Now let's put a resistor, capacitor and inductor in series. At any given time, the voltage across the three components in series, vseries(t), is the sum of these:
vseries(t) = vR(t) + vL(t) + vC(t),
The current i(t) we shall keep sinusoidal, as before. The voltage across the resistor, vR(t), is in phase with the current. That across the inductor, vL(t), is 90° ahead and that across the capacitor, vC(t), is 90° behind.
Once again, the time-dependent voltages v(t) add up at any time, but the RMS voltages V do not simply add up. Once again they can be added by phasorsrepresenting the three sinusoidal voltages. Again, let's 'freeze' it in time for the purposes of the addition, which we do in the graphic below. Once more, be careful to distinguish v and V.
Look at the phasor diagram: The voltage across the ideal inductor is antiparallel to that of the capacitor, so the total reactive voltage (the voltage which is 90° ahead of the current) is VL - VC, so Pythagoras now gives us:
V2series = V2R + (VL - VC)2
Now VR = IR, VL = IXL = ωL and VC = IXC= 1/ωC. Substituting and taking the common factor I gives:
where Zseries is the series impedance: the ratio of the voltage to current in an RLC series ciruit. Note that, once again, reactances and resistances add according to Pythagoras' law:
Zseries2 = R2 + Xtotal2 = R2 + (XL- XC)2.
Remember that the inductive and capacitive phasors are 180° out of phase, so their reactances tend to cancel.Now let's look at the relative phase. The angle by which the voltage leads the current is
φ = tan-1 ((VL - VC)/VR).
Substiting VR = IR, VL = IXL = ωL and VC = IXC= 1/ωC gives:
The dependence of Zseries and φ on the angular frequency ω is shown in the next figure. The angular frequency ω is given in terms of a particular value ωo, the resonant frequency* (ωo2 = 1/LC), which we meet below.
* Again, let's check the units: L (flux per unit current) is in henrys = tesla.m2/ampere, and Faraday's law gives volts = tesla.metre2/second, so L is in volt.second/ampere = volt.second2/coulomb. C (charge per unit voltage) is in farads (coulomb/volt), so LC has the units of (volt.second2/coulomb)(coulomb/volt) = second2.
(Setting the inductance term to zero gives back the equations we had above for RC circuits, though note that phase is negative, meaning (as we saw above) that voltage lags the current. Similarly, removing the capacitance terms gives the expressions that apply to RL circuits.)The next graph shows us the special case where the frequency is such that VL = VC.
Because vL(t) and vC are 180° out of phase, this means that vL(t) = - vC(t), so the two reactive voltages cancel out, and the series voltage is just equal to that across the resistor. This case is called series resonance, which is our next topic.
The power p converted in a resistor (ie the rate of conversion of electrical energy to heat) is
p(t) = iv = v2/R = i2R.
We use lower case p(t) because this is the expression for the instantaneous power at time t. Usually, we are interested in the mean power delivered, which is normally written P. P is the total energy converted in one cycle, divided by the period T of the cycle, so:
In the last line, we have used a standard trigonometrical identity that cos(2A) = 1 - 2 sin2A. Now the sinusoidal term averages to zero over any number of complete cycles, so the integral is simple and we obtain
This last set of equations are useful because they are exactly those normally used for a resistor in DC electricity. However, one must remember that P is the average power, and V = Vm/21/2 and I= Im/21/2. Looking at the integral above, and dividing by R, we see that I is equal to the square root of the mean value of i2, so I is called the root-mean-square or RMS value. Similarly, V = Vm/21/2 ~ 0.71*Vm is the RMS value of the voltage.
When talking of AC, RMS values are so commonly used that, unless otherwise stated, you may assume that RMS values are intended*. For instance, normal domestic AC in Australia is 240 Volts AC with frequency 50 Hz. The RMS voltage is 240 volts, so the peak value Vm= V.21/2 = 340 volts. So the active wire goes from +340 volts to -340 volts and back again 50 times per second. (This is the answer to the teaser question at the top of the page: rectification of the 240 V mains can give both + 340 Vdc and -340 Vdc.)
* An exception: manufacturers and sellers of HiFi equipment sometimes use peak values rather than RMS values, which makes the equipment seem more powerful than it is. Power in a resistor. In a resistor R, the peak power (achieved instantaneously 100 times per second for 50 Hz AC) is Vm2/R = im2*R. As discussed above, the voltage, current and so the power pass through zero volts 100 times per second, so the average power is less than this. The average is exactly as shown above: P = Vm2/2R = V2/R. Power in inductors and capacitors. In ideal inductors and capacitors, a sinusoidal current produces voltages that are respecively 90° ahead and behind the phase of the current. So if i = Imsin wt, the voltages across the inductor and capacitor are Vmcos wt and -Vmcos wt respectively. Now the integral of cos*sin over a whole number of cycles is zero. Consequently, ideal inductors and capacitors do not, on average, take power from the circuit.
When we connect components together, Kirchoff's laws apply at any instant. So the voltage v(t) across a resistor and capacitor in series is just
vseries(t) = vR(t) + vC(t)
however the addition is complicated because the two are not in phase. The next animation makes this clear: they add to give a new sinusoidal voltage, but the amplitude is less than VmR(t) + VmC(t). Similarly, the AC voltages (amplitude times √2) do not add up. This may seem confusing, so it's worth repeating:
vseries = vR + vC but Vseries < VR + VC.
