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I previously explained inductive reactance, now consider an ac circuit with a capacitor connected across the terminals of an ac generator. Applying Kirchhoff's loop rule, we get
v = v(c) = 0
Where v is the instantaneous voltage and v(c) is the instantaneous voltage across the capacitor.
So that
v = v(c) = V(max)sin wt (1)
Where V(max) is the maximum voltage of the source and w = angular frequency = 2 * Pi * frequency.
Since v(c) = Q / C, substitution of v(c) into expression (1) yields,
Q / C = V(max) sin wt or
Q = CV(max) sin wt (2)
Since i = dQ / dt, we obtain the instantaneous current in the circuit by differentiating expression (2) giving us
i(c) = dQ / dt = wCV(max) cos wt (3)
Applying the trig rule cos wt = sin (wt + Pi/2) we can express equation (3) in the form
i(c) = wCV(max) sin (wt + Pi/2) (4)
Comparison of expression (4) and expression (1) shows, as in the case of the inductor, the current and voltage are again out of phase. In this case, the current leads the voltage across the capacitor by Pi/2 radians or 90 degrees so that the current reaches its maximum value a quarter of an oscillation period before the voltage reaches its maximum value.
The current reaches its maximum value when cos wt = 1 therefore
I(max) = wCV(max) = V(max) / X(c) (5)
Where X(c) = 1 / wC (6)
Where X(c) is the capacitive reactance in units of ohms.
Combining expression (1) with expression (5) gives us the instantaneous voltage drop across the capacitor as
v(c) = V(max) sin wt = I(max) X(c) sin wt (7)
We see that as the frequency of the circuit increases, the maximum current increases as the reactance decreases. For a given maximum applied voltage, we see that the current increases as the frequency increases. On the other hand, as the frequency approaches zero, the capacitive reactance tends to infinity and the current approaches zero. This is to be expected because as w tends to zero, the circuit approaches steady state dc conditions whereby no current passes through the capacitor.
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