Electromagnetic Induction, AC Circuits, and Electrical Technologies

56 Reactance, Inductive and Capacitive

Learning Objectives

  • Describe how current and voltage are related in simple AC circuits containing inductors, capacitors, and resistors.
  • Calculate inductive and capacitive reactance.
  • Calculate the current or voltage in simple AC circuits containing inductors, capacitors, or resistors.

In previous chapters, we studied how capacitors and inductors behave when connected to a direct current (DC) source. A capacitor gradually charges until its current becomes zero, while an inductor resists sudden changes in current before eventually behaving like an ordinary wire.

Most electrical power systems, however, operate using alternating current (AC), in which both the voltage and current continuously change direction. Because the current is always changing, capacitors and inductors never reach the steady conditions observed in DC circuits. Instead, they continuously store and release electrical or magnetic energy.

In this section, we examine how inductors, capacitors, and resistors respond to a sinusoidal AC voltage. We will introduce the concept of reactance, which plays a role similar to resistance in AC circuits, and explain why voltage and current are not always synchronized. These ideas are essential for understanding many medical devices, including electrocardiographs (ECGs), electroencephalographs (EEGs), magnetic resonance imaging (MRI) systems, and the electronic filters used to remove unwanted electrical noise from physiological signals.

Inductors and Inductive Reactance

Consider an ideal inductor connected directly to an AC voltage source, as shown in Figure 56.1. We assume that the resistance of the inductor is negligible, so its behavior is determined almost entirely by its inductance. The figure also shows how the voltage across the inductor and the current through it vary with time.

An ideal inductor connected to an alternating-current source. A graph compares voltage and current as functions of time, showing that the voltage reaches its maximum one-quarter of a cycle before the current.
Figure 56.1. (a) An AC voltage source connected to an ideal inductor. (b) Voltage and current versus time, illustrating that the voltage reaches its maximum one-quarter of a cycle before the current.

Notice that the graph begins with the voltage at its maximum value, while the current is initially zero. As the voltage drives current through the inductor, the current increases gradually rather than immediately reaching its maximum value. This behavior is similar to what happens when a DC voltage is first applied to an inductor—the changing current generates a back emf that opposes the change.

As the applied voltage becomes negative (point a), the current begins to decrease. At point b, the voltage reaches its most negative value while the current passes through zero. The current then reverses direction and becomes negative. Later, when the voltage becomes positive again (point c), the current becomes less negative until it again reaches zero at point d. The cycle then repeats.

AC Voltage in an Inductor

When a sinusoidal voltage is applied to an ideal inductor, the voltage leads the current by one-quarter of a cycle, corresponding to a phase difference of 90°. Equivalently, the current lags the voltage by 90°.

The current lags because an inductor opposes changes in current. Whenever the current changes, the inductor produces a back emf given by

[latex]V=-L\frac{\Delta I}{\Delta t}.[/latex]

This back emf limits the current in much the same way that electrical resistance limits current in a DC circuit. In AC circuits, however, this opposition depends on both the inductance and the frequency of the applied voltage. For this reason, we introduce a quantity called the inductive reactance.

The rms current through an inductor is given by an AC version of Ohm's law:

[latex]I=\frac{V}{X_L},[/latex]

where [latex]V[/latex] is the rms voltage across the inductor and [latex]X_L[/latex] is the inductive reactance, defined as

[latex]X_L=2\pi fL.[/latex]

Here, [latex]f[/latex] is the frequency of the AC source in hertz and [latex]L[/latex] is the inductance. The inductive reactance has units of ohms, making it directly comparable to electrical resistance. Unlike ordinary resistance, however, reactance does not dissipate energy as heat. Instead, it arises because energy is continually stored in and released from the magnetic field of the inductor.

The expression for [latex]X_L[/latex] also explains how inductors behave at different frequencies. Increasing the inductance increases the reactance because larger inductors produce stronger opposing magnetic fields. Likewise, increasing the frequency increases the reactance because the current must change more rapidly, producing a larger back emf. As a result, inductors oppose high-frequency signals much more strongly than low-frequency signals.

