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### Topic: Problem in internal resistor in digital pin (Arduino UNO) (Read 795 times)previous topic - next topic

#### Khanhdautroc

##### Jun 10, 2014, 01:27 pmLast Edit: Jun 10, 2014, 02:07 pm by Khanhdautroc Reason: 1
When I read tutorials about internal resistor of input pin, I encountered many issues. But the main problem is here. If I can explain this one, all will be cleared.
As we know, V=E-IR (*) ( Ohm'law applied in whole circuit)

When no current flow(open circuit)   U and E will be the same
But I made 2 experiments  and realized that that is not true
4 pictures following are my result.
First:  Experiment on lipo (test voltage with and without 1K resistor)
Second: Experiment on 9v battery(with and without 1k resistor, too)
According to (*), voltage will be the same before and after i add 1k reistor, but the result is not my desire.
So explain to me, thanks in advance

#### Paul__B

#1
##### Jun 13, 2014, 04:10 am
whether you had mis-identified the resistor since the photos, whilst posted in a ridiculously excessive resolution (should be no more than 1024 by 768), are too poorly lit and way too poorly focused to be of any use whatsoever!

Then I realised you have the meter set to "battery test".  On either of these settings, it is no longer reading open circuit voltage, but deliberately putting a load resistor across its terminals.  This means that when you put a resistor in series, you are now measuring the voltage across a voltage divider.  What we can approximate, is the load resistance of the test resistor in the meter on the 9V "battery test" setting, as 1.5k.

#### krupski

#2
##### Jun 15, 2014, 01:08 am

First:  Experiment on lipo (test voltage with and without 1K resistor)
Second: Experiment on 9v battery(with and without 1k resistor, too)
According to (*), voltage will be the same before and after i add 1k reistor, but the result is not my desire.
So explain to me, thanks in advance

You have your meter in "battery test" mode which purposely places a load on the battery in order to test it. The meter does this because even a dead battery can read "good" with no load, so they apply a load to give you a more realistic test.

You are seeing a voltage drop between your series resistor and the built in load resistor of the meter.

Place your meter on "20 DC Volts" instead and you will see little to no difference between a battery with or without a series resistor.

Gentlemen may prefer Blondes, but Real Men prefer Redheads!

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