AC-1 nductors, Transformers, and AC circuits nductors An inductor is simply a coil of wire. nductors are used in circuits to store energy in the form of magnetic field energy. mportant point: The magnetic flux Φ through any loop is proportional to the current making the flux. All our formulas for -field show : µ ˆ d r iot-avart: d = Ampere: d = 4π r Φ Φ. o the ratio Φ/ is independent of. µ thru Definition: elf-inductance of a coil of wire: Φ The inductance = Φ / is independent of. Current makes, which makes Φ. units of inductance [] = [Φ] / [] = T m / A = 1 henry (H). An inductor is a coil of wire. One or a few centimeter-sized loops of wire has 1 µh (usually insignificant). A coil with many thousands of turns has 1 H (big!). o, why do we care about inductors? An inductor acts like a "current regulator". An inductor helps to maintain constant current. How is that? o, E d = dφ d Φ = = = E Faraday The induced emf is often called a "back emf". Changing the current in an inductor creates an emf which opposes the change in (by enz's aw). t is difficult (requires a big external voltage) to change quickly the current in an inductor. The current in an inductor cannot change instantly. f it did, there would be an infinite back emf, an infinite E-field to fight the change. ast update: 1/3/9 Dubson hys11 otes, University of Colorado
AC- Computing the inductance of a single turn coil (or a few turn coil) is quite messy because the - field in a loop of wire is non-uniform. The non-uniform makes computing the magnetic flux Φ = da quite difficult. n practice, one determines the inductance of a coil by measuring it, using d E = : put in a known d/, measure emf, compute. Computing the inductance of a long solenoid is easy, because the -field is uniform: elf-inductance of a solenoid: length z f the coil is very long, = µ o n inside, so total flux is Φ = A = µ na Φ µ µ inductance = = n A = n A z (We use z for length here, to avoid confusion with turns for inductance). area A n = / z Magnetic Energy Density Recall that for a capacitor, the stored electrostatic potential energy is U 1 = C. This energy is in the electric field, and the energy density (energy per volume) is u U = = vol. ε 1 E E For an inductor, the stored energy is U 1 =. This energy is stored in the magnetic field (so we call it magnetostatic potential energy) and the energy density is u U 1 vol. µ = =. roof of U 1 = : t takes work to get a current flowing in an inductor. The battery which make the current flow in an inductor must do work against the back emf, which opposes any du d change in current. Watch closely: power = = = d t, so du = = d, 1 so U= du = d = d = ast update: 1/3/9 Dubson hys11 otes, University of Colorado
AC-3 Exercise for the motivated student: how that u 1 µ = for the case of a long solenoid. tart with U 1 =, and use the previously found expressions for and for a solenoid. R circuits (circiuts with 's and R's) 3 things to remember about inductors in circuits: An inductor acts like a battery when its current is changing: of the battery voltage is such as to fight any change in the current. d E =. The direction The current through an inductor cannot change instantly (because that would cause an infinite E). n the steady-state (after a long time), when the current is constant, = const E = the inductor acts like a short (a zero-resistance wire). Example: imple R circuit. E A switch R E witch at position A for a long time: = constant, so E =, = E / R. At t =, switch, and the circuit becomes: The emf in the inductor keeps the current going. Apply oop aw: E d = R = R (ote on signs: d/ < so E >. ) d R = This is a differential equation with an expontial solution. R t t/(/r) t/ τ (t) = e = e = e, τ = R = time constant of R circuit = time for anything in circuit to change by factor of e ast update: 1/3/9 Dubson hys11 otes, University of Colorado
