Voltages are not divided as per ohms law

[Sai teja]

Hello all,
             I connected two resistors (R1 and R2) in series.
             R1 : 2M ohm
             R2 : 1M ohm 
My set up is as follows.
            First terminal of R1 is connected to 9 volts from wall adapter
            Second terminal of R1 is connected to first terminal of R2
            Second terminal of R2 is connected to ground.

 According to ohms law ,
             voltage across R1 should be 6V
             voltage across R2 should be 3V
 But my observations are as follows.
            voltage across R1 is around 3 volts and
            voltage across R2 is around 2 volts.
            voltage across R1 and R2 is not summing up to 9 V.
           
What is happening here? Where am I missing? Please help me.             

[matseng]

I've got two guesses.

1) You don't actually have 9 volts from the power supply

2) You're using a cheap analog meter that have low internal resistance that are affecting the resistances you measure across. (The internal resistance of the is paralleled with the 1 or 2 megohm resistor you're measuring across)

[RoGeorge]

Your voltmeter have it's own internal resistance.

When you measured the voltage across one of the resistors, the voltmeter's internal resistance was put in parallel with the resistor from the measured circuit, thus affecting the readings.

[Benta]

Yep, the voltmeter is influencing the measurement. Some meters have 1 Mohm input impedance, some have 10 Mohms. In both cases they influence the measurement.
For doing measurements on such high resistances, you need a special high-impedance buffer/amplifier before the voltmeter, or a special voltmeter.

[rstofer]

Repeat your experiment with 2k and 1k resistors.  This will help verify the fact that the meter influences the circuit when you use high values of resistors.

[Tomorokoshi]

But my observations are as follows.
            voltage across R1 is around 3 volts and
            voltage across R2 is around 2 volts.

Go back and write the Ohm's Law equations for the two circuit cases with an unknown resistor in parallel with the one being measured, using the two voltages you list. Assume it has the same resistance for each measurement.

Solve each equation, and you will get different answers for the two solutions, even though one would expect them to be the same.

Why is that?

Using how the two answers differ, make a guess about the actual value of the unknown resistor. Using that value, attempt to calculate (or re-measure the voltages) to derive what should be the actual voltages shown on the meter. You can also solve for the resistance of the meter.

[Sai teja]

Yeah!
With lower values the readings are appropriate.
Thank you for your advice.

[Brumby]

That's good - and expected.

Now that you know there is a straightforward reason for what you originally observed, work out the resistance of the meter.  It's just using Ohm's Law again - plus a little bit of algebra.

Do that - and you will have made an important contribution to your knowledge.  Much more than a simple voltage divider example.

[kosine]

A related oversight is holding component leads or wires against your meter probes when trying to measure things. You'll add your own body resistance in parallel and again skew the result. Try it with a 1M resistor and it'll read about 900k, maybe less.

[Shock]

Another oversight is measuring an unregulated dc supply, or a battery that has lost capacity, the circuit under test will "load down" the supply and lower the voltage across it, while a multimeter on an open circuit may show a higher voltage until those components are added.