Capacitors have a special way of opposing alternating current (AC) which is called capacitive reactance. This is like an internal resistance in the capacitor which changes based on the frequency of the electricity flowing through it.
Unlike normal resistance which stays the same, no matter how fast the electricity changes (frequency), capacitive reactance is affected by this frequency. Both are measured in ohms (Ω).
Inductors, and capacitors have opposite effects on AC current based on frequency. Inductive reactance (XL) increases as the frequency goes up while capacitive reactance (XC) decreases.
Remember from the RC Network tutorial, when you apply a steady voltage (DC) to a capacitor it initially pulls current to charge up. Once charged it stops conducting current. Likewise when you decrease the voltage, the capacitor releases its stored charge and conducts current again.
In an AC circuit the voltage is not steady, instead it constantly flips between positive and negative, quite like a seesaw, with a speed determined by the frequency of the electricity.
You can compare it with a sine wave, that is the shape of this changing voltage. Because of this, the capacitor is constantly being charged and discharged following the rhythm of the AC signal.
As the capacitor goes through these charge and discharge cycles, current flows through the capacitor. This flow is not unlimited rather it is restricted by the capacitors internal opposition, which we call capacitive reactance (XC) and this is measured in ohms (Ω).
Here is the main difference from regular resistance. Capacitive reactance is not fixed, for example like 100 ohms, 1 kilohm, or 10 kilohms (remember all those values follow Ohm's Law). But instead, this changes depending on the frequency of the AC signal. So, any adjustments in the frequency will significantly impact the capacitor's capacitive reactance value.
Things gets interesting, when we change the frequency of the AC signal flowing through the capacitor. Here is why:
Higher Frequency Lower Resistance: As the frequency increases, the capacitors resistance (reactance) actually decreases (measured in ohms). It is like the capacitor is letting more current flow through it with ease.
Lower Frequency Higher Resistance: On the other hand if the frequency slows down, the capacitor's resistance (reactance) increases. It is like the capacitor is putting up more fight against the current flow. This dependence on frequency is what we call the capacitor's complex impedance.
This basically means that the capacitor's opposition to current is not a simple fixed value. Here is why this happens, compare the electrons on the capacitor plates like tiny dancers.
At higher frequencies they seem to jump back and forth between the plates faster allowing more current to flow. It is like they are following a faster beat.
This makes the capacitor appear to have less internal resistance.
So a capacitor in a circuit with changing frequencies is truly frequency dependent. Its resistance (reactance) which is symbolized by XC (in ohms, is just like regular resistance - R), changes based on the oscillations (frequency) of the AC signal.
The formula to calculate this changing resistance (reactance) is given as below:
Formula for Calculating Capacitive Reactance
XC = 1 / 2πfC
- In the above formula
- XC represents the Capacitive Reactance in Ohms, (Ω)
- π (pi) is equal to 3.142 (decimal) or as 22 / 7 (fraction)
- ƒ indicates the Frequency in Hertz, (Hz)
- C denotes the Capacitance in Farads, (F)
Solving a Capacitance Reactance Problem#1
We are able to determine the resistance that a capacitor provides to AC (alternating current) at a certain frequency. Measured in ohms (Ω), this resistance is known as capacitive reactance and is dependent on the frequency of the current as well as the value of the capacitor.
Calculating Capacitive Reactance
Given a 100 nanofarad (nF) capacitor, we have to calculate its capacitive reactance at two different frequencies: 1 kHz (kilohertz) and 10 kHz.
The formula for capacitive reactance (XC) is:
XC = 1 / (2 * π * f * C)
Calculating Reactance at 1 kHz:
- f = 1 kHz = 1000 Hz (convert kilohertz to hertz)
- C = 100 nF = 100 * 10-9 F (convert nanofarads to farads)
Plug the values into the formula:
XC = 1 / (2 * π * 1000 Hz * 100 * 10-9 F) XC ≈ 1591.55 ohms (round to two decimal places)
Therefore the capacitive reactance of the 100 nF capacitor at 1 kHz is approximately 1591.55 ohms.
Calculating Reactance at 10 kHz:
- f = 10 kHz = 10000 Hz (convert kilohertz to hertz)
Substituting the new frequency value into the formula, keeping the capacitance (C) the same:
XC = 1 / (2 * π * 10000 Hz * 100 * 10-9 F) XC ≈ 159.15 ohms (round to two decimal places)
Therefore the capacitive reactance of the 100 nF capacitor at 10 kHz is approximately 159.15 ohms.
As expected the capacitive reactance decreases as the frequency increases. This confirms the capacitor's frequency-dependent behavior.
Our experiment with the 100 nanofarad capacitor confirms what we expected:
As we increased the frequency from 1 kHz to 10 kHz the capacitor's resistance (reactance, XC) dropped significantly from about 1592 ohms to just 159 ohms. This always happens because capacitive reactance and frequency have an inverse relationship.
The greater the frequency, the more current the capacitor can withstand at the same voltage.
Graph Showing Capacitive Reactance against Frequency
This frequency dependency may be illustrated by plotting the capacitor's reactance (XC in ohms) versus frequency. The reactance decreases with increasing frequency.
