The purpose of a capacitor is to store energy in the form of an electrical charge, on its plates. When a capacitor is connected across a DC supply voltage it charges up to the value of the applied voltage at a rate determined by its time constant.
A capacitor will maintain or hold this charge indefinitely as long as the supply voltage is present. During this charging process, a charging current, i flows into the capacitor opposed by any changes to the voltage at a rate which is equal to the rate of change of the electrical charge on the plates. A capacitor therefore has an opposition to current flowing onto its plates.
The relationship between this charging current and the rate at which the capacitors supply voltage changes can be defined mathematically as: i = C(dv/dt), where C is the capacitance value of the capacitor in farads and dv/dt is the rate of change of the supply voltage with respect to time. Once it is “fullycharged” the capacitor blocks the flow of any more electrons onto its plates as they have become saturated and the capacitor now acts like a temporary storage device.
A pure capacitor will maintain this charge indefinitely on its plates even if the DC supply voltage is removed. However, in a sinusoidal voltage circuit which contains “AC Capacitance”, the capacitor will alternately charge and discharge at a rate determined by the frequency of the supply. Then capacitors in AC circuits are constantly charging and discharging respectively.
When an alternating sinusoidal voltage is applied to the plates of an AC capacitor, the capacitor is charged firstly in one direction and then in the opposite direction changing polarity at the same rate as the AC supply voltage. This instantaneous change in voltage across the capacitor is opposed by the fact that it takes a certain amount of time to deposit (or release) this charge onto the plates and is given by V = Q/C. Consider the circuit below.
AC Capacitance with a Sinusoidal Supply
When the switch is closed in the circuit above, a high current will start to flow into the capacitor as there is no charge on the plates at t = 0. The sinusoidal supply voltage, V is increasing in a positive direction at its maximum rate as it crosses the zero reference axis at an instant in time given as 0^{o}. Since the rate of change of the potential difference across the plates is now at its maximum value, the flow of current into the capacitor will also be at its maximum rate as the maximum amount of electrons are moving from one plate to the other.
As the sinusoidal supply voltage reaches its 90^{o} point on the waveform it begins to slow down and for a very brief instant in time the potential difference across the plates is neither increasing nor decreasing therefore the current decreases to zero as there is no rate of voltage change. At this 90^{o} point the potential difference across the capacitor is at its maximum ( V_{max} ), no current flows into the capacitor as the capacitor is now fully charged and its plates saturated with electrons.
At the end of this instant in time the supply voltage begins to decrease in a negative direction down towards the zero reference line at 180^{o}. Although the supply voltage is still positive in nature the capacitor starts to discharge some of its excess electrons on its plates in an effort to maintain a constant voltage. This results in the capacitor current flowing in the opposite or negative direction.
When the supply voltage waveform crosses the zero reference axis point at instant 180^{o}, the rate of change or slope of the sinusoidal supply voltage is at its maximum but in a negative direction, consequently the current flowing into the capacitor is also at its maximum rate at that instant. Also at this 180^{o} point the potential difference across the plates is zero as the amount of charge is equally distributed between the two plates.
Then during this first half cycle 0^{o} to 180^{o}, the applied voltage reaches its maximum positive value a quarter (1/4ƒ) of a cycle after the current reaches its maximum positive value, in other words, a voltage applied to a purely capacitive circuit “LAGS” the current by a quarter of a cycle or 90^{o} as shown below.
Sinusoidal Waveforms for AC Capacitance
During the second half cycle 180^{o} to 360^{o}, the supply voltage reverses direction and heads towards its negative peak value at 270^{o}. At this point the potential difference across the plates is neither decreasing nor increasing and the current decreases to zero. The potential difference across the capacitor is at its maximum negative value, no current flows into the capacitor and it becomes fully charged the same as at its 90^{o} point but in the opposite direction.
As the negative supply voltage begins to increase in a positive direction towards the 360^{o} point on the zero reference line, the fully charged capacitor must now loose some of its excess electrons to maintain a constant voltage as before and starts to discharge itself until the supply voltage reaches zero at 360^{o} at which the process of charging and discharging starts over again.
From the voltage and current waveforms and description above, we can see that the current is always leading the voltage by 1/4 of a cycle or π/2 = 90^{o} “outofphase” with the potential difference across the capacitor because of this charging and discharging process. Then the phase relationship between the voltage and current in an AC capacitance circuit is the exact opposite to that of an AC Inductance we saw in the previous tutorial.
