Why does capacitive reactance decrease with the increase of the frequency of the applied signal?
It is easy to prove why capacitive reactance decreases with increased capacitance. The more we increase the capacitance of a capacitor -> for the same charge at the plates of the capacitor we get less voltage which resists current from the AC source.
But why is reactance decreased with the increase of the frequency of the applied signal?
Best Answer
First, let's look at how the capacitive reactance is obtained.
The relationship between electrical charge and current is:
$$ dq = i\ dt $$
where \$q\$ is the electrical charge, \$i\$ is the current and \$t\$ is the time.
The change of electrical charge stored by the capacitor is:
$$ dq = C\ dV $$
where \$C\$ is the capacitance and \$V\$ is the voltage across the capacitor.
From these two equations, we obtain the famous capacitor current-voltage relationship:
$$ i\ dt = C\ dV \\ \Rightarrow i_C = C \ \frac{dV_C}{dt} $$
We can replace \$d/dt\$ with \$j\omega\$ (or \$2\pi f\$), so finally we obtain: $$ i_C = V_C \ j\omega\ C \\ \Rightarrow V_C = i_C \ \frac{1}{j\omega \ C} $$
The impedance of the capacitor is
$$ Z_C=\frac{i_C}{V_C}=\frac{1}{j\omega \ C}=\frac{1}{j\ 2\pi f \ C} $$
and the capacitive reactance is the magnitude of the impedance:
$$ X_C=|Z_C|=\frac{1}{2\pi f \ C} $$
So,