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Capacitors that are connected to a sinusoidal supply produce reactance from the effects of supply frequency and capacitor size
Capacitance in AC Circuits results in a time-dependent current which is shifted in phase by 90o with respect to the supply voltage producing an effect known as capacitive reactance.
When capacitors are connected across a direct current DC supply voltage, their plates charge-up until the voltage value across the capacitor is equal to that of the externally applied voltage. The capacitor will hold this charge indefinitely, acting like a temporary storage device as long as the applied voltage is maintained.
During this charging process, an electric current ( i ) flows into the capacitor which results in its plates beginning to hold an electrostatic charge. This charging process is not instantaneous or linear as the strength of the charging current is at its maximum when the capacitors plates are uncharged, decreasing exponentially over time until the capacitor is fully-charged.
This is because the electrostaic field between the plates opposes any changes to the potential difference across the plates at a rate that is equal to the rate of change of the electrical charge on the plates. The property of a capacitor to store a charge on its plates is called its capacitance, (C).
Thus a capacitors charging current can be defined as: i = CdV/dt. Once the capacitor is “fully-charged” the capacitor blocks the flow of any more electrons onto its plates as they have become saturated. However, if we apply an alternating current or AC supply, the capacitor will alternately charge and discharge at a rate determined by the frequency of the supply. Then the Capacitance in AC circuits varies with frequency as the capacitor is being constantly charged and discharged.
We know that the flow of electrons onto the plates of a capacitor is directly proportional to the rate of change of the voltage across those plates. Then, we can see that for capacitance in AC circuits they like to pass current when the voltage across its plates is constantly changing with respect to time such as in AC signals.
However, they do not like to pass current when the applied voltage is of a constant steady state value such as in DC signals. Consider the circuit below.

In the purely capacitive circuit above, the capacitor is connected directly across the AC supply voltage. As the supply voltage increases and decreases, the capacitor charges and discharges with respect to this change. We know that the charging current is directly proportional to the rate of change of the voltage across the plates with this rate of change at its greatest as the supply voltage crosses over from its positive half cycle to its negative half cycle or vice versa at points, 0o and 180o along the sine wave.
Consequently, the least voltage rate-of-change occurs when the AC sine wave crosses over at its maximum positive peak ( +VMAX ) and its minimum negative peak, ( -VMAX ). At these two positions within the cycle, the sinusoidal voltage is constant, therefore its rate-of-change is zero, so dv/dt is zero, resulting in zero current change within the capacitor. Thus when dv/dt = 0, the capacitor acts as an open circuit, so i = 0 and this is shown below.

At 0o the rate of change of the supply voltage is increasing in a positive direction resulting in a maximum charging current at that instant in time. As the applied voltage reaches its maximum peak value at 90o for a very brief instant in time the supply voltage is neither increasing or decreasing so there is no current flowing through the circuit.
As the applied voltage begins to decrease to zero at 180o, the slope of the voltage is negative so the capacitor discharges in the negative direction. At the 180o point along the line the rate of change of the voltage is at its maximum again so maximum current flows at that instant and so on.
Then we can say that for capacitors in AC circuits the instantaneous current is at its minimum or zero whenever the applied voltage is at its maximum and likewise the instantaneous value of the current is at its maximum or peak value when the applied voltage is at its minimum or zero.
From the waveform above, we can see that the current is leading the voltage by 1/4 cycle or 90o as shown by the vector diagram. Then we can say that in a purely capacitive circuit the alternating voltage lags the current by 90o.
We know that the current flowing through the capacitance in AC circuits is in opposition to the rate of change of the applied voltage. But just like resistors, capacitors also offer some form of resistance against the flow of current. For capacitors in AC circuits opposition is known as Reactance, and as we are dealing with capacitor circuits, it is therefore known as Capacitive Reactance. Thus capacitance in AC circuits suffer from Capacitive Reactance.
Capacitive Reactance in a purely capacitive circuit is the opposition to current flow in AC circuits only. Like resistance, reactance is also measured in Ohm’s but is given the symbol X to distinguish it from a purely resistive value. As reactance is a quantity that can also be applied to Inductors as well as Capacitors, when used with capacitors it is more commonly known as Capacitive Reactance.
For capacitors in AC circuits, capacitive reactance is given the symbol Xc. Then we can actually say that Capacitive Reactance is a capacitors resistive value that varies with frequency. Also, capacitive reactance depends on the capacitance of the capacitor in Farads as well as the frequency of the AC waveform and the formula used to define capacitive reactance is given as:

Where: F is in Hertz and C is in Farads. 2πƒ can also be expressed collectively as the Greek letter Omega, ω to denote an angular frequency.
From the capacitive reactance formula above, it can be seen that if either of the Frequency or Capacitance where to be increased the overall capacitive reactance would decrease. As the frequency approaches infinity the capacitors reactance would reduce to zero acting like a perfect conductor.
However, as the frequency approaches zero or DC, the capacitors reactance would increase up to infinity, acting like a very large resistance. This means then that capacitive reactance is “Inversely proportional” to frequency for any given value of Capacitance and this shown below:

The capacitive reactance of the capacitor decreases as the frequency across it increases therefore capacitive reactance is inversely proportional to frequency.
The opposition to current flow, the electrostatic charge on the plates (its AC capacitance value) remains constant as 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 through the capacitor increases in value because the rate of voltage change across its plates increases.
Then we can see that at DC a capacitor has infinite reactance (open-circuit), at very high frequencies a capacitor has zero reactance (short-circuit).
Find the rms current flowing in an AC capacitive circuit when a 4μF capacitor is connected across a 880V, 60Hz supply.

