# AC Capacitance and Capacitive Reactance

The contrast to the current flow through an AC capacitor is called Capacitive Reactance and itself is inversely proportional to the feeding frequency.

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Capacitorsstore energy in the form of electric charge on conductor plates.When a capacitor is connected to the DC supply voltage, it charges up to the value of the voltage applied at a rate determined by the time constant.

A capacitor will maintain or hold this load indefinitely as long as the supply voltage is present. During this charging process, a charging current flows to the capacitor against any change in voltage at a rate equal to the speed of change of the electric charge on the plates. Therefore, a capacitor has a state of defiance against the current flowing to its plates.

The relationship between this charging current and the rate at which the feed voltage of the capacitors changes can be defined mathematically as follows: i = C (dv/dt), where C is the capacitance value of the capacitor in farce, and dv/dt is the rate at which the supply voltage changes over time. After "fully charged", the capacitor prevents more electrons from flowing to its plates as it becomes saturated, and the capacitor now acts as a temporary storage device.

A pure capacitor will maintain this load indefinitely on its plates, even if the DC supply voltage is removed. However, in a sinusoidal voltage circuit containing "AC Capacitance", the capacitor alternately charges and discharges at a rate determined by the feeding frequency.

When an alternative sinusoidal voltage is applied to the plates of an AC capacitor, the capacitor is charged first in one direction and then in the opposite direction, changing the polarity at the same rate as the AC supply voltage. This momentary change in voltage in the capacitor is faced with a certain period of time for this load to be deposited (or released) on the plates and given with V = Q/C.

### Sinusoidal Feed AC Capacitance

When the switch is turned off in the above circuit, a high current will begin to flow into the capacitor, as there is no load on the plates *at t = 0.*Sinusoidal supply voltage increases at maximum speed in positive direction when passing zero reference axis at a time given as V , 0^{o.}Since the exchange rate of the potential difference between the plates is now at its maximum value, the maximum amount of electrons will move from one plate to another, while the current flow to the capacitor will be at its maximum speed.

When the sinusoidal supply voltage reaches that point of 90 o in the wave^{form, it} begins to slow down, and over time the potential difference between the plates increases or decreases for a very short moment, so the current decreases to zero because there is no voltage ratio.It is 90^{to the} maximum potential difference on the capacitor (as it is where the V_{max} capacitor is fully charged and, as it is), a current into the capacitor is also saturated with electrons plates.

At the end of this moment, the supply voltage begins to decrease negatively towards the zero reference line at 180^{o.} Although the feeding voltage is still positive in nature, the capacitor begins to discharge some of its excess electrons into its plates to maintain a constant voltage. This causes the capacitor current to flow in the opposite or negative direction.

When the feed voltage waveform exceeds the zero reference axis point at 180 ^{o instantly,} the rate or inclination of the sinusoidal supply voltage is at the maximum but in the negative direction, as a result, the current flowing into the capacitor is at its maximum speed.Also ^{at} this point of 180, the potential difference between the plates is zero, since the amount of load is evenly distributed between the two plates.

Then, during this first half cycle between 0 ^{o} and 180 ^{o,} after the current reaches its maximum positive value, the applied voltage reaches the maximum positive value of a cycle (1/4ε), that is, a voltage current applied to a fully capacitive circuit reaches 90^{o} "PASS" as shown below or a quarter of a cycle.

### Sinusoidal WaveForms for AC Capacitance

In the second half cycle between 180^{o} and 360^{o,} the supply voltage changes direction and heads towards the negative peak value of 270^{o.}At this point, the potential difference between the plates neither decreases nor increases, and the current drops to zero.The potential difference on the capacitor is of the maximum negative value, the current does not flow into the capacitor and charges exactly as at^{that} point of 90, but in the opposite direction.

Since the negative supply voltage begins to increase positively towards the point 360^{on} the zero reference line, the full-charge capacitor must now lose some of its excess electrons and start self-discharge until the supply is provided to maintain a constant voltage as before.At 360^{o,} where the charging and discharge process restarts, the voltage reaches zero.

From the above voltage and current waveforms and description, we can see that the current always directs the voltage with 1/4 of a cycle or π / 2 = 90o "out of phase". The potential difference throughout the capacitor due to this charging and discharge process is the phase relationship between voltage and current in an AC capacitance circuit, the opposite of that of an AC Inductanc that we saw in the previous tutorial.

This effect can also be represented by a phaser diagram in which the voltage "delays" the current of 90^{o (LAGGING)} in a fully capacitive circuit. However, using the voltage as a reference, we can also say that the voltage of the current "prevents"leading" a quarter of a cycle or 90^{o} as shown in the vector diagram below.