This frequency dependence makes inductors useful in many electronic systems. For example, inductors are commonly used to filter unwanted high-frequency electrical noise from power supplies, medical instruments, and communication systems. MRI scanners and other medical imaging equipment also rely on carefully designed inductive circuits to generate and control rapidly changing magnetic fields.

Example 56.1: Calculating Inductive Reactance and Current

An inductor with an inductance of 3.00 mH is connected to an AC voltage source. Calculate:

  1. the inductive reactance when the source frequency is 60.0 Hz and when it is 10.0 kHz, and
  2. the rms current at each frequency if the applied rms voltage is 120 V.

Strategy

First calculate the inductive reactance using

[latex]X_L=2\pi fL.[/latex]

Then use the AC version of Ohm's law,

[latex]I=\frac{V}{X_L},[/latex]

to determine the rms current at each frequency.

Solution

Step 1: Calculate the inductive reactance at 60.0 Hz.

[latex]X_L=2\pi(60.0\,\text{Hz})(3.00\times10^{-3}\,\text{H}) =1.13\,\Omega.[/latex]

Step 2: Calculate the inductive reactance at 10.0 kHz.

[latex]X_L=2\pi(1.00\times10^4\,\text{Hz})(3.00\times10^{-3}\,\text{H}) =188\,\Omega.[/latex]

Step 3: Calculate the rms current at 60.0 Hz.

[latex]I=\frac{V}{X_L} =\frac{120\,\text{V}}{1.13\,\Omega} =106\,\text{A}.[/latex]

Step 4: Calculate the rms current at 10.0 kHz.

[latex]I=\frac{V}{X_L} =\frac{120\,\text{V}}{188\,\Omega} =0.637\,\text{A}.[/latex]

Discussion

This example illustrates one of the defining characteristics of an inductor: its opposition to current depends strongly on frequency. At the household power frequency of 60 Hz, the reactance is only 1.13 Ω, so the inductor allows a very large current to flow. At 10.0 kHz, however, the reactance increases to 188 Ω, reducing the current by more than two orders of magnitude.

This behavior explains why inductors are commonly used as high-frequency filters. They allow low-frequency signals, such as the 50–60 Hz power used in electrical systems, to pass relatively easily while blocking rapidly changing electrical signals. In healthcare and biomedical electronics, inductors help suppress high-frequency electrical noise in power supplies and sensitive instruments. Similar filtering techniques improve the quality of signals recorded by devices such as ECGs, EEGs, and other physiological monitoring systems by reducing unwanted electrical interference.

Notice that even though the resistance of the circuit is assumed to be negligible, the AC current does not become infinitely large. As the current continually changes direction, the inductor continuously generates a back emf that opposes those changes, limiting the current. At higher frequencies the current changes more rapidly, producing a larger back emf and a larger inductive reactance. As a result, the current decreases as the frequency increases.

Capacitors and Capacitive Reactance

We have already seen that a capacitor connected to a DC source charges until its voltage equals the source voltage. Once it is fully charged, the current stops because no additional charge can accumulate on the plates. When a capacitor is connected to an alternating-current source, however, the voltage continually reverses direction. The capacitor therefore charges, discharges, and then charges again with the opposite polarity during every AC cycle.

Consider the capacitor connected directly to an AC voltage source in Figure 56.2. We assume that the resistance of the circuit is negligible so that the capacitor determines the behavior of the circuit. The figure also shows the voltage across the capacitor and the current through it as functions of time.

An ideal capacitor connected to an alternating-current source. A graph shows the capacitor voltage and current as functions of time, with the current reaching each maximum one-quarter of a cycle before the voltage.
Figure 56.2: (a) An AC voltage source connected to a capacitor with negligible resistance. (b) The current and voltage as functions of time. In a purely capacitive circuit, the current leads the voltage by one-quarter of a cycle.

Suppose the voltage across the capacitor begins at its maximum positive value. At that instant, the capacitor is fully charged and the current is zero. As the applied voltage decreases, the capacitor begins to discharge and current flows in the negative direction. At point a, the capacitor is fully discharged, so [latex]Q=0[/latex] and the voltage across it is zero. However, the current has its maximum negative value.