This effect can also be represented by a phasor diagram where in a purely capacitive circuit the voltage “LAGS” the current by 90^{o}. But by using the voltage as our reference, we can also say that the current “LEADS” the voltage by one quarter of a cycle or 90^{o} as shown in the vector diagram below.
Phasor Diagram for AC Capacitance
So for a pure capacitor, V_{C} “lags” I_{C} by 90^{o}, or we can say that I_{C} “leads” V_{C} by 90^{o}.
There are many different ways to remember the phase relationship between the voltage and current flowing in a pure AC capacitance circuit, but one very simple and easy to remember way is to use the mnemonic expression called “ICE”. ICE stands for current I first in an AC capacitance, C before Electromotive force. In other words, current before the voltage in a capacitor, I, C, E equals “ICE”, and whichever phase angle the voltage starts at, this expression always holds true for a pure AC capacitance circuit.
Capacitive Reactance
So we now know that capacitors oppose changes in voltage with the flow of electrons onto the plates of the capacitor being directly proportional to the rate of voltage change across its plates as the capacitor charges and discharges. Unlike a resistor where the opposition to current flow is its actual resistance, the opposition to current flow in a capacitor is called Reactance.
Like resistance, reactance is measured in Ohm’s but is given the symbol X to distinguish it from a purely resistive R value and as the component in question is a capacitor, the reactance of a capacitor is called Capacitive Reactance, ( X_{C} ) which is measured in Ohms.
Since capacitors charge and discharge in proportion to the rate of voltage change across them, the faster the voltage changes the more current will flow. Likewise, the slower the voltage changes the less current will flow. This means then that the reactance of an AC capacitor is “inversely proportional” to the frequency of the supply as shown.
Capacitive Reactance
Where: X_{C} is the Capacitive Reactance in Ohms, ƒ is the frequency in Hertz and C is the AC capacitance in Farads, symbol F.
When dealing with AC capacitance, we can also define capacitive reactance in terms of radians, where Omega, ω equals 2πƒ.
From the above formula we can see that the value of capacitive reactance and therefore its overall impedance ( in Ohms ) decreases towards zero as the frequency increases acting like a short circuit. Likewise, as the frequency approaches zero or DC, the capacitors reactance increases to infinity, acting like an open circuit which is why capacitors block DC.
The relationship between capacitive reactance and frequency is the exact opposite to that of inductive reactance, ( X_{L} ) we saw in the previous tutorial. This means then that capacitive reactance is “inversely proportional to frequency” and has a high value at low frequencies and a low value at higher frequencies as shown.
Capacitive Reactance against Frequency

Capacitive reactance of a capacitor decreases as the frequency across its plates increases. Therefore, capacitive reactance is inversely proportional to frequency. Capacitive reactance opposes current flow but the electrostatic charge on the plates (its AC capacitance value) remains constant.
This means it becomes easier for the capacitor to fully absorb the change in charge on its plates during each half cycle. Also as the frequency increases the current flowing into the capacitor increases in value because the rate of voltage change across its plates increases.

We can present the effect of very low and very high frequencies on the reactance of a pure AC Capacitance as follows:
In an AC circuit containing pure capacitance the current (electron flow) flowing into the capacitor is given as:
and therefore, the rms current flowing into an AC capacitance will be defined as:
Where: I_{C} = V/(1/ωC) (or I_{C} = V/X_{C}) is the current magnitude and θ = + 90^{o} which is the phase difference or phase angle between the voltage and current. For a purely capacitive circuit, Ic leads Vc by 90^{o}, or Vc lags Ic by 90^{o}.
Phasor Domain
In the phasor domain the voltage across the plates of an AC capacitance will be:
and in Polar Form this would be written as: X_{C}∠90^{o} where:
AC Across a Series R + C Circuit
We have seen from above that the current flowing into a pure AC capacitance leads the voltage by 90^{o}. But in the real world, it is impossible to have a pure AC Capacitance as all capacitors will have a certain amount of internal resistance across their plates giving rise to a leakage current.
Then we can consider our capacitor as being one that has a resistance, R in series with a capacitance, C producing what can be loosely called an “impure capacitor”.