This should be clear on the animation and the still graphic below: check that the voltages v(t) do add up, and then look at the magnitudes. The amplitudes and the RMS voltages V do not add up in a simple arithmetical way.
Here's where phasor diagrams are going to save us a lot of work. Play the animation again (click play), and look at the projections on the vertical axis. Because we have sinusoidal variation in time, the vertical component (magnitude times the sine of the angle it makes with the x axis) gives us v(t). But the y components of different vectors, and therefore phasors, add up simply: if
rtotal = r1 + r2, then ry total = ry1 + ry2.
So v(t), the sum of the y projections of the component phasors, is just the y projection of the sum of the component phasors. So we can represent the three sinusoidal voltages by their phasors. (While you're looking at it, check the phases. You'll see that the series voltage is behind the current in phase, but the relative phase is somewhere between 0 and 90°, the exact value depending on the size of VR and VC. We'll discuss phase below.)Now let's stop that animation and label the values, which we do in the still figure below. All of the variables (i, vR, vC, vseries) have the same frequency f and the same angular frequency ω, so their phasors rotate together, with the same relative phases. So we can 'freeze' it in time at any instant to do the analysis. The convention I use is that the x axis is the reference direction, and the reference is whatever is common in the circuit. In this series circuit, the current is common. (In a parallel circuit, the voltage is common, so I would make the voltage the horizontal axis.) Be careful to distinguish v and V in this figure!
(Careful readers will note that I'm taking a shortcut in these diagrams: the size of the arrows on the phasor diagrams are drawn the same as the amplitudes on the v(t) graphs. However I am just calling them VR, VC etc, rather than VmR, VmR etc. The reason is that the peak values (VmR etc) are rarely used in talking about AC: we use the RMS values, which are peak values times 0.71. Phasor diagrams in RMS have the same shape as those drawn using amplitudes, but everything is scaled by a factor of 0.71 = 1/√2.)The phasor diagram at right shows us a simple way to calculate the series voltage. The components are in series, so the current is the same in both. The voltage phasors (brown for resistor, blue for capacitor in the convention we've been using) add according to vector or phasor addition, to give the series voltage (the red arrow). By now you don't need to look at v(t), you can go straight from the circuit diagram to the phasor diagram, like this:
From Pythagoras' theorem:
V2mRC = V2mR + V2mC
If we divide this equation by two, and remembering that the RMS value V = Vm/√2, we also get:
Now this looks like Ohm's law again: V is proportional to I. Their ratio is the series impedance, Zseries and so for this series circuit,
Note the frequency dependence of the series impedance ZRC: at low frequencies, the impedance is very large, because the capacitive reactance 1/ωC is large (the capacitor is open circuit for DC). At high frequencies, the capacitive reactance goes to zero (the capacitor doesn't have time to charge up) so the series impedance goes to R. At the angular frequency ω = ωo = 1/RC, the capacitive reactance 1/ωC equals the resistance R. We shall show this characteristic frequency* on all graphs on this page.
* Let's check the units: it should be quick. First, a radian is the ratio of an arc to a radius, so it is a pure number. R is in ohms, and we can use Ohm's law (V = IR) and the definition of current (I = dQ/dt) to write ohm = volt/ampere = volt/(coulomb/second) = volt*second/coulomb. C (charge per unit voltage) is in farads (coulomb/volt), so RC has the units of (volt*second/coulomb)(coulomb/volt) = second. See, it was quick: it only took us 30 ohm.farads!
Remember how, for two resistors in series, you could just add the resistances: Rseries = R1 + R2 to get the resistance of the series combination. That simple result comes about because the two voltages are both in phase with the current, so their phasors are parallel. Because the phasors for reactances are 90° out of phase with the current, the series impedance of a resistor R and a reactance X are given by Pythagoras' law:
Zseries2 = R2 + X2 .
Ohm's law in AC. We can rearrange the equations above to obtain the current flowing in this circuit. Alternatively we can simply use the Ohm's Law analogy and say that I = Vsource/ZRC. Either way we get
where the current goes to zero at DC (capacitor is open circuit) and to V/R at high frequencies (no time to charge the capacitor).So far we have concentrated on the magnitude of the voltage and current. We now derive expressions for their relative phase, so let's look at the phasor diagram again.
From simple trigonometry, the angle by which the current leads the voltage is
However, we shall refer to the angle φ by which the voltage leads the current. The voltage is behind the current because the capacitor takes time to charge up, so φ is negative, ie
φ = -tan-1 (1/ωRC) = tan-1 (1/2πfRC).
At low frequencies, the impedance of the series RC circuit is dominated by the capacitor, so the voltage is 90° behind the current. At high frequencies, the impedance approaches R and the phase difference approaches zero. The frequency dependence of Z and φ are important in the applications of RC circuits. The voltage is mainly across the capacitor at low frequencies, and mainly across the resistor at high frequencies. Of course the two voltages must add up to give the voltage of the source, but they add up as vectors.
V2RC = V2R + V2C.
At the frequency ω = ωo = 1/RC, the phase φ = 45° and the voltage fractions are VR/VRC = VC/VRC = 1/2V1/2 = 0.71.
