not add up. As per above reply there will be some voltage loss in the ameter and so effectively lower voltage on the resistors. the ameter also has a resistance.

[Seekonk]

This is an important lesson, Never fully trust what instruments read.  For a good explanation read Dave's explanation of uCURRENT operation.  Using such low currents will induce a lot of errors.  Read the voltage at the resistors.

[naragon1]

Doh. 

It turns out I was not properly breaking the circuit when taking the current measurement!

Thanks everyone for the replies. Everything works as it should, carry on ...

[alsetalokin4017]

For a crosscheck, and as an interesting exercise, you might try eliminating the ammeter and, instead, look at the voltage drop across the resistors using a voltmeter, and computing the current from that, using Ohm's Law.

[babysitter]

... which might lead to errors due to the instrument impedance in parallel to one of the resistors if done right (big enough resistors).  Predictable as your meter impedance is known at least as a ballpark figure (most times its 10 MOhm).
Very educating, too