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How to show that current leads voltage by 90° in capacitor by experiment?

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Edward Yuen:

Many books told us that current leads voltage by 90° in capacitor.
How to show that it is true by an experiment with an oscilloscope?

Edward Yuen

A series RC network with a function generator.  Measure the voltage across the capacitor and measure the current by looking at the voltage across the resistor.

Simple setup for two-channel (dual-trace) oscilloscope (analog or digital).
Choose frequency and capacitor such that the reactance is, say, 1000 ohms.
Connect generator to capacitor and other end of capacitor to small resistor, say, 20 ohms.
Other end of resistor to ground.
One channel input to generator output, which is approximately the voltage across the capacitor.
Other channel input is proportional to current through the capacitor.
Trigger the oscilloscope from channel 1 (voltage).

It's a lot easier if you have floating scope inputs.  Then you use one channel to measure the voltage across the resistor, giving current, and one channel across the capacitor giving voltage.

The easiest way I know to do this is with the Digilent Analog Discovery 2.  If I were a student, this would be my first piece of test equipment. 

I don't like the new and improved pricing (a lot more than I paid) but there is a sale.  If you buy an FPGA board > $100, the AD2 is discounted substantially.  Still, a lot of money flying out of pocket.  The student price of $279 is a lot better than the list price.

Here's a video and there are MANY more.  I haven't tried any of them...

Edward Yuen:
I used a series RC circuit in the experiment.

In circuit one, I had sine signal to resistor to capacitor to ground.
Then, I had Vs and Vc in scope. I got the phase difference,p1.

In circuit two, I exchanged the positions of resistor and capacitor.
So, I had sine signal to capacitor to resistor to ground.
Then, I had Vs and Vr in scope. I got the phase difference,p2.

p1+ p2 is approximate to 90 degrees.


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