So, by chosing to look at the voltage across the resistor, you select mainly the high frequencies, across the capacitor, you select low frequencies. This brings us to one of the very important applications of RC circuits, and one which merits its own page: where we use sound files as examples of RC filtering.
Single phase is used in domestic applications for low power applications but it has some drawbacks. One is that it turns off 100 times per second (you don't notice that the fluorescent lights flicker at this speed because your eyes are too slow: even 25 pictures per second on the TV is fast enough to give the illusion of continuous motion.) The second is that it makes it awkward to produce rotating magnetic fields. For this reason, some high power (several kW) domestic devices may require three phase installation. Industrial applications use three phase extensively, and the three phase induction motor is a standard workhorse for high power applications. The three wires (not counting earth) carry three possible potential differences which are out of phase with each other by 120°, as shown in the animation below. Thus three stators give a smoothly rotating field.
If one puts a permanent magnet in such a set of stators, it becomes a synchronous three phase motor. The animation shows a squirrel cage, in which for simplicity only one of the many induced current loops is shown. With no mechanical load, it is turning virtually in phase with the rotating field. The rotor need not be a squirrel cage: in fact any conductor that will carry eddy currents will rotate, tending to follow the rotating field. This arrangement can give an induction motor capable of high efficiency, high power and high torques over a range of rotation rates.
A simple DC motor has a coil of wire that can rotate in a magnetic field. The current in the coil is supplied via two brushes that make moving contact with a split ring. The coil lies in a steady magnetic field. The forces exerted on the current-carrying wires create a torque on the coil.
The force F on a wire of length L carrying a current i in a magnetic field B is iLB times the sine of the angle between B and i, which would be 90° if the field were uniformly vertical. The direction of F comes from the right hand rule*, as shown here. The two forces shown here are equal and opposite, but they are displaced vertically, so they exert a torque. (The forces on the other two sides of the coil act along the same line and so exert no torque.)
* A number of different nmemonics are used to remember the direction of the force. Some use the right hand, some the left. For students who know vector multiplication, it is easy to use the Lorentz force directly: F = q vXB , whence F = i dLXB . That is the origin of the diagram shown here.
The coil can also be considered as a magnetic dipole, or a little electromagnet, as indicated by the arrow SN: curl the fingers of your right hand in the direction of the current, and your thumb is the North pole. In the sketch at right, the electromagnet formed by the coil of the rotor is represented as a permanent magnet, and the same torque (North attracts South) is seen to be that acting to align the central magnet.Note the effect of the brushes on the split ring. When the plane of the rotating coil reaches horizontal, the brushes will break contact (not much is lost, because this is the point of zero torque anyway – the forces act inwards). The angular momentum of the coil carries it past this break point and the current then flows in the opposite direction, which reverses the magnetic dipole. So, after passing the break point, the rotor continues to turn anticlockwise and starts to align in the opposite direction. In the following text, I shall largely use the 'torque on a magnet' picture, but be aware that the use of brushes or of AC current can cause the poles of the electromagnet in question to swap position when the current changes direction.
The torque generated over a cycle varies with the vertical separation of the two forces. It therefore depends on the sine of the angle between the axis of the coil and field. However, because of the split ring, it is always in the same sense. The animation below shows its variation in time, and you can stop it at any stage and check the direction by applying the right hand rule.
Some notes about AC and DC motors for high power applications
This site was originally written to help high school students and teachers in New South Wales, Australia, where a new syllabus concentrating on the history and applications of physics, at the expense of physics itself, has been introduced. The new syllabus, in one of the dot points, has this puzzling requirement: "explain that AC motors usually produce low power and relate this to their use in power tools".
AC motors are used for high power applications whenever it is possible. Three phase AC induction motors are widely used for high power applications, including heavy industry. However, such motors are unsuitable if multiphase is unavailable, or difficult to deliver. Electric trains are an example: it is easier to build power lines and pantographs if one only needs one active conductor, so this usually carries DC, and many train motors are DC. However, because of the disadvantages of DC for high power, more modern trains convert the DC into AC and then run three phase motors.Single phase induction motors have problems for applications combining high power and flexible load conditions. The problem lies in producing the rotating field. A capacitor could be used to put the current in one set of coils ahead, but high value, high voltage capacitors are expensive. Shaded poles are used instead, but the torque is small at some angles. If one cannot produce a smoothly rotating field, and if the load 'slips' well behind the field, then the torque falls or even reverses.
Power tools and some appliances use brushed AC motors. Brushes introduce losses (plus arcing and ozone production). The stator polarities are reversed 100 times a second. Even if the core material is chosen to minimise hysteresis losses ('iron losses'), this contributes to inefficiency, and to the possibility of overheating. These motors may be called 'universal' because they can operate on DC. This solution is cheap, but crude and inefficient. For relatively low power applications like power tools, the inefficiency is usually not economically important.
If only single phase AC is available, one may rectify the AC and use a DC motor. High current rectifiers used to be expensive, but are becoming less expensive and more widely used. If you are confident you understand the principles, it's time to go to How real electric motors work by John Storey. Or else continue here to find out about loudspeakers and transformers.