If the capacitor has some “internal” resistance then we need to represent the total impedance of the capacitor as a resistance in series with a capacitance and in an AC circuit that contains both capacitance, C and resistance, R the voltage phasor, V across the combination will be equal to the phasor sum of the two component voltages, V_{R} and V_{C}.
This means then that the current flowing into the capacitor will still lead the voltage, but by an amount less than 90^{o} depending upon the values of R and C giving us a phasor sum with the corresponding phase angle between them given by the Greek symbol phi, Φ.
Consider the series RC circuit below where an ohmic resistance, R is connected in series with a pure capacitance, C.
Series ResistanceCapacitance Circuit
In the RC series circuit above, we can see that the current flowing into the circuit is common to both the resistance and capacitance, while the voltage is made up of the two component voltages, V_{R} and V_{C}. The resulting voltage of these two components can be found mathematically but since vectors V_{R} and V_{C} are 90^{o} outofphase, they can be added vectorially by constructing a vector diagram.
To be able to produce a vector diagram for an AC capacitance a reference or common component must be found. In a series AC circuit the current is common and can therefore be used as the reference source because the same current flows through the resistance and into the capacitance. The individual vector diagrams for a pure resistance and a pure capacitance are given as:
Vector Diagrams for the Two Pure Components
Both the voltage and current vectors for an AC Resistance are in phase with each other and therefore the voltage vector V_{R} is drawn superimposed to scale onto the current vector. Also we know that the current leads the voltage ( ICE ) in a pure AC capacitance circuit, therefore the voltage vector V_{C} is drawn 90^{o} behind ( lagging ) the current vector and to the same scale as V_{R} as shown.
Vector Diagram of the Resultant Voltage
In the vector diagram above, line OB represents the horizontal current reference and line OA is the voltage across the resistive component which is inphase with the current. Line OC shows the capacitive voltage which is 90^{o} behind the current therefore it can still be seen that the current leads the purely capacitive voltage by 90^{o}. Line OD gives us the resulting supply voltage.
As the current leads the voltage in a pure capacitance by 90^{o} the resultant phasor diagram drawn from the individual voltage drops V_{R} and V_{C} represents a right angled voltage triangle shown above as OAD. Then we can also use Pythagoras theorem to mathematically find the value of this resultant voltage across the resistor/capacitor ( RC ) circuit.
As V_{R} = I.R and V_{C} = I.X_{C} the applied voltage will be the vector sum of the two as follows.
The quantity represents the impedance, Z of the circuit.
The Impedance of an AC Capacitance
Impedance, Z which has the units of Ohms, Ω’s is the “TOTAL” opposition to current flowing in an AC circuit that contains both Resistance, ( the real part ) and Reactance ( the imaginary part ). A purely resistive impedance will have a phase angle of 0^{o} while a purely capacitive impedance will have a phase angle of 90^{o}.
However when resistors and capacitors are connected together in the same circuit, the total impedance will have a phase angle somewhere between 0^{o} and 90^{o} depending upon the value of the components used. Then the impedance of our simple RC circuit shown above can be found by using the impedance triangle.
The RC Impedance Triangle
Then: ( Impedance )^{2} = ( Resistance )^{2} + ( j Reactance )^{2} where j represents the 90^{o} phase shift.
This means then by using Pythagoras theorem the negative phase angle, θ between the voltage and current is calculated as.
Phase Angle
AC Capacitance Example No1
A singlephase 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 phasor diagram.
The voltage across the capacitor will be the same as the supply voltage. Converting this time domain value into polar form gives us: V_{C} = 240 ∠20^{o} (v). The capacitive reactance will be: X_{C} = 1/( ω.200uF ). Then the current flowing into the capacitor can be found using Ohms law as:
With the current leading the voltage by 90^{o} in an AC capacitance circuit the phasor diagram will be.
AC Capacitance Example No2
A capacitor which has an internal resistance of 10Ω’s and a capacitance value of 100uF is connected to a supply voltage given as V_{(t)} = 100 sin (314t). Calculate the current flowing into the capacitor. Also construct a voltage triangle showing the individual voltage drops.
The capacitive reactance and circuit impedance is calculated as:
Then the current flowing into the capacitor and the circuit is given as:
The phase angle between the current and voltage is calculated from the impedance triangle above as:
Then the individual voltage drops around the circuit are calculated as:
Then the resultant voltage triangle will be.