Between points a and b, the current continues in the negative direction and charges the capacitor with the opposite polarity. At point b, the current is zero and the voltage has reached its most negative value. The current then becomes positive, removing the negative charge and bringing the voltage back to zero at point c. At this point, the positive current reaches its maximum value. Between points c and d, the current gradually decreases to zero while the voltage rises to its positive maximum. The cycle then repeats.

AC Voltage in a Capacitor

When a sinusoidal voltage is applied to a capacitor, the current leads the voltage by one-quarter of a cycle, corresponding to a phase difference of [latex]90^\circ[/latex]. Equivalently, the voltage lags behind the current by [latex]90^\circ[/latex].

A capacitor affects the current because it can stop current completely once it becomes fully charged. In an AC circuit, however, the voltage reverses before the capacitor remains fully charged for long. The repeated charging and discharging allow an rms current to flow.

The opposition that a capacitor presents to alternating current is called capacitive reactance. The rms current through a circuit containing only a capacitor is given by an AC form of Ohm's law:

[latex]I=\frac{V}{X_C},[/latex]

where [latex]V[/latex] is the rms voltage and [latex]X_C[/latex] is the capacitive reactance. Capacitive reactance is defined as

[latex]X_C=\frac{1}{2\pi fC},[/latex]

where [latex]f[/latex] is the frequency of the AC source and [latex]C[/latex] is the capacitance. Capacitive reactance has units of ohms, so it plays a role similar to resistance when determining the current in an AC circuit. Unlike ordinary resistance, however, an ideal capacitor does not continuously convert electrical energy into thermal energy. It stores energy in its electric field and then returns that energy to the circuit.

The equation for capacitive reactance shows two important relationships:

  • Increasing the capacitance decreases the reactance, allowing more current to flow.
  • Increasing the frequency also decreases the reactance because the capacitor has less time to become fully charged during each cycle.

Example 56.2: Calculating Capacitive Reactance and Current

A [latex]5.00\ \mu\text{F}[/latex] capacitor is connected to a 120 V rms AC source.

  1. Calculate the capacitive reactance at 60.0 Hz and at 10.0 kHz.
  2. Calculate the rms current at each frequency.

Strategy

First, calculate the capacitive reactance using

[latex]X_C=\frac{1}{2\pi fC}.[/latex]

Once the reactance has been found at each frequency, use

[latex]I=\frac{V}{X_C}[/latex]

to calculate the rms current.

Solution for (a)

At 60.0 Hz,

[latex]\begin{aligned} X_C &=\frac{1}{2\pi fC}\\ &=\frac{1}{2\pi(60.0\ \text{Hz})(5.00\times10^{-6}\ \text{F})}\\ &=531\ \Omega. \end{aligned}[/latex]

At 10.0 kHz,

[latex]\begin{aligned} X_C &=\frac{1}{2\pi(1.00\times10^4\ \text{Hz})(5.00\times10^{-6}\ \text{F})}\\ &=3.18\ \Omega. \end{aligned}[/latex]

Solution for (b)

At 60.0 Hz,

[latex]\begin{aligned} I &=\frac{V}{X_C}\\ &=\frac{120\ \text{V}}{531\ \Omega}\\ &=0.226\ \text{A}. \end{aligned}[/latex]

At 10.0 kHz,

[latex]\begin{aligned} I &=\frac{120\ \text{V}}{3.18\ \Omega}\\ &=37.7\ \text{A}. \end{aligned}[/latex]

Discussion

The capacitor behaves in the opposite way from an inductor. At 60.0 Hz, its capacitive reactance is relatively large, so the current is small. At 10.0 kHz, its reactance is much smaller, allowing a much larger current to flow. Capacitors therefore impede low-frequency signals more strongly than high-frequency signals.

This frequency dependence allows capacitors to be used in electronic filters. In biomedical instruments, filters help separate useful physiological signals from unwanted electrical interference. For example, ECG and EEG systems use carefully designed capacitor and resistor networks to reduce noise while preserving the frequencies associated with biological activity.

Although a capacitor behaves like an open circuit after it becomes fully charged by a DC source, current can continue to flow in an AC circuit because the voltage is continually reversing and the capacitor is repeatedly charging and discharging.