A set of coils can be used to create a magnetic field that translates, rather than rotates. The pair of coils in the animation below are pulsed on, from left to right, so the region of magnetic field moves from left to right. A permanent or electromagnet will tend to follow the field. So would a simple slab of conducting material, because the eddy currents induced in it (not shown) comprise an electromagnet. Alternatively, we could say that, from Faraday's law, an emf in the metal slab is always induced so as to oppose any change in magnetic flux, and the forces on the currents driven by this emf keep the flux in the slab nearly constant. (Eddy currents not shown in this animation.)
Alternatively, we could have sets of powered coils in the moving part, and induce eddy currents in the rail. Either case gives us a linear motor, which would be useful for say maglev trains. (In the animation, the geometry is, as usual on this site, highly idealised, and only one eddy current is shown.)
To build this simple but strange motor, you need two fairly strong magnets (rare earth magnets about 10 mm diameter would be fine, as would larger bar magnets), some stiff copper wire (at least 50 cm), two wires with crocodile clips on either end, a six volt lantern battery, two soft drink cans, two blocks of wood, some sticky tape and a sharp nail.
Make the coil out of stiff copper wire, so it doesn't need any external support. Wind 5 to 20 turns in a circle about 20 mm in diameter, and have the two ends point radially outwards in opposite directions. These ends will be both the axle and the contacts. If the wire has lacquer or plastic insulation, strip it off at the ends
The supports for the axle can be made of aluminium, so that they make electrical contact. For example poke holes in a soft drink cans with a nail as shown. Position the two magnets, north to south, so that the magnetic field passes through the coil at right angles to the axles. Tape or glue the magnets onto the wooden blocks (not shown in the diagram) to keep them at the right height, then move the blocks to put them in position, rather close to the coil. Rotate the coil initially so that the magnetic flux through the coil is zero, as shown in the diagram.Now get a battery, and two wires with crocodile clips. Connect the two terminals of the battery to the two metal supports for the coil and it should turn.
Note that this motor has at least one 'dead spot': It often stops at the position where there is no torque on the coil. Don't leave it on too long: it will flatten the battery quickly.The optimum number of turns in the coil depends on the internal resistance of the battery, the quality of the support contacts and the type of wire, so you should experiment with different values.
As mentioned above, this is also a generator, but it is a very inefficient one. To make a larger emf, use more turns (you may need to use finer wire and a frame upon which to wind it.) You could use eg an electric drill to turn it quickly, as shown in the sketch above. Use an oscilloscope to look at the emf generated. Is it AC or DC?
This motor has no split ring, so why does it work on DC? Simply put, if it were exactly symmetrical, it wouldn't work. However, if the current is slightly less in one half cycle than the other, then the average torque will not be zero and, because it spins reasonably rapidly, the angular momentum acquired during the half cycle with greater current carries it through the half cycle when the torque is in the opposite direction. At least two effects can cause an asymmetry. Even if the wires are perfectly stripped and the wires clean, the contact resistance is unlikely to be exactly equal, even at rest. Also, the rotation itself causes the contact to be intermittent so, if there are longer bounces during one phase, this asymmetry is sufficient. In principle, you could partially strip the wires in such a way that the current would be zero in one half cycle.
An alternative relisation of the simple motor, by James Taylor.
An even simpler motor (one that is also much simpler to understand!) is the homopolar motor.
Consider a circuit for a single-phase AC power system, where a 120 volt, 60 Hz AC voltage source is delivering power to a resistive load: (Figure below)
Ac source drives a purely resistive load.
In this example, the current to the load would be 2 amps, RMS. The power dissipated at the load would be 240 watts. Because this load is purely resistive (no reactance), the current is in phase with the voltage, and calculations look similar to that in an equivalent DC circuit. If we were to plot the voltage, current, and power waveforms for this circuit, it would look like Figure below.
Current is in phase with voltage in a resistive circuit.
Note that the waveform for power is always positive, never negative for this resistive circuit. This means that power is always being dissipated by the resistive load, and never returned to the source as it is with reactive loads. If the source were a mechanical generator, it would take 240 watts worth of mechanical energy (about 1/3 horsepower) to turn the shaft.
Also note that the waveform for power is not at the same frequency as the voltage or current! Rather, its frequency is double that of either the voltage or current waveforms. This different frequency prohibits our expression of power in an AC circuit using the same complex (rectangular or polar) notation as used for voltage, current, and impedance, because this form of mathematical symbolism implies unchanging phase relationships. When frequencies are not the same, phase relationships constantly change.
As strange as it may seem, the best way to proceed with AC power calculations is to use scalar notation, and to handle any relevant phase relationships with trigonometry.
For comparison, let's consider a simple AC circuit with a purely reactive load in Figure below.
AC circuit with a purely reactive (inductive) load.
Power is not dissipated in a purely reactive load. Though it is alternately absorbed from and returned to the source.
Note that the power alternates equally between cycles of positive and negative. (Figure above) This means that power is being alternately absorbed from and returned to the source. If the source were a mechanical generator, it would take (practically) no net mechanical energy to turn the shaft, because no power would be used by the load. The generator shaft would be easy to spin, and the inductor would not become warm as a resistor would.
Now, let's consider an AC circuit with a load consisting of both inductance and resistance in Figure below.
AC circuit with both reactance and resistance.