AC-4 /e τ t Another R circuit: E switch R Close switch at t =. At t = +, = (since cannot change instantly), Apply oop aw, inductor acts like second battery: d E = R d d nitially, =, so E =, = E t= As t,, R = R, E = d / As t, E, E = R E / R E (t) = 1 e R t/(/r) t AC oltage and Current atteries produce voltage that is constant in time, DC voltage. The wall socket produces sinusoidally-varying voltage, AC voltage. (DC originally stood for "direct current" but now it just means "constant in time". AC is short for "alternating current" but now means "sinusoidally-varying".) ast update: 1/3/9 Dubson hys11 otes, University of Colorado
AC-5 = constant o wall socket (AC) battery (DC) time o period T = 1 / 6 s time t Wall socket voltage: = (t) = osin π = osin(π f t) = osin( ωt) T n the U, the frequency of "line voltage" is f = 6 Hz = 6 cycles per second (Recall f = 1 / T, period T = 1/6 s) 1 AC symbol for AC voltage AC voltage causes AC current in resistor. Current actually flows back and forth, 6 times a second. = = ω = n( ω t) R R sin( t) o si The instantaneous voltage is (+) as often as ( ), so = avg =, but avg. Electrical engineers always report AC voltage using a kind of average called "root-mean-square" or rms average. AC = "volts AC" = rms = = 1 (in U) The average voltage rms is less than the peak voltage o by a factor of : rms = o o rms t o Why? sin( ωt), sin ( ωt) sin varies from +1 to 1 (sin = +1 1 +1 1.. ) ast update: 1/3/9 Dubson hys11 otes, University of Colorado
AC-6 sin varies from to +1 (sin = +1 +1 +1 +1.. ) The average of sin is ½. 1 o rms = = o sin ( ω t) = o sin ( ω t) = o = Wall socket voltage or "line voltage" : rms = 1, peak = = rms 17 Average vs. instantaneous quantities: ower (AC) = = sin( o ωt)sin( o ω t) = sin( o o ωt) ince sin alternates between and +1, the power alternates between and max = o o. The average value of sin 1 o o = ½ = = = avg o o rms rms The old formula = works OK with AC quantities if we use avg, rms, and rms. All the old DC formulas, = R, = = / R = R, still work fine for AC if we use rms, rms, and avg. Transformers The entire electrical power distribution system in the civilized world depends on a simple device called a transformer. A transformer is a device for transforming AC voltage from one value (say 1 AC) to another value (like 1 AC or AC). A transformer is made of coils of wire, usually wrapped around an iron core. t is a simple device with no moving parts. rimary coil = input coil, with turns in = econdary coil = output coil, with turns out = Fe core ast update: 1/3/9 Dubson hys11 otes, University of Colorado
AC-7 We will show below that = = or = out This "Transformer Equation" says that, for AC voltage, the voltage ratio is equal to the turns ratio. OTE: the transformer only works for AC voltage. f in is DC, then out =. = > 1 "step-up transformer" = < 1 "step-down transformer" (A step-down transformer gives a smaller, but a larger current.) Transformers work because of Faraday's aw: (AC) (AC) (AC) (AC) + Faraday E = roof of the Transformer Equation: f we apply Faraday's aw to the primary and secondary coils, we get: dφ = (1) dφ = () same Φ = A in each turn of primary and secondary because the iron core "guides flux" from to. (1) () = (End of proof.) f a transformer is well-designed, only 1 to 5% power in is lost to heating of coils and eddy currents in the iron core. out in = = = A step-down transformer produces a smaller voltage, but a bigger current (same = ). ast update: 1/3/9 Dubson hys11 otes, University of Colorado
AC-8 ight bulbs and appliances with motors (vacuum cleaners, blenders) use AC voltage to operate. ut devices with electronic circuits (T's, computers, phones, etc) need DC voltage to function. The "power supply" in computers and T's converts the AC voltage from the wall socket into DC voltage (usually 1-15 ) that the electronic circuitry needs. Example of use of transformers: uppose you want to melt a nail by putting a big current through it. What happens if you try to melt the nail by putting 1 AC (from your wall socket) across the nail? Answer: you will blow a fuse or trip a breaker. The resistance of a nail is quite small: R nail 1 3 Ω. The current produced by a 1 voltage difference across the nail is = = 1 = 1 A. This will never happen since your breaker will Ω huge: nail 3 Rnail 1 trip when the current exceeds 15 A. (Here's an experiment you should never try at home: end a nail into a U shape and plug it into your wall socket. Watch the lights go out.) o how do we melt that nail? olution: Use a 1-to-1 step-down transformer. 