Ben Schaeffer said:

Didn't mean to confuse things - I meant negative when I

said ground.

Circuit is:

9v positive---resistor---capacitor---9v negative

Measuring voltage with a 9v battery with 100k resistor

and .01uf Cap

black meter probe to 9v negative and red meter probe to

9v positive = 8.47v black meter probe to 9v negative to

red meter probe to node between resistor and capacitor =

7.71

black meter probe to node between resistor and capacitor

and red meter probe to 9v positive = 0

By the way, have you tried this?

You have 0 drop across the resistor because there is almost no current

flowing through the cap once it has charged. This agrees with the math fine.

The reason you measure less than full battery voltage across the capacitor

is because some current is flowing through the meter itself, causing a

voltage drop to develop across the resistor. If your meter has an impedance

of 1M, then the math works out again:

8.47V * 1M / (1M + 100k) = 7.71V (approx)

This is because the 100k resistor and the meter form a voltage divider, and

the cap basically does nothing. Try taking the cap out of the circuit, and

measuring with 9V <-> resistor <-> meter <-> 0V

and you should get the same value.

Most newer, digital multimeters have a 10M impedance, so either

A) your meter is unusual

B) your 100k resistor is actually 1M or

C) your battery is REALLY dead.

Since you tried with different voltage sources and resistors, we can

probably rule out B and C, and say that A is the case. Try looking for the

specs in the manual.

If you did try the experiment again with a 1M resistor, I think your would

measure very close to 50% of the battery voltage across the cap. A 10M

resistor would give you about 1/11 the battery voltage, or a 91% drop across

the resistor.