At a frequency of 60 Hz, the 160 millihenrys of inductance gives us 60.319 Ω of inductive reactance. This reactance combines with the 60 Ω of resistance to form a total load impedance of 60 + j60.319 Ω, or 85.078 Ω ∠ 45.152o. If we're not concerned with phase angles (which we're not at this point), we may calculate current in the circuit by taking the polar magnitude of the voltage source (120 volts) and dividing it by the polar magnitude of the impedance (85.078 Ω). With a power supply voltage of 120 volts RMS, our load current is 1.410 amps. This is the figure an RMS ammeter would indicate if connected in series with the resistor and inductor.
We already know that reactive components dissipate zero power, as they equally absorb power from, and return power to, the rest of the circuit. Therefore, any inductive reactance in this load will likewise dissipate zero power. The only thing left to dissipate power here is the resistive portion of the load impedance. If we look at the waveform plot of voltage, current, and total power for this circuit, we see how this combination works in Figure below.
A combined resistive/reactive circuit dissipates more power than it returns to the source. The reactance dissipates no power; though, the resistor does.
As with any reactive circuit, the power alternates between positive and negative instantaneous values over time. In a purely reactive circuit that alternation between positive and negative power is equally divided, resulting in a net power dissipation of zero. However, in circuits with mixed resistance and reactance like this one, the power waveform will still alternate between positive and negative, but the amount of positive power will exceed the amount of negative power. In other words, the combined inductive/resistive load will consume more power than it returns back to the source.
Looking at the waveform plot for power, it should be evident that the wave spends more time on the positive side of the center line than on the negative, indicating that there is more power absorbed by the load than there is returned to the circuit. What little returning of power that occurs is due to the reactance; the imbalance of positive versus negative power is due to the resistance as it dissipates energy outside of the circuit (usually in the form of heat). If the source were a mechanical generator, the amount of mechanical energy needed to turn the shaft would be the amount of power averaged between the positive and negative power cycles.
Mathematically representing power in an AC circuit is a challenge, because the power wave isn't at the same frequency as voltage or current. Furthermore, the phase angle for power means something quite different from the phase angle for either voltage or current. Whereas the angle for voltage or current represents a relative shift in timing between two waves, the phase angle for power represents a ratio between power dissipated and power returned. Because of this way in which AC power differs from AC voltage or current, it is actually easier to arrive at figures for power by calculating with scalar quantities of voltage, current, resistance, and reactance than it is to try to derive it from vector, or complex quantities of voltage, current, and impedance that we've worked with so far.
REVIEW:
In a purely resistive circuit, all circuit power is dissipated by the resistor(s). Voltage and current are in phase with each other.
In a purely reactive circuit, no circuit power is dissipated by the load(s). Rather, power is alternately absorbed from and returned to the AC source. Voltage and current are 90o out of phase with each other.
In a circuit consisting of resistance and reactance mixed, there will be more power dissipated by the load(s) than returned, but some power will definitely be dissipated and some will merely be absorbed and returned. Voltage and current in such a circuit will be out of phase by a value somewhere between 0o and 90o.
With AC currents, we can reverse field directions without having to use brushes. This is good news, because we can avoid the arcing, the ozone production and the ohmic loss of energy that brushes can entail. Further, because brushes make contact between moving surfaces, they wear out.The first thing to do in an AC motor is to create a rotating field. 'Ordinary' AC from a 2 or 3 pin socket is single phase AC--it has a single sinusoidal potential difference generated between only two wires--the active and neutral. (Note that the Earth wire doesn't carry a current except in the event of electrical faults.) With single phase AC, one can produce a rotating field by generating two currents that are out of phase using for example a capacitor. In the example shown, the two currents are 90° out of phase, so the vertical component of the magnetic field is sinusoidal, while the horizontal is cosusoidal, as shown. This gives a field rotating counterclockwise.
(* I've been asked to explain this: from simple AC theory, neither coils nor capacitors have the voltage in phase with the current. In a capacitor, the voltage is a maximum when the charge has finished flowing onto the capacitor, and is about to start flowing off. Thus the voltage is behind the current. In a purely inductive coil, the voltage drop is greatest when the current is changing most rapidly, which is also when the current is zero. The voltage (drop) is ahead of the current. In motor coils, the phase angle is rather less than 90¡, because electrical energy is being converted to mechanical energy.)
In this animation, the graphs show the variation in time of the currents in the vertical and horizontal coils. The plot of the field components Bx and By shows that the vector sum of these two fields is a rotating field. The main picture shows the rotating field. It also shows the polarity of the magnets: as above, blue represents a North pole and red a South pole.
If we put a permanent magnet in this area of rotating field, or if we put in a coil whose current always runs in the same direction, then this becomes a synchronous motor. Under a wide range of conditions, the motor will turn at the speed of the magnetic field. If we have a lot of stators, instead of just the two pairs shown here, then we could consider it as a stepper motor: each pulse moves the rotor on to the next pair of actuated poles. Please remember my warning about the idealised geometry: real stepper motors have dozens of poles and quite complicated geometries!
