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# Capacitive Impedance | Circuit Theory

Capacitive impedance is the ratio of phasor voltage to phasor current in a capacitor. It also means the opposition of a capacitor to an alternating current. Impedance of a capacitor is dependent on the frequency of the voltage or current across it.

### What is Impedance?

When electrical current flows through a material, it meets resistance primarily due to the nature of the material. Conductors will resist less and insulators will resist more. Voltage pushes the current; the stronger the voltage, the higher the current. And since the higher the resistance, the lesser the current, we have this math relationship:

$I=\frac{V}{R}$

This is popularly known as Ohm’s Law. Here, voltage and current are direct and have magnitude only.

However, alternating current and voltages both have magnitude and phase since they are sine/cosine quantities. The equation for household AC mains, for example, is :

$v(t) = 325.27sin(377t)$

Phasors are invented to avoid exhausting calculations of sinusoidal equations. In circuit theory,  sinusoids are functions of time and are therefore time-domain quantities.

Phasors are complex numbers and can be written in rectangular or polar form.

$X + jY \Rightarrow A\angle \theta$

Here, X is the real part and Y is the imaginary part.  j is the imaginary number equal to √-1.

It is customary to use j instead of i in circuits to avoid confusion with the symbol for current.

In polar form, A is the magnitude and ϴ is the phase.

Conversion between two phasor forms are as follows:

Rectangular to Polar:

$X = Acos \theta$

$Y = Asin \theta$

Polar to Rectangular:

$A=\sqrt{X^2+Y^2}$

$\theta = tan{-1}\frac{Y}{X}$

A cosine equation easily converts to polar phasor:

$Acos(\omega t + \theta) \Rightarrow A\angle \theta$

And so instead of writing AC voltage like this:

$v=12cos(10t+10^{ \circ })$

We can just have it like this:

$\mathbf{V}=12\angle 10^{\circ }$

Note that phasor quantities are often in bold, capital letters.

Also instead of dealing with this:

$p=12cos(10t+10^{\circ})*10cos(10t-30^{\circ})$

We can just use phasor multiplication which is much simpler:

$\boldsymbol{P}=12 \angle 10^{\circ}*10 \angle-30^{\circ}$

$\boldsymbol{P}=12*10 \angle (10^{\circ}-30^{\circ})$

$\boldsymbol{P}=120 \angle -20^{\circ}$

But if voltage and current are now phasors, does Ohm’s Law’s equation hold true? The answer is no since you now have both magnitude and phase:

$\frac{\boldsymbol{V}}{\boldsymbol{I}}=\frac{V\angle \theta_{V}}{I\angle \theta_{I}}=\frac{V}{I}\angle (\theta_V - \theta_I)$

So, what do we now call the ratio of phasor voltage and current? Not resistance but impedance (Z).

$\boldsymbol{Z}=\frac{\boldsymbol{V}}{\boldsymbol{I}}=\frac{V}{I}\angle (\theta_V - \theta_I)$

### Impedance and Reactance

If an impedance has no phase difference:

$\boldsymbol{Z}=\frac{V}{I}\angle (0)$

In rectangular form,

$\boldsymbol{Z}=\frac{V}{I}cos \theta + j\frac{V}{I}sin \theta$

$\boldsymbol{Z}=\frac{V}{I}cos 0 + j\frac{V}{I}sin 0$

$\boldsymbol{Z}=\frac{V}{I}$

$\boldsymbol{Z}=R$

As you can see, impedance becomes equal to resistance. In fact, in a resistor, there is no phase difference between the current and the voltage.

What if the phase is non-zero? Remember that the imaginary component sinϴ will only be zero if ϴ is zero (or 180 but it’s supplementary to 0). Thus, there will always be an imaginary component in impedance if the phase is not zero.

$\boldsymbol{Z}=R + jX$

This imaginary component of impedance is called reactance. Moreover, the real part of impedance is the resistance.

### Capacitive Impedance

Now with a capacitor, the voltage and current have a relationship of:

$i_c=C \frac{dv_c}{dt}$

The equation is due to the ability of the capacitor to hold electric charge. This means that the current in a capacitor is proportional to the change of voltage in a given point in time. In short, a steady voltage produces no current across a capacitor.

Now besides sine/cosine equations, you can also convert differential equations to phasor too.

$\frac{dv_c}{dt}\Rightarrow j\omega \boldsymbol{V_C}$

Here, ω is the radian frequency of the voltage or current.

So, the current-voltage differential equation for capacitors now becomes:

$\boldsymbol{I_C}=C(j\omega\boldsymbol{V_C})$

Solving for the ratio of phasor voltage and current:

$\frac{\boldsymbol{V_C}}{\boldsymbol{I_C}}=\frac{1}{j\omega C}$

Hence, the impedance of the capacitor is:

$\boldsymbol{Z_C}=\frac{1}{j\omega C}$

As you can see, this value changes when the radian frequency changes. The higher the value of ω, the lower the Z and vice versa.

Finally, since the impedance of a capacitor is an imaginary number, it is also called capacitive reactance.

You can also write the above equation like this:

$\boldsymbol{Z_C}=\frac{-j}{\omega C}$

Converting to polar form,

$\boldsymbol{Z_C}=\sqrt{0^2+({\frac{-1}{\omega C}})^2} \angle tan^{-1} \frac{\frac{-1}{\omega C}}{0}$

$\boldsymbol{Z_C}=\frac{1}{\omega C} \angle 90^{\circ}$

$\boldsymbol{Z_C}=\frac{V_C}{I_C} \angle 90^{\circ}$

This means that in a capacitor, the voltage is ahead of current in phase by 90°.

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