### Re: Measuring capacitance with equation

Date: Wed Mar 21 05:28:19 2001
Posted By: Bruno Putzeys, Staff, Electroacoustics and Analog Electronics, Philips ITCL
Area of science: Physics
ID: 984093390.Ph
Message:
```
Hi,

"Q" or charge is a measure for the amount of electrons that has been
displaced. The unit of current, the ampere is defined as being the current
needed to displace 1 Coulomb of charge in one second.

Suppose you set your multimeter to "ohms" and that it uses a constant test
current to measure the resistance - as is with digital multimeters. The
multimeter measures the voltage across the test object when this current
is flowing through it. The reading will be R=V/I.
Now you connect a capacitor instead of a resistor.

Then:

Q=I*t   and   Q=C*V

then

I*t=C*V

or the reading as it will appear on the multimeter will be

V/I=t/C

So, this means the resistance reading will increase linearly with
time. Hardly a surprise, since in the long run a capacitor is an insulator
and has infinite resistance.
Given the slow response time of both analog and digital multimeters it
will not be practical to stand by with a stop-watch and determine

A more practical way is the following:
Since the capacitor can pass current only when voltage is changing (see
above), it means that when subjected to an AC voltage the capacitor will
pass an AC current (peaking when the rate-of-change of the voltage is
maximal).
Mathematically:
dV(t)
I=C*-------
dt
(Current equals capacitance multiplied by voltage derived for time).
Going to a derivative allows one to calculate with complex voltage and
current shapes.

From that you can calculate (I hope you already know some basic calculus)
that if the voltage is a sinewave of amplitude Vt at a frequency f:

V(t)=Vt*Sin(f*2*pi*t)

The current will be

I(t)=C*Vt*f*2*pi*Cos(60*2*pi*t)

The voltage source can be a low-output-voltage transformer connected to
mains. Using a multimeter you can measure the amplitudes of these two
values (Volts RMS and Amperes RMS - The RMS factor is constant and will
factor out)

|V|    1
---=--------
|I| C*f*2*pi

So the capacitor will have an apparent resistance called impedance of
1/2*pi*f*C

So, measure AC volts across the capacitor, AC current through the
capacitor and calculate
|I|
C=----------
|V|*2*pi*f

So, it's more complicated than you might have initially thought but it
isn't too hard.

Regards,

Bruno

```

Current Queue | Current Queue for Physics | Physics archives