Note that the expression for the series impedance goes to infinity at high frequency because of the presence of the inductor, which produces a large emf if the current varies rapidly. Similarly it is large at very low frequencies because of the capacitor, which has a long time in each half cycle in which to charge up. As we saw in the plot of Zseriesω above, there is a minimum value of the series impedance, when the voltages across capacitor and inductor are equal and opposite, ie vL(t) = - vC(t) so VL(t) = VC, so
ωL = 1/ωC so the frequency at which this occurs is
where ωo and fo are the angular and cyclic frequencies of resonance, respectively. At resonance, series impedance is a minimum, so the voltage for a given current is a minimum (or the current for a given voltage is a maximum).
This phenomenon gives the answer to our teaser question at the beginning. In an RLC series circuit in which the inductor has relatively low internal resistance r, it is possible to have a large voltage across the the inductor, an almost equally large voltage across capacitor but, as the two are nearly 180° degrees out of phase, their voltages almost cancel, giving a total series voltage that is quite small. This is one way to produce a large voltage oscillation with only a small voltage source. In the circuit diagram at right, the coil corresponds to both the inducance L and the resistance r, which is why they are drawn inside a box representing the physical component, the coil. Why are they in series? Because the current flows through the coil and thus passes through both the inductance of the coil and its resistance.You get a big voltage in the circuit for only a small voltage input from the power source. You are not, of course, getting something for nothing. The energy stored in the large oscillations is gradually supplied by the AC source when you turn on, and it is then exchanged between capacitor and inductor in each cycle. For more details about this phenomenon, and a discussion of the energies involved.
In an RL series circuit, the voltage across the inductor is aheadof the current by 90°, and the inductive reactance, as we saw before, is XL = ωL. The resulting v(t) plots and phasor diagram look like this.
It is straightforward to use Pythagoras' law to obtain the series impedance and trigonometry to obtain the phase. We shall not, however, spend much time on RL circuits, for three reasons. First, it makes a good exercise for you to do it yourself. Second, RL circuits are used much less than RC circuits. This is because inductors are always* too big, too expensive and the wrong value, a proposition you can check by looking at an electronics catalogue. If you can use a circuit involving any number of Rs, Cs, transistors, integrated circuits etc to replace an inductor, one usually does. The third reason why we don't look closely at RL circuits on this site is that you can simply look at RLC circuits (below) and omit the phasors and terms for the capacitance.* Exceptions occur at high frequencies (~GHz) where only small value Ls are required to get substantial ωL. In such circuits, one makes an inductor by twisting copper wire around a pencil and adjusts its value by squeezing it with the fingers.
Let's recap what we now know about voltage and curent in linear components. The impedance is the general term for the ratio of voltage to current. Resistance is the special case of impedance when φ = 0, reactance the special case when φ = ± 90°. The table below summarises the impedance of the different components. It is easy to remember that the voltage on the capacitor is behind the current, because the charge doesn't build up until after the current has been flowing for a while.
The same information is given graphically below. It is easy to remember the frequency dependence by thinking of the DC (zero frequency) behaviour: at DC, an inductance is a short circuit (a piece of wire) so its impedance is zero. At DC, a capacitor is an open circuit, as its circuit diagram shows, so its impedance goes to infinity.
An inductor is usually a coil of wire. In an ideal inductor, the resistance of this wire is negligibile, as is its capacitance. The voltage that appears across an inductor is due to its own magnetic field and Faraday's law of electromagnetic induction. The current i(t) in the coil sets up a magnetic field, whose magnetic flux φB is proportional to the field strength, which is proportional to the current flowing. (Do not confuse the phase φ with the flux φB.) So we define the (self) inductance of the coil thus:
φB(t) = L.i(t)
Faraday's law gives the emf EL = - dφB/dt. Now this emf is a voltage rise, so for the voltage drop vL across the inductor, we have:
Again we define the inductive reactance XL as the ratio of the magnitudes of the voltage and current, and from the equation above we see that XL = ωL. Again we note the analogy to Ohm's law: the voltage is proportional to the current, and the peak voltage and currents are related by
Vm = XL.Im.
Remembering that the derivative is the local slope of the curve (the purple line), we can see in the next animation why voltage and current are out of phase in an inductor.
Again, there is a difference in phase: the derivative of the sinusoidal current is a cos function: it has its maximum (largest voltage across the inductor) when the current is changing most rapidly, which is when the current is intantaneously zero. The animation should make this clear. The voltage across the ideal inductor is 90° ahead of the current, (ie it reaches its peak one quarter cycle before the current does). Note how this is represented on the phasor diagram.
Again we note that the reactance is frequency dependent XL = ωL. This is shown in the next animation: when the frequency is halved but the current amplitude kept constant, the current is varying only half as quickly, so its derivative is half as great, as is the Faraday emf. For an inductor, the ratio of voltage to current increases with frequency, as the next animation shows.
The voltage on a capacitor depends on the amount of charge you store on its plates. The current flowing onto the positive capacitor plate (equal to that flowing off the negative plate) is by definition the rate at which charge is being stored. So the charge Q on the capacitor equals the integral of the current with respect to time. From the definition of the capacitance,
vC = q/C, so
Now remembering that the integral is the area under the curve (shaded blue), we can see in the next animation why the current and voltage are out of phase.
Once again we have a sinusoidal current i = Im . sin (ωt), so integration gives
(The constant of integration has been set to zero so that the average charge on the capacitor is 0).