As the frequency approaches zero, corresponding to DC, the capacitive reactance becomes extremely large:

[latex]X_C=\frac{1}{2\pi fC}\longrightarrow\infty\qquad\text{as}\qquad f\longrightarrow0.[/latex]

Once the capacitor is charged, the current therefore becomes zero. At very high frequencies, the capacitive reactance approaches zero, so the capacitor offers very little opposition to the current and behaves almost like a conducting wire.

Capacitors and inductors have opposite effects in AC circuits. Inductors strongly oppose high-frequency signals, whereas capacitors strongly oppose low-frequency signals.

Resistors in an AC Circuit

For comparison, consider the resistor connected to an AC voltage source in Figure 56.3. Unlike inductors and capacitors, an ideal resistor does not create a phase difference between current and voltage. The current responds immediately to the changing voltage, so both quantities reach zero, their positive maxima, and their negative maxima at the same times.

An AC voltage source connected to a resistor. A graph shows that the resistor voltage and current reach their maxima, minima, and zero values at the same times, meaning they are exactly in phase.
Figure 56.3: (a) An AC voltage source connected to a resistor. (b) The current and voltage across the resistor vary together and are exactly in phase.

AC Voltage in a Resistor

When a sinusoidal voltage is applied to a resistor, the voltage and current are exactly in phase. Their phase difference is [latex]0^\circ[/latex].

The rms current through a resistor is determined by the ordinary form of Ohm's law:

[latex]I=\frac{V}{R}.[/latex]

For an ideal resistor, the resistance does not depend on frequency. The resistor also converts electrical energy into thermal energy, unlike ideal capacitors and inductors, which alternately store and return energy to the circuit.

Section Summary

  • When a sinusoidal AC voltage is applied to an inductor, the voltage leads the current by one-quarter of a cycle, corresponding to a phase difference of [latex]90^\circ[/latex].
  • An inductor opposes changes in current through a quantity called the inductive reactance, which acts like an effective resistance in an AC circuit.
  • The rms current through an ideal inductor is given by
    [latex]I=\frac{V}{X_L},[/latex]

    where [latex]V[/latex] is the rms voltage across the inductor.

  • The inductive reactance is
    [latex]X_L=2\pi fL,[/latex]

    where [latex]f[/latex] is the frequency of the AC source and [latex]L[/latex] is the inductance.

  • Inductive reactance has units of ohms and increases with frequency. Inductors therefore oppose high-frequency signals more strongly than low-frequency signals.
  • When a sinusoidal AC voltage is applied to a capacitor, the current leads the voltage by one-quarter of a cycle (equivalently, the voltage lags the current by [latex]90^\circ[/latex]).
  • A capacitor opposes changes in voltage through a quantity called the capacitive reactance. The rms current through an ideal capacitor is
    [latex]I=\frac{V}{X_C},[/latex]

    where [latex]V[/latex] is the rms voltage across the capacitor.

  • The capacitive reactance is
    [latex]X_C=\frac{1}{2\pi fC}.[/latex]
  • Capacitive reactance also has units of ohms but decreases with frequency. Capacitors therefore oppose low-frequency signals more strongly than high-frequency signals.
  • Resistors behave differently from inductors and capacitors because their resistance is independent of frequency. In an ideal resistor, the voltage and current remain exactly in phase.

Conceptual Questions

  1. Presbycusis is an age-related hearing loss that progressively reduces sensitivity to high-frequency sounds. Suppose a hearing aid amplifier increases all frequencies equally. To better compensate for presbycusis, would you place a capacitor in series with the hearing aid speaker or in parallel with it? Explain your reasoning.
  2. Would you use a large inductance or a large capacitance connected in series with a circuit to filter out low-frequency noise, such as the 100 Hz hum in a sound system? Explain.
  3. High-frequency electrical noise in AC power lines can damage computers and other sensitive electronic equipment. Would a power filter connected in series with the computer use a large inductance or a large capacitance to reduce these high-frequency signals? Explain.
  4. Does the inductance of an inductor depend on the current flowing through it, the frequency of the applied voltage, or both? How does this compare with inductive reactance?
  5. Explain why the capacitor in Figure 56.4(a) acts as a low-frequency filter between the two circuits, whereas the capacitor in Figure 56.4(b) acts as a high-frequency filter.
Two capacitor filter circuits. In the first circuit, the capacitor is connected in series between two grounded circuits. In the second circuit, the capacitor is connected from the signal line to ground. These two configurations preferentially transmit different frequency ranges.
Figure 56.4: Two common capacitor filter configurations. Depending on how the capacitor is connected, the circuit can preferentially transmit either low-frequency or high-frequency signals.
  1. If the capacitors in Figure 56.4 were replaced by inductors, which circuit would act as a low-frequency filter and which would act as a high-frequency filter? Explain your answer.

