|MadSci Network: Physics|
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 Ohms_Reading/Time. 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
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