Now we define the capacitive reactance XC as the ratio of the magnitude of the voltage to magnitude of the current in a capacitor. From the equation above, we see that XC = 1/ωC. Now we can rewrite the equation above to make it look like Ohm's law. The voltage is proportional to the current, and the peak voltage and current are related by
Vm = XC.Im.
Note the two important differences. First, there is a difference in phase: the integral of the sinusoidal current is a negative cos function: it reaches its maximum (the capacitor has maximum charge) when the current has just finished flowing forwards and is about to start flowing backwards. Run the animation again to make this clear. Looking at the relative phase, the voltage across the capacitor is 90°, or one quarter cycle, behind the current. We can see also see how the φ = 90° phase difference affects the phasor diagrams at right. Again, the vertical component of a phasor arrow represents the instantaneous value of its quanitity. The phasors are rotating counter clockwise (the positive direction) so the phasor representing VC is 90° behind the current (90° clockwise from it).
Recall that reactance is the name for the ratio of voltage to current when they differ in phase by 90°. (If they are in phase, the ratio is called resistance.) Another difference between reactance and resistance is that the reactance is frequency dependent. From the algebra above, we see that the capacitive reactance XCdecreases with frequency . This is shown in the next animation: when the frequency is halved but the current amplitude kept constant, the capacitor has twice as long to charge up, so it generates twice the potential difference. The blue shading shows q, the integral under the current curve (light for positive, dark for negative). The second and fourth curves show VC = q/C . See how the lower frequency leads to a larger charge (bigger shaded area before changing sign) and therefore a larger VC.
Thus for a capacitor, the ratio of voltage to current decreases with frequency. We shall see later how this can be used for filtering different frequencies.
The voltage v across a resistor is proportional to the current i travelling through it.
Further, this is true at all times: v = Ri. So, if the current in a resistor is
i = Im . sin (ωt) , we write: v = R.i = R.Im sin (ωt) v = Vm. sin (ωt) where Vm = R.Im
So for a resistor, the peak value of voltage is R times the peak value of current. Further, they are in phase: when the current is a maximum, the voltage is also a maximum. (Mathematically, φ = 0.) The first animation shows the voltage and current in a resistor as a function of time.
The rotating lines in the right hand part of the animation are a very simple case of a phasor diagram (named, I suppose, because it is a vector representation of phase). With respect to the x and y axes, radial vectors or phasors representing the current and the voltage across the resistance rotate with angular velocity ω. The lengths of these phasors represent the peak current Im and voltage Vm. The y components are Im sin (ωt) = i(t) and voltage Vm sin (ωt)= v(t). You can compare i(t) and v(t) in the animation with the vertical components of the phasors. The animation and phasor diagram here are simple, but they will become more useful when we consider components with different phases and with frequency dependent behaviour.
Single phase AC has the advantage that it only requires 2 wires. Its disadvantage is seen in the graph at the top of this page: twice every cycle V goes to zero. If you connect a phototransistor circuit to an oscilloscope, you will see that fluorescent lights turn off 100 times per second (or 120, if you are on 60 Hz supply). What if you need a more even supply of electricity? One can store energy in capacitors, of course, but with high power circuits this would require big, expensive capacitors. What to do?An AC generator may have more than one coil. If it has three coils, mounted at relative angles of 120°, then it will produce three sinusoidal emfs with relative phases of 120°, as shown in the upper figure at right. The power delivered to a resistive load by each of these is proportional to V2. The sum of the three V2 terms is a constant. We saw above that the average of V2 is half the peak value, so this constant is 1.5 times the peak amplitude for any one circuit, as is shown in the lower figure at right.
Do you need four wires? In principle, no. The sum of the three V terms is zero so, provided that the loads on each phase are identical, the currents drawn from the three lines add to zero. In practice, the current in the neutral wire is usually not quite zero. Further, it should be the same guage as the other wires because, if one of the loads were to fail and form an open circuit, the neutral would carry a current similar to that in the remaining two loads.
The voltage (top) and square of the voltage (bottom) in the three active lines of 3 phase supply.
The photograph shows a transformer designed for demonstration purposes: the primary and secondary coils are clearly separated, and may be removed and replaced by lifting the top section of the core. For our purposes, note that the coil on the left has fewer coils than that at right (the insets show close-ups).
The sketch and circuit show a step-up transformer. To make a step-down transformer, one only has to put the source on the right and the load on the left. (Important safety note: for a real transformer, you could only 'plug it in backwards' only after verifying that the voltage rating were appropriate.) So, how does a transformer work?The core (shaded) has high magnetic permeability, ie a material that forms a magnetic field much more easily than free space does, due to the orientation of atomic dipoles. (In the photograph, the core is laminated soft iron.) The result is that the field is concentrated inside the core, and almost no field lines leave the core. If follows that the magnetic fluxes φ through the primary and secondary are approximately equal, as shown. From Faraday's law, the emf in each turn, whether in the primary or secondary coil, is −dφ/dt. If we neglect resistance and other losses in the transformer, the terminal voltage equals the emf. For the Np turns of the primary, this gives
Vp = − Np.dφ/dt .
For the Ns turns of the secondary, this gives
Vs = − Ns.φ/dt
Dividing these equations gives the transformer equation
Vs/Vp = Ns/Np = r.
where r is the turns ratio. What about the current? If we neglect losses in the transformer (see the section below on efficiency), and if we assume that the voltage and current have similar phase relationships in the primary and secondary, then from conservation of energy we may write, in steady state:
Power in = power out, soVpIp = VsIs, whence
Is/Ip = Np/Ns = 1/r.
