WEBVTT

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These experiments will use this generator BF which will be partly out of scope ...

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And our oscilloscope on tablet (view on the Maïdo in the island of Reunion 974 :-))

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The BF generator supplies in series: capacitor (red curve) and resistance (blue)

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The common point of the oscilloscope is thus at the center of the assembly series RC

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The oscilloscope uses an inverting probe to compensate for wiring reversal

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In low frequency, the voltage in the resistor (image intensity) is very low

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Normal: in BF the capacitor opposes a high impedance (Zc = 1 / C * 2 * pi * f)

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The green multimeter is used as a frequency meter: 30 hertz currently

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When the frequency increases, the blue amplitude, UR and thus the intensity, increases

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Normal: the impedance Z of the capacitor decreases as f increases!

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I have set identical gauges on the red and blue channels of the oscilloscope!

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The amplitudes will soon be identical!

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We just passed! We're stuck on equality ...

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At 630 Hertz, the amplitude in my resistance and my capacitor are identical

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The phase shift is 1/4 of a period (90 °)

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Who is ahead?

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The voltage in the capacitor in red is late, I in 'advance', one is capacitive!

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The voltage in the capacitor is lagging because it requires intensity to charge

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The value of the capacitor

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Marked 1 μFarad, we will apply C = 1 / (Z * 2 * pi * f)

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Here is displayed: Z * 2 * pi. But C = 1 / Z!

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1.011 * 10 ^ -6 Farad so 1.011 μ Farad! As marked!

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At this frequency Z of the capacitor = Resistance since the amplitudes are identical

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By increasing the frequency, it is found that the impedance of C continues to decrease

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I now use a capacitor 8 μFarad

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Here, impedance Z of the capacitor = Z and R of the resistance ('perfect')

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I find 8.2 * 10 ^ -6 or 8.2 μFarad, for 8 marked

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I'll check the capacitance!

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With the capacitance 1.008 instead of 1.011, very good! Beautiful precision :-)

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8.238 read for 8.2 seen on the oscilloscope and calculated: very correct too!