Problems & Exercises

  1. At what frequency will a 30.0 mH inductor have an inductive reactance of [latex]100\ \Omega[/latex]?
  2. What inductance is required to produce an inductive reactance of [latex]20.0\ \text{k}\Omega[/latex] at a frequency of 500 Hz?
  3. What capacitance is required to produce a capacitive reactance of [latex]2.00\ \text{M}\Omega[/latex] at 60.0 Hz?
  4. At what frequency will an 80.0 mF capacitor have a capacitive reactance of [latex]0.250\ \Omega[/latex]?
  5. A 0.500 H inductor is connected to an AC source.
    1. Find the rms current when the source has a frequency of 60.0 Hz and an rms voltage of 480 V.
    2. What would the rms current be if the frequency were increased to 100 kHz while the rms voltage remained 480 V?
  6. A [latex]0.250\ \mu\text{F}[/latex] capacitor is connected to a 480 V rms AC source.
    1. What rms current flows when the source frequency is 60.0 Hz?
    2. What would the rms current be at 25.0 kHz?
  7. A 20.0 kHz, 16.0 V rms source connected to an inductor produces an rms current of 2.00 A. What is the inductance?
  8. A 20.0 Hz, 16.0 V rms source produces an rms current of 2.00 mA when connected to a capacitor. What is the capacitance?
  9. An inductor designed to filter high-frequency noise from the power supplied to a computer is connected in series with the computer.
    1. What minimum inductance is needed to produce an inductive reactance of [latex]2.00\ \text{k}\Omega[/latex] for 15.0 kHz noise?
    2. What is the inductive reactance of this inductor at the 60.0 Hz power frequency?
  10. The capacitor in Figure 56.4(a) is designed to impede low-frequency signals passing between two circuits.
    1. What capacitance is required to produce a capacitive reactance of [latex]100\ \text{k}\Omega[/latex] at 120 Hz?
    2. What would the capacitive reactance be at 1.00 MHz?
    3. Discuss what these results imply about the ability of the capacitor to transmit low- and high-frequency signals.
  11. The capacitor in Figure 56.4(b) is used to divert high-frequency signals to earth/ground.
    1. What capacitance is required to produce a capacitive reactance of [latex]10.0\ \text{m}\Omega[/latex] at 5.00 kHz?
    2. What would the capacitive reactance be at 3.00 Hz?
    3. Discuss what these results imply about the behavior of the filter at high and low frequencies.
  12. Unreasonable Results. During an electroencephalogram (EEG), a 10.0 mV signal with a frequency of 0.500 Hz is applied to a capacitor, producing a current of 100 mA. Assume the resistance is negligible.
    1. What capacitance would be required?
    2. What is unreasonable about this result?
    3. Which assumption or premise is responsible for the unreasonable result?
  13. Construct Your Own Problem. Consider an inductor connected in series with a computer operating from a 60 Hz AC power source. Construct a problem in which you calculate the relative reduction of an incoming high-frequency noise signal compared with the 60 Hz power signal. Choose realistic values for the inductance and noise frequency, and consider the maximum acceptable reactance of the inductor at 60 Hz.

Glossary

inductive reactance
The opposition that an inductor presents to alternating current. It is measured in ohms and is calculated using
[latex]X_L=2\pi fL[/latex].
Inductive reactance increases as either the frequency or the inductance increases.
capacitive reactance
The opposition that a capacitor presents to alternating current. It is measured in ohms and is calculated using
[latex]X_C=\frac{1}{2\pi fC}[/latex].
Capacitive reactance decreases as either the frequency or the capacitance increases.
definition

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Introductory Physics for the Health and Life Sciences II Copyright © 2012 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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