Capacitive Reactance: Calculate Xc & V=I*Xc

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In an AC circuit with a pure capacitor, the voltage and current can be expressed as Vc = V0 * sin(w*t) and I = wCV0 * sin(wt), leading to the capacitive reactance formula Xc = 1/(wC). The discussion clarifies that the equation V = I * Xc applies to amplitude or RMS values rather than instantaneous values. It also highlights that complex numbers are not strictly necessary unless considering phase differences, where the introduction of imaginary numbers helps describe the phase relationship using e^(i*theta). The conversation emphasizes the distinction between real and complex representations in AC circuit analysis. Understanding these concepts is essential for accurate calculations in capacitive circuits.
Devil Moo
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Hi,

A pure capacitor C is in an a.c. circuit.

Vc = V0 * sin(w*t)
Q = C * Vc
= C * V0 * sin(w*t)
I = dQ/dt
= wCV0 sin(wt)
then I0 = wCV0
Xc = V0 / I0 = 1/(wC)

So why people would say V = I * Xc?

Is it a must to include complex number?
 
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The derivative of sin is cos.

You don't need complex numbers.
 
Whoops!
I = I0 cos(wt)
Then V = I * Xc * tan(wt)
not V = I * Xc
 
The relationship V=I*Xc (with real numbers) is between the amplitudes or rms values and not between instantaneous values.
Otherwise you need to consider Xc as a complex number, to take into account the phase difference between v and i.
 
But why do we introduce imaginary root in it?
Is it used to describe the phase based on the e^(i*theta)?
 
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