### Phaser Diagram for AC Capacitance

So for a pure capacitor, we can say that V _{C} "delays" I _{C by} 90 ^{o} or I _{C}from V _{C}to 90 ^{o} "leadership".

There are many different ways to remember the phase relationship between flowing voltage and current in a pure AC capacitance circuit, but a very simple and easy-to-remember way is to use the evvocate phrase called "ICE".ICE current expansion I AC capacitances, E lectromotive force before the first.In other words, the current before the voltage in a capacitor equals I , C , E "ICE", and no matter what phase angle the voltage starts at, this expression always applies to the pure AC capacitance circuit.

## Capacitive Reactance

Therefore, as the capacitors are loaded and discharged,We know that electrons on capacitor plates oppose changes in voltage because the flow is directly proportional to the voltage change rate along the plates.Unlike a resistance where resistance to current flow is the actual resistance, aThe contrast to the current flow in the capacitor is called **Reassurance.**

Like resistance, reactance is measured in Ohm, but to distinguish it from a fully resistant R value, the X symbol is given, and the component in question issince it is a capacitor, aThe recessor is represented by **capacitive reassurance** (X _{C).}

Since capacitors charge and discharge proportionally to the voltage change rate between them, the faster the voltage changes, the more current will flow.Likewise, the slower the voltage changes, the less current will flow.This means that the recess of an AC capacitor is "inversely proportional" to the feeding frequency, as shown.

### Capacitive Reactance

Where: X _{C} is Capacitive Reactance in Ohm, frequency in ε Hertz, and C is AC capacitance in Farad.

When dealing with AC capacitance, we can also define capacitive reassurance in radians, where Omega is equal to ω 2nε.

From the formula above, we can see that the value of capacitive reassurance and therefore its general impedance (in Ohms) decreases towards zero by acting as a short circuit as the frequency increases.Similarly, as the frequency approaches zero or DC, the colorance of capacitors rises to infinity, acts like an open circuit, so capacitors block dc.

The relationship between capacitive reassurance and frequency is the opposite of the inductive rectal (X _{L)} we saw in the previous lesson.This means that capacitive recess is "inversely proportional to frequency" and has a high value at low frequencies and a low value at high frequencies, as shown.

### Capacitive Reactance Against Frequency

Capacitive reassurance of a capacitor decreases as the frequency on its plates increases.Therefore, capacitive reassurance is inversely proportional to the frequency.Capacitive reassurance is against current flow, but the electrostatic load (AC capacitance value) on the plates remains constant.

This means that it becomes easier for the capacitor to completely absorb the load change in its plates during each half cycle.In addition, as the frequency increases, the value of the current flowing into the capacitor increases because the voltage change rate increases along the plates.

We can present the effect of very low and very high frequencies on the reassurance of a pure AC Capacitance as follows:

In an AC circuit containing pure capacitance, the current flowing into the capacitor (electron flow) is given as follows:

and therefore, the rms current flowing into an AC capacitor will be defined as follows:

Where: IC = V / (1 / wC) (or IC = V/XC) is the current size and ε = + 90^{o}is the phase difference or phase angle between voltage and current. For a fully capacitive circuit, it opens the Ic Vc with 90^{o} or delays the Vc Ic by 90^{o.}

## Phaser Domain

In the phaser area, the voltage between the plates of an AC capacitor will be as follows:

and in polar form, this is written as follows: X _{C} ∠-90 ^{o}

## AC in a Series R + C Circuit

We saw from above that the current flowing into a pure AC capacitor^{prevents }the voltage from 90 o.But in the real world, it is impossible to have a pure **AC Capacitance,** since all capacitors will have a certain amount of internal resistance that leads to a leakage current in their plates.

Then we can think of our capacitor as a capacitor with R resistance serially with a capacitance.

If the capacitor has an "internal" resistance, then we need to represent the total impedance of the capacitor in series with a capacitance and in an AC circuit that contains both capacitance, C and resistance, the R voltage phaser in an AC circuit containing V. This combination will equate the voltage of the two components to the sum of phasers V _{R} and V _{C.}

This means that the current flowing into the capacitor will still direct the voltage, but depending on the R and C values it will be less than 90 ^{o}and gives us a phaser sum with the corresponding phase angle given between them by the Greek symbol phi.

Consider the following series RC circuit, where an omic resistance is connected serially with R , pure capacitance, C.

### Serial Resistance-Capacitance Circuit

In the RC series circuit above, we can see that the current flowing into the circuit is common for both resistance and capacitance, while the voltage consists of two component voltages, V _{R} and V _{C.}The resulting voltage of these two components can be found mathematically, but since the V _{R} and V _{C} vectors are out of ^{phase} 90, they can be added vectorically by creating a vector diagram.