We may also determine the frequency at which a capacitor will have a specific capacitive reactance (XC) value, simply by rearranging the above reactance formula.
Solving another Capacitive Reactance Problem#2
Let's say, we have a capacitor with a capacitance of 1 microfarad (uF). We want to calculate at what frequency will this capacitor resist current flow with a reactance of 100 ohms (Ω)?
Solution:
The formula that links a capacitor's reactance (Xc), capacitance (C), and frequency (f) is:
XC = 1 / (2 * π * f * C)
We can rearrange the equation to solve for the frequency (f) like this:
f = 1 / (2 * π * XC * C)
Now, we can input the known values:
- Xc = 100 Ω (given)
- C = 1 uF = 1 x 10-6 F (convert microfarads to farads)
Calculation:
f = 1 / (2 π * 100 Ω * 1 x 10-6 F)
f ≈ 1591.55 Hz
Therefore the frequency at which the 1uF capacitor may have a reactance of 100 Ω is approximately is 1591.55 Hz.
Alternatively, by knowing the applied frequency and the reactance value of the capacitor at that frequency, we may calculate the capacitor's Farad value.
Solving another Capacitive Reactance Problem#3
Now, imagine we have a 60 Hz AC power supply. If a capacitor connected to this supply exhibits a reactance of 100 ohms (Ω), let's calculate what capacitance value will the capacitor have?
Solution:
The formula for capacitive reactance (XC) of a capacitor is:
XC = 1 / (2 * π * f * C)
We are given the values for XC and f, and want to solve for C. Let's rearrange the formula to isolate C:
C = 1 / (2 * π * f * XC)
Now, we can substitute the known values:
- XC = 100 Ω (given)
- f = 60 Hz (given)
Calculation:
C = 1 / (2 * π * 60 Hz * 100 Ω)
C ≈ 2.652 x 10-6 F
Therefore the capacitor must have a capacitance of approximately 2.652 microfarads (uF) (it is because a 1 microfarad is equal to 10-6 farads).
The above examples explain how a capacitor coupled to a frequency-varying power source acts like a resistor whose resistance changes with frequency. This is due to the inverse connection between frequency and reactance (X) of the capacitor. A capacitor, for example, has a high reactance value at very low frequencies, acting as an open circuit.
On the other hand the capacitor's reactance drastically decreases at extremely high frequencies simulating a short circuit. As a result our capacitor has infinite reactance at zero frequency, or in a steady-state DC situation. This means that it acts as a "open circuit" between its plates, which totally prevents current passage.
Revising the Voltage Divider
We will now investigate voltage divider circuits, by extending on our knowledge of resistors in series where the voltage drop across each resistor (R) is dependent on its resistance value.
In order to split the applied supply voltage (Vs), as per the ratio of R2 to the total of R1 and R2 (R2 / (R1 + R2)), these circuits use two resistors (R1 and R2) connected in series.
As a result the output voltage (Vo) will equal half of the source voltage (Vs) when R1 and R2 have identical values. On the other hand, the output voltage will vary proportionately to the input voltage for any value of R2 that is larger or lower than R1. This idea is demonstrated by the circuit below.
Voltage Divider Circuit
VOUT = VIN[R2 / (R1 + R2)]
We have shown that the reactance (Xc) of a capacitor changes with applied frequency and adds to its total impedance. The voltage drop across these two components will now vary in response to frequency fluctuations if we swap out resistor R2 in the earlier circuit with a capacitor. This is so, because the reactance of the capacitor affects its impedance directly.
However the impedance of resistor R1 does not vary as frequency does. Frequency variations have little effect on fixed-value resistors like R1. The capacitive reactance of the capacitor at a certain frequency, thus, determines the voltage dropped across R1 and, in turn, the output voltage.
This phenomena leads to the formation of a frequency-dependent RC voltage divider circuit. Through the application of this principle, passive high-pass and low-pass filters may be constructed by simply substituting a selected capacitor for one of the voltage divider resistors, as shown in the schematics below.
Low pass filter
High Pass Filter
Capacitors are ideal for the implementation in DC power supply smoothing circuits and AC filter circuits due to their capacitive reactance. Unwanted frequencies could be bypassed by the capacitor thanks to its capacitive reactance, which minimizes their effect on the output voltage. This indicates that the capacitor can readily filter out high-frequency "ripples" or variations in the signal, producing a smoother and cleaner output.
Conclusions
Consider a capacitor, which functions similarly to a special resistor, whose resistance varies depending on the frequency (or speed) at which electricity passes through it.
The capacitor exhibits strong reactance, or extreme resistance to current, at very low frequencies.
This looks similar to an open circuit where current finds it difficult to flow.
Oppositely, a capacitor with low reactance makes it extremely simple for electricity to flow through it, at very high frequencies.
This resembles a short circuit where electricity flows freely.
This behavior is depicted in the graph you saw previously.
Important Things to Bear in Mind:
Capacitors function similar to frequency-controlled resistors.
Low frequency = High reactance (open circuit effect)
High frequency = Low reactance (short circuit effect)
In our next discussion we will investigate how passive low-pass filters employ capacitive reactance to obstruct undesirable high-frequency signals and permit only low frequencies to go through.
References: Capacitive Reactance