In AC circuits, the sinusoidal current through a capacitor, which leads the voltage by 90o, varies with frequency as the capacitor is being constantly charged and discharged by the applied voltage. The AC impedance of a capacitor is known as Reactance and as we are dealing with capacitor circuits, more commonly called Capacitive Reactance, XC
When a parallel plate capacitor was connected to a 60Hz AC supply, it was found to have a reactance of 390 ohms. Calculate the value of the capacitor in micro-farads.

This capacitive reactance is inversely proportional to frequency and produces the opposition to current flow around a capacitive AC circuit as we looked at in the AC Capacitance tutorial in the AC Theory section.
I have really enjoyed the lesson, it has a clear flow of information
Thanks
I have blips (minute power loss that is NOT power outage) in my AC household line, from time to time, and I also have a 1930’s clock with a Hammond spin-to-start motor. Every time a blip comes along, it stops the clock, and I have to get a ladder and climb the wall to take it down, and spin to start. I need a capacitor, or something, to boost the power along during these small blips. Is there anything available?
Please do a search for an uninterruptible power supply (UPS)
Wayne, thank you. I’m not too familiar with these kinds of things, but I saw an APC that read as if it may do what I need. I have contacted APC company, and awaiting their reply. Again. I appreciate your kindness. Bill
Thank u for the contribution
How does a start capacitor charge if it passes A/C current? Or does it assist the starting via the Capacitive reactance shifting the phase ?
Perform the experiment for Capacitive reactance and show the curves with a capacitor value of 47 micro Farads and show the current curve of simulation.
Hello.Thank you very much for such information you’re filling us with.I would like to know the formula for getting a bleeder resistor for a 105k 250v capacitor.I shall appreciate your feedback.
A bleed resistor, or “bleeder” resistor is placed across a capacitor to help safely discharge it when the supply voltage is removed. If the load resistance is very high, it can take seconds or even minutes for the capacitor’s terminal voltage to quickly discharge down to a safe level reducing the risk of electric shock.
The rate of discharge will depend on the RC time constant of the parallel combination which is simply R*C. The voltage across the capacitor is therefore given as: V(t) = Vp*e-(t/RC) where Vp is the start voltage, and t is the required time. Thus the resistive value of any parallel connected bleed resistor chosen will depend on how quickly the capacitor must discharge one power is removed, and the leakage current through the resistor when the capcitor is fully charged.
“During this charging process, a charging current, ( i ) will flow into the capacitor opposing any changes to the voltage at a rate that is equal to the rate of change of the electrical charge on the plates.”
Can you clarify which of the voltage(Source voltage or voltage between the capacitor terminals) is opposed by the charging current?
AC Capacitance Example No2. Should show 10^6 in the numerator to give the answer in mF.
For AC Capacitance Example No2,
C = 1/(2pi*60*390) = 1/(147046) = 0.0000068 Farads or 6.8 uF (micro-farads) as given in the tutorial, and not milli-farads (mF) as you have suggested.
I have two capacitors in parallel in Power factor correction panel each 3×96 uF.When i check the capacitance between any two lines,I get the value of 300 uF.can you please explain me how is this calculated?
I like discussions very much
I have a refrigeration compressor operating at 220V AC/50Hz mains supply. It was supplied with a start capacitor, which has failed, with markings as follows: 50uF +-10%, 330V, 50/60Hz
My local electrical supply shop does not stock this item. I need to purchase an equivalent capacitor that is going to provide same function as OEM part.
They say that the replacement item will not work because the voltage parameter is not he same. Their part is a: 50uF +-5%, 450V, 50/60HZ.
My question is: If it will be fitted into a 220V AC circuit what does the “max voltage” rating have to do with whether it will work for my compressor or not?
Is the supplier correct in their assertion that it will not work for me?
they are thieves. Any higher rated cap will work, it is just more expensive and will not break, and will not make more paid work for them.
The essential ratings for a start or run capacitor are: Capacitance (uF) and Nominal Voltage (VAC). The capacitance value of 50uF is exactly the same and the 450VAC rated capacitor has a better tolerance. 330VAC implies its a “Starting” capacitor, 450VAC implies its a “Run” capacitor. As the nominal voltage of 450V is higher than 330V, you can use a 450V rated capacitor inplace of a 330V rated one, although the physical size may be different.
Many thanks for the reply. Concurs with my own thoughts.
I will try the part and see if it has any problems.
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sir give me suggestion how make the C2 circuit in capacitor divider. the rating of divider is 300kv of c1 and c2 approximately 300 v so please advice me the design of c2 for divider.
Am really enjoying your ideas and content you have.
very useful lessons about capacitors and formulas for electronics engineering , diploma students and D I Y project students. publish any advance and latest research topics about electronics……thank you…..all the best….
How frequency is inversly related to capacitive reactance