To produce a vector diagram for AC capacitance, a reference or common component must be present.In a series of AC circuits, the current is common and therefore can be used as a reference source as the same current flows through the resistance towards the capacitansa.Separate vector diagrams for pure resistance and pure capacitance are given as follows:

### Vector Diagrams for Two Pure Components

For an AC Resistance, both voltage and current vectors are in the same phase, and therefore the voltage vector V _{R} is drawn on top of each other to scale on the current vector.We also know that in a pure AC capacitance circuit, the current directs voltage (ICE), so the voltage vector V _{C} is drawn on the back of the current vector 90 ^{o} (delayed) and on the same scale as the V _{R }as shown.

### Vector Diagram of The Resulting Voltage

In the vector diagram above, the OB line represents the horizontal current reference, and the OA line represents the voltage on the resistant component, which is in the same phase as the current.The OC line indicates capacitive voltage 90 o behind the current, so^{it} can still be seen^{that }the current directs the fully capacitive voltage by 90 o.Line OD gives us the resulting supply voltage.

Since the current directs the voltage at a pure capacitance by 90^{o, }the resulting phaser diagram V _{R} and V _{C,} drawn from individual voltage drops, represent a right-angle voltage triangle shown above as OAD.Next, we can also use pythagorean theorem to mathematically find the value of this voltage obtained during the resistance/capacitor (RC) circuit.

The two vector sums of V _{, R} = IR and V _{C} = IX _{C} will be as follows.

## Impedance of AC Capacitance

Empedans, which have Ohm units, Z , Ω, is the "TOTAL" contrast of the current flowing in an AC circuit containing both Resistance (real part) and Reassurance (imaginary part).A fully resistant impedance 0 will have ^{that} phase angle, while a fully capacitive impedance -90 will have ^{that} phase angle.

However, when the resistors and capacitors are connected in the same circuit, the total impedance will have a phase angle between 0 ^{o} and 90 ^{o,} depending on the value of the components used.Then the impedance of our simple RC circuit shown above can be found using the impedance triangle.

### RC Impedance Triangle

( Impedance ) ^{2} = ( Resistance ) ^{2} + ( *j* Reactance ) ^{2} where *j* 90 represents ^{that} phase shift.

This means that the negative phase angle is calculated as ε between voltage and current using pythagorean theorem.

### Phase Angle

### AC Capacitance Question Example 1

A single-phase sinusoidal AC supply voltage defined as V _{(t)} = 240 sin(314t – 20 ^{o)} is connected to a pure AC capacitance of 200uF.Determine the value of the current flowing into the capacitor and draw the resulting phaser diagram.

The voltage on the capacitor will be the same as the feed voltage.Converting this time zone value to a polar form gives us: V _{C} = 240 ∠-20 ^{o} (v) .Capacitive reassurance will be as follows: X _{C} = 1/( ω.200uF ) .Then the current flowing into the capacitor can be found using the Ohm law as follows:

In an AC capacitance circuit, there will be a phaser diagram with the current holding the voltage 90 ^{o} in front.

### AC Capacitance Question Example 2

A capacitor with an internal resistance of 10Ω and capacitance of 100uF is connected to the supply voltage given as V _{(t)} = 100 sin (314t).Calculate the peak current flowing into the capacitor.Also create a voltage triangle that shows individual voltage drops.

Capacitive recess and circuit impedance are calculated as follows:

Then the current and circuit that flows into the capacitor are given as follows:

The phase angle between current and voltage is calculated from the impedance triangle above as follows:

Then individual voltage drops around the circuit are calculated as follows:

The voltage triangle obtained for the peak values calculated later will be as follows:

## AC Capacitance Summary

In the pure **AC Capacitance** circuit, the voltage and current are both "out of phase" with the current leading to 90 ^{o's} voltage, and we can remember this using the phrase "ICE".The AC resistance value of a capacitor called impedance, ( Z ) the reactive value of a capacitor called "capacitive reactax" is associated with the frequency with X _{C.}In an *AC Capacity* circuit, this capacitive reacquance value is equal to 1/( 2πεC ) or 1/( jωC ).

So far, we have found that the relationship between voltage and current is not the same, and changes in all three pure passive components. The phase angle in resistance is 0^{o,}inductansta is +90^{o,}and capacitance is -90^{o.}

In the next tutorial about Serial RLC Circuits, we will look at the voltage-current relationship of three of these passive components when connected in the same series circuit when a fixed-state sinusoidal AC waveform is applied with the corresponding phaser diagram representation.