# What is capacitive reactance and how to calculate it?

The **capacitive reactance** is a capacitor load resistance element loop flow regulator which alternating current opposes the passage of current.

In a circuit consisting of a capacitor and activated by an alternating current source, the capacitive reactance X _{C} can be defined as follows:

X _{C} = 1 / ωC

Or also:

X _{C} = 1/2πfC

Where C is the capacitance of the capacitor and ω is the angular frequency of the source, related to the frequency f by:

ω = 2πf

Capacitive reactance depends on the inverse of the frequency; therefore, at high frequencies it tends to be small, while at low frequencies the reactance is large.

The International System unit for measuring capacitive reactance is ohm (Ω), since the capacitance of capacitor C is far away, (abbreviated F) and the frequency is expressed in inverse seconds (s ^{-1} ).

While the load lasts, an alternating voltage and current are established through the capacitor, whose amplitudes or maximum values, denoted respectively as V _{C} and I _{C} , are related by capacitive reactance in a manner analogous to Ohm’s law:

V _{C} = I _{C} ⋅ X _{C}

In a capacitor, the voltage is 90° behind the current, or 90° ahead of the current, whichever is preferred. Anyway, the frequency is the same.

When X _{C} is very large, the current tends to be small and, making the value of X _{C} tend to infinity, the capacitor behaves like an open circuit and the current is zero.

__How to Calculate Capacitive Reactance__

__How to Calculate Capacitive Reactance__

An example of how to calculate capacitive reactance, suppose uF capacity 6 is connected to an alternating current of 40V and frequency *f* of 60Hz.

To find the capacitive reactance, the definition given at the beginning is used. The angular frequency ω is given by:

ω = 2πf = 2π x 60 Hz = 377 s ^{-1}

So this result is replaced in the definition:

X _{C} = 1 / ωC = 1 / (377 s ^{-1} x 6 x10 ^{-6} F) = 442.1 ohm

Now let’s see the amplitude of the current flowing in the circuit. As the source offers a voltage amplitude V _{C} = 40 V, we use the ratio of capacitive reactance, current and voltage to calculate the maximum current or current amplitude:

I _{C} = V _{C} / X _{C} = 40 V / 442.1 ohm = 0.09047 A = 90.5 m A.

If the frequency becomes too large, the capacitive reactance becomes small, but if the frequency becomes 0 and we have a forward current, the reactance will tend to be infinite.

**Capacitor current and voltage**

When a capacitor is connected to an alternating current source, as it oscillates and changes its polarity, the capacitor charges and discharges alternately.

For a frequency of 60 Hz, as in the example, the voltage is positive 60 times a second and negative another 60 times a second.

As the voltage increases, it drives current in one direction, but if the capacitor is discharging, reverse current is produced that opposes the former.

If v _{C} (t) = V _{m} sin ωt, knowing that the capacity is the ratio between the load and the voltage, we have the load:

C = q / V → q (t) = CV = CV _{m} sin ωt

And, having the charge as a function of time, we have the current, which is the derivative of this:

i _{C} (t) = CV _{m} ω cos ωt

But sine and cosine are related by: cos α = sin (α + π / 2), so:

i _{C} (t) = CV _{m} ω sin (ωt + π / 2) = I _{C} sin (ωt + π / 2)

With I _{C} = CV _{C} ω

As you can see, there is a difference of 90º in the advance of the current in relation to the voltage, as mentioned in the beginning.

In describing this type of circuit, the concept of *phasor* is used , which is very similar to a vector and allows any alternating quantity, such as current, voltage or impedance, to be represented in the complex plane.

The following figure shows, on the right, the voltage and current phasors in the capacitor, which form a 90º angle to each other, which is the phase change between the two.

On the left are the respective graphs, of different amplitudes, but with the same frequency. Over time, the current advances to voltage and when it is maximum, current is zero and when voltage is zero, current is maximum, but with the polarity reversed.

__Complex Capacitor Impedance__

__Complex Capacitor Impedance__

In a circuit with resistors, capacitors, and inductances, reactance is the imaginary part of impedance Z, a complex quantity that, in AC circuits, plays a role similar to electrical resistance for direct current circuits.

In fact, the impedance of a circuit is defined as the ratio of voltage to current:

Z = V / I

For a capacitor or capacitor, its impedance is given by the quotient:

Z _{C} = v (t) / i (t) = V _{C} sin ωt / I _{C} sin (ωt + π / 2)

One way to express voltage and current as phasors is to indicate the amplitude and phase angle (polar shape):

v (t) = V _{C} ∠ 0º

i (t) = I _{C} ∠ 90º

Therefore:

Z _{C} = V _{C} ∠ 0º / I _{C} ∠ 90º = (V _{C} / I _{C} ) ∠ 0º -90º =

= V _{C } / CV _{C} ω ∠ -90º = (1 / ωC) ∠ -90º =

Z _{C} = (-j) X _{C}

That is, the capacitor’s impedance is its capacitive reactance multiplied by the negative of the imaginary unit.

**Impedance of a series RC circuit**

The impedance of an alternating current circuit with resistors, capacitors and inductors can also be binomial represented by:

Z = R + jX

In this equation, R represents the resistance, which corresponds to the real part, j is the imaginary unit, and X is the reactance, which can be either capacitive or inductive or a combination of both, if these elements are present at the same time in the circuit.

If the circuit contains a resistor and capacitor in series, its impedance is:

Z = Z _{R} + Z _{C }

Since voltage and current are in phase across the resistance, the resistive impedance is simply the value of resistance R.

In the case of capacitive impedance, we have already seen that Z _{C} = -jX _{C} , therefore, the impedance of the RC circuit is:

Z = R – jX _{C} = R – j (1 / ωC)

For example, in the circuit shown below, whose source is of the form:

100 V ⋅ sin (120πt)

Observing that ω = 120π, the impedance is:

Z = 83.0 – j [(1 / (120π ⋅ 6 x 10 ^{-6} )] ohm = 83.0 – 442.1 j ohm.

**Capacitive reactance applications**

High-pass filters, low-pass filters, bridge-type circuits for measuring capacitances and inductances, and phase shift circuits are among the main applications for circuits that contain capacitive reactances, in combination with electrical inductances and resistances.

For sound equipment, some speakers come with separate *woofer* (larger) speakers for low frequencies and a *tweeter* or small speaker for high frequencies. This improves performance and audio quality.

They use capacitors that prevent low frequencies from reaching the tweeter, while an inductor is added to the woofer to avoid high frequency signals, since the inductance has a reactance proportional to frequency: X _{L} = 2πfL .