So you don't get something for nothing: if you increase the voltage, you decrease the current by (at least) the same factor. Note that, in the photograph, the coil with more turns has thinner wire, because it is designed to carry less current than that with fewer turns.In some cases, decreasing the current is the aim of the exercise. In power transmission lines, for example, the power lost in heating the wires due to their non-zero resistance is proportional to the square of the current. So it saves a lot of energy to transmit the electrical power from power station to city at very high voltages so that the currents are only modest.
Finally, and again assuming that the transformer is ideal, let's ask what the resistor in the secondary circuit 'looks like' to the primary circuit. In the primary circuit:
Vp = Vs/r and Ip = Is.r soVp/Ip = Vs/r2Is = R/r2.
R/r2 is called the reflected resistance. Provided that the frequency is not too high, and provided that there is a load resistance (conditions usually met in practical transformers), the inductive reactance of the primary is much smaller than this reflected resistance, so the primary circuit behaves as though the source were driving a resistor of value R/r2.
Efficiency of transformers
In practice, real transformers are less than 100% efficient.
First, there are resistive losses in the coils (losing power I2.r). For a given material, the resistance of the coils can be reduced by making their cross section large. The resistivity can also be made low by using high purity copper. (See Drift velocity and Ohm's law)
Second, there are some eddy current losses in the core. These can be reduced by laminating the core. Laminations reduce the area of circuits in the core, and so reduce the Faraday emf, and so the current flowing in the core, and so the energy thus lost.
Third, there are hysteresis losses in the core. The magentisation and demagnetisation curves for magnetic materials are often a little different (hysteresis or history depedence) and this means that the energy required to magnetise the core (while the current is increasing) is not entirely recovered during demagnetisation. The difference in energy is lost as heat in the core.
Finally, the geometric design as well as the material of the core may be optimised to ensure that the magnetic flux in each coil of the secondary is nearly the same as that in each coil of the primary.
More about transformers: AC vs DC generators
Transformers only work on AC, which is one of the great advantages of AC. Transformers allow 240V to be stepped down to convenient levels for digital electronics (only a few volts) or for other low power applications (typically 12V). Transformers step the voltage up for transmission, as mentioned above, and down for safe distribution. Without transformers, the waste of electric power in distribution networks, already high, would be enormous. It is possible to convert voltages in DC, but more complicated than with AC. Further, such conversions are often inefficient and/or expensive. AC has the further advantage that it can be used on AC motors, which are usually preferable to DC motors for high power applications.
A loudspeaker is a linear motor with a small range. It has a single moving coil that is permanently but flexibly wired to the voltage source, so there are no brushes.
The coil moves in the field of a permanent magnet, which is usually shaped to produce maximum force on the coil. The moving coil has no core, so its mass is small and it may be accelerated quickly, allowing for high frequency motion. In a loudspeaker, the coil is attached to a light weight paper cone, which is supported at the inner and outer edges by circular, pleated paper 'springs'. In the photograph below, the speaker is beyond the normal upward limit of its travel, so the coil is visible above the magnet poles.For low frequency, large wavelength sound, one needs large cones. The speaker shown below is 380 mm diameter. Speakers designed for low frequencies are called woofers. They have large mass and are therefore difficult to accelerate rapidly for high frequency sounds. In the photograph below, a section has been cut away to show the internal components.
Tweeters - loudspeakers designed for high frequencies - may be just speakers of similar design, but with small, low mass cones and coils. Alternatively, they may use piezoelectric crystals to move the cone.
Speakers are seen to be linear motors with a modest range - perhaps tens of mm. Similar linear motors, although of course without the paper cone, are often used to move the reading and writing head radially on a disc drive.
Loudspeakers as microphones
In the picture above, you can see that a cardboard diaphragm (the loudspeaker cone) is connected to a coil of wire in a magnetic field. If a soundwave moves the diaphragm, the coil will move in the field, generating a voltage. This is the principle of a dynamic microphone – though in most microphones, the diaphragm is rather smaller than the cone of a loudspeaker. So, a loudspeaker should work as a microphone. This is a nice project: all you need is a loudspeaker and two wires to connect it to the input of an oscilloscope or the microphone input of your computer. Two questions: what do you think the mass of the cone and coil will do to the frequency response? What about the wavelength of sounds your use?
The power p converted in a resistor (ie the rate of conversion of electrical energy to heat) is
The coil moves in the field of a permanent magnet, which is usually shaped to produce maximum force on the coil. The moving coil has no core, so its mass is small and it may be accelerated quickly, allowing for high frequency motion. In a loudspeaker, the coil is attached to a light weight paper cone, which is supported at the inner and outer edges by circular, pleated paper 'springs'. In the photograph below, the speaker is beyond the normal upward limit of its travel, so the coil is visible above the magnet poles.For low frequency, large wavelength sound, one needs large cones. The speaker shown below is 380 mm diameter. Speakers designed for low frequencies are called woofers. They have large mass and are therefore difficult to accelerate rapidly for high frequency sounds. In the photograph below, a section has been cut away to show the internal components.