AC Capacitance Summary
In a pure AC Capacitance circuit, the voltage and current are both “outofphase” with the current leading the voltage by 90^{o} and we can remember this by using the mnemonic expression “ICE”. The AC resistive value of a capacitor called impedance, ( Z ) is related to frequency with the reactive value of a capacitor called “capacitive reactance”, X_{C}. In an AC Capacitance circuit, this capacitive reactance value is equal to 1/( 2πƒC ) or 1/( jωC )
Thus far we have seen that the relationship between voltage and current is not the same and changes in all three pure passive components. In the Resistance the phase angle is 0^{o}, in the Inductance it is +90^{o} while in the Capacitance it is 90^{o}.
In the next tutorial about Series RLC Circuits we will look at the voltagecurrent relationship of all three of these passive components when connected together in the same series circuit when a steady state sinusoidal AC waveform is applied along with the corresponding phasor diagram representation.
Just simplified am able to understand well
good morning plz accept of a student for you
Greetings! Sir, Ic=V/(Xc) & Ic=V/(wc) is this both equal.? Because Xc=1/wc..Thanks for your work.
Capacitive Reactance, Xc = 1/2pifC = 1/wC and Ic = Vc/Xc
C = V/(ωC)
is this right?
The tutorial does not state that equation, but does state: Ic = V/(1/ωC) = V/Xc
Xc = 1/wc . so, Ic=v/(1/wc)…Ic=v.wc…..am i wrong anywhere.?sorry am a slow learner…
Nice Explanation bro. But, the formula to find degree(ɸ) you have mentioned wrongly. There should be, tan inverse of (Xc/R), instead of tan inverse of (ɸ).
thank you.
which is the correct way to represent the capacitive reactance ? xc= 1/ wc or xc = 1/jwc. for example ,in some cases the xc is represented with magnitude only and in some cases it is represented with magnitude and imaginary part. so which is correct? magnitude or magnitude with imaginary part.
Capacitive reactance, Xc is defined as 1/(2pifC) in the frequency domain. As the radian is simply a unit of angular measurement, like the degree, we can correctly say that one cycle (1Hz) is equal to 2pi radians, and therefore angular velocity of a waveform can be written as 2pif radians per second. In Electrical & Electronic Engineering it is common to represent this quantity “2pif ” by the small Greek letter omega, (w). Then capacitive reactance can therefore be correctly defined and used as 1/wC or 1/(2pifC) ohms.
As the current leads the voltage (ICE) in a capacitor by 90 degs, we can define this phase angle by the use of the joperator, representing any angle between 0 degs and 90 degs. Then for a pure capacitance an angle of 90 degs or j is used with Xc. For example, Xc = 10 ohms is j10 ohms. Reactance in complex form is generally used when calculating capacitor voltages and currents within a circuit to give a magnitude and angle of the resulting complex impedance. Then either magnitude or magnitude with imaginary part can be used depending upon what it is you are calculating.
Hello, Any procedure to counter this phase shift? is it adding series of inductor would do? Because capacitor have 90deg phase angle while inductor hava 90deg phase angle in theoretically.
Hi again,
Just a correction on some typos in previous post.
Why do we need a rate of change to give rise to current flow and not just the potential diffrence between the plate and and ac supply. I’m finding it hard to accept that when the maximum emf is present this doesn’t push the most current. Do I have to think of it as an accelerating force and if so why?
Many thanks
For a capacitor, the stored charge Q is given by CV in coulombs. The flow of charge in coulombs per second is given the name “amperes”. The voltage to current ratio v/i of a capacitor connected to an AC supply is not linear as it is for a resistor, R. The current flowing into the capacitor is equal to the rate of change of the charge across the capacitors plates giving the flow of current as: i(t) = Cdv/dt.
Then the current flowing through the capacitor is not proportional to the voltage but to its “rate of change” giving a higher current flow for higher variations or change of the voltage. For a capacitor the opposition to change is not resistance (as for the resistor), but reactance X and, particularly, capacitive reactance Xc. So the higher the frequency, the higher the variations of change in voltage with respect to time, the lower the capacitive reactance, Xc and the higher the current, i for a given voltage supply.
This article does not mention about Real Power, Reactive Power, etc. therefore it is not thorough knowledge. Do not follow is my recommendation.
The tutorial is about Capacitive Reactance and not Power. There are other tutorials covering that.