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Week 12 - Alternating Current November 21, 2012 The city lights of Prince George, British Columbia viewed in a motion blurred exposure. The AC blinking causes the lines to be dotted rather than continuous. Exercise 12.1: Discussion Questions 1 a) The current in an ac power line changes direction 120 times per second, and its average value is zero. Explain how it’s possible for power to be transmitted in such a system. Answer: Power can be transmitted because the power delivered to another circuit is p = iv, (1) I.e. the product between the curent and the voltage, and the average power is not necessarily zero. For example if the circuit to which the ac power line is connected is a pure resistor, then v = iR and the instantaneous power is p = i2 R. 1 Figure is from http://en.wikipedia.org/wiki/File:City_lights_in_motion.jpg 1 (2) This is always positive which means that power is dissipated in the resistor at any moment. The average power is given by Pav = 1 2 I R. 2 (3) b) Fluorescent lights often use an inductor, known as ballast to limit current through the tubes. Why is it better to use inductor than a resistor for this purpose? Answer: One reason that the inductor is preferable over the resistor is that the average power dissipated in the inductor is practically zero. This means that we dont waste precious energy on just limiting the current so that everything can go into producing visible light. The applied voltage does work on the current as the back emf resists a rising current, this energy is then stored in the magnetic fields of the inductor and given back when the emf of the inductor does work on a decreasing current. c) A light bulb and a parallel-plate capacitor with air between the plates are connected in series to an ac source. What happens to the brightness of the bulb when a dielectric is inserted between the plates? Explain. Answer: The impedance of the circuit goes down as the dielectric is inserted so the current amplitude goes up and more power is delivered to the bulb. Therefore the bulb glows brighter. Mathematically this is seen by looking at the expression for the impedance of the circuit s Z= q R2 + XC2 = R2 + 1 ωC0 2 (4) Now as the dielectric is inserted the capacitane of the goes to C = KC0 whereK is the dielecric constant and since K > 1 the capacitive reactance goes down. Physically what happens is that when the dielectric is inserted, the dipoles in the dielectric is aligned in such a way that they create an additional elecric field which opposes the original. The electric field weakens which allows more charge to accumulate on the plates for the voltage provided by the ac source. The consequence of the capacitor allowing for a greater amount of charge to accumulate is that a greater current can flow in the circuit. d) Can a transformer be used with dc? Explain. What happens when a transformer designed for 230 V ac is conneced to a 230 V dc line? Answer: No it can not! One of the great advantages of ac over dc is that voltages can be stepped up and down using a transformer. This allows for high voltage transmission which very effectively reduces energy loss due to heating of transmission wires. If you connect a dc line to any transformer it will not do anything for you. The primary will not even deliver an emf to the secondary. This is because the emf in the secondary was created by a 2 flux change trough it’s coil. With no changing current we won’t have changing magnetic fields and therefore no changing flux. Exercise 12.2: Off to the USA! You take your hair blower off the USA, where the electrical outlets puts out 120 V (rms) instead of the 230 V (rms) found here in Norway. The blower puts out 1600 W (average) at 230 V. a) What could you do to operate your blower via the 120 V line in USA? Solution: We have to get our hands at a step down transformer. It will need to have a ratio V2 230 V 23 N2 = = = N1 V1 120 V 12 (5) of windings in the primary and secondary. Answer: Get a transformer with the ratio V2 23 N2 = = . N1 V1 12 (6) b) What current will your blower draw from an outlet in USA? Solution: The current it will draw in USA is determined by the average power, I= Pav 1600 A = 13.3 A = V1 120 (7) Answer: I = 13.3 A c) What resistance will your blower appear to have when operated at 120 V? Solution: The original resistance of the hair blower is 3 (8) R= V12 (230V )2 = Ω = 33.1 Ω, Pav 1600 (9) and the transformed ressitance of hair blower is given by equation (31.37) in the book. Ref f = 33.1 R = Ω = 9 Ω. 2 (N2 /N1 ) (23/12)2 (10) Answer: Ref f = 9 Ω. (11) Figure 1 Exercise 12.3: The L-R-C Parallel Circuit A resistor, inductor, and capacitor are connected in parallel to an ac source with voltage amplitude V and angular frequency ω. This is shown in figure 1. Let the source voltage be given by v = V cos ωt. a) Show that the instantaneous voltages vR , vL and vC at any instant are equal to v and that i = iR + iL + iC , where i is the current through the source and iR , iL and iC are the currents trough the resistor, inductor and capacitor, respectively. Solution: Applying Kirchoff’s loop rule to the three loops including the source voltage v we find that v = vR = vL = vC , and applying the junction rule to one of the leftmost junctions we get i = iR + iL + iC . b) What are the phases of iR , iL and iC with respect to v? Use current phasors to represent i, iR , iL and iC . In a phasor diagram, show the phases of these four currents with respect to v. Solution: The resistor voltage has a zero phase compared to the voltage, so since iR = vR /R, this also holds for the current iR . Similarly the inductor voltage leads the current by 90◦ , so this means that the current iL lags behind the voltage by 90◦ . Finally the capacitor voltage lags behind it’s current by 90◦ , so the capacitor current iC leads the voltage by 90◦ . The phasor diagram of this situation is shown in figure 3. 4 Answer: Figure 2 Figure 3 c) Use the phasor diagram in (b) to show that the current amplitude I for the current i trough the source is given by q 2 + (I − I )2 . I = IR (12) C L Solution: The phasor diagram and the pythagorean theorem implies that the current amplitude is I= q 2 + (I − I )2 IR C L 5 (13) d) Show that the result of part (d) can be written as I = V /Z with s 1 = Z 2 1 1 + ωC − . R2 ωL (14) Solution: The current amplitudes is related to their respective reactances by IR = V , R IL = V V = , XL ωL IC = V = ωCV XC (15) so a substitution into equation 12 then yields s I= 2 1 1 + ωC − V R2 ωL (16) where we have factored out the voltage V from under the square root sign. Now from the definition of impedance I = V /Z so therefore s I= 2 V 1 1 V = + ωC − R2 ωL Z (17) implying 1 = Z s 2 1 1 + ωC − . R2 ωL (18) 2 1 1 + ωC − . R2 ωL (19) Answer: 1 = Z s Exercise 12.4: An L-R-C parallel circuit is connected to an ac source of