1 = =, = in = 1 AC 1 out 1 = = = 1 = 1. AC 1 1. = = = = 1 A Ω out 3 Rnail 1 (enough to melt the nail) How much current will this draw from the wall socket? Recall that in = out or =. 1 = = = = = (ot enough to blow the fuse.) in (1 A) 1 A 1 Circuit diagram: = 1 = 1 A 1:1 R nail.1 Ω = 1 A ast update: 1/3/9 symbol for transformer Dubson hys11 otes, University of Colorado
AC-9 ower dissipated in nail = R = (1) (1 3 ) = 144 W will melt nail. Transformers and ower Distribution Economical power distribution is only possible because of transformers. Electrical power is transmitted from the power plant to the city by big aluminum cables (power lines). ome energy is inevitably wasted because the power lines have a resistance, and so they get hot: lost = R cable. n order to minimize this waste, the power must be transmitted from the plant to the city at very high voltage (typically 3 k). A high voltage allows a small current, at a given power (since = ). And a small current means small R losses in the cable. When the high-voltage power is delivered to the city, step-down transformers are used to transform the very dangerous high voltage down to the not-so-dangerous 1 before it enters your home. The voltage is stepped down in stages as it is distributed throughout the city. k step-up step-down step-down 3 k 8 k 1 Home power plant transmission line substation neighborhood A very simplified model of power distribution (transformers not shown): R cable ( made as small as possible) ower plant (t's really AC) R city ast update: 1/3/9 Dubson hys11 otes, University of Colorado
AC-1 ome typical numbers: ower output of plant = out = 1 MW to 1 GW = 1 7 to 1 8 W (fixed by demands of the city) out =, lost = R cable Using = out /, we get lost Fraction of power wasted = f R cable 1Ω and out = 1 8 W, then out out = R cable lost out = R cable f = 5, : f =, : 1 = 1 =.4 8 lost out 5 1 4 ( ) 1 = 1 =.5 8 lost out 1 5 ( ) (4% lost!) (.5% lost) oosting the voltage at which the power is transmitted makes the losses acceptably small. Household Wiring Wall socket = 3-prong plug cold The short slot is the dangerous high-voltage one; short (1-5 ) slot is harder to stick your finger in. tandard electrical wiring colors: black = hot (1 ) "charred black" ground( ) white = cold (few ) "white ice cold" green = ground ( ) "green grass" ever assume the wiring colors are correct! Always check with a voltmeter. hot (1AC) The resistance of copper wires in the walls of your home is less than.1 Ω. o R wire << R bulb 1 Ω. R wire is small, but not zero wires get hot if too much current fire hazard. o all circuits in your house have fuses or circuit breakers which automatically break the circuit if the current exceeds 15 A. ast update: 1/3/9 Dubson hys11 otes, University of Colorado
AC-11 fuse hot line =1 switch 1 AC 1 R R 3 R 1 cold line few volts ground = earth connection Example of voltage drop along a wire: What is the resistance of copper wire, length = 1 m, diameter = 1 mm, ρ = 1.7 1-8 Ω m (typical of wires in the walls of your house.) 8 ρ ρ (1.7 1 )(1) wire R = = = =.54Ω A πr π(.1) f the current through this wire is = 15 A (close to tripping the breaker), what is the voltage drop along this wire? wire = R wire = (15 A) (.54 Ω) =.81 Cost of electricity ower company charges for total energy used. energy = power time ( = W / t, W = t) Unit of energy = kilowatt-hour (kw h) = two hairdryers on for 1 hour. 1 kw h costs about 1 cents (varies). Example of energy cost. What's the bill for a 5 W hairdryer left on for 1 year? d h 1year 365 4 876 hours y d =, $.1 876 h.5kw $438 kw h = (yikes!) ast update: 1/3/9 Dubson hys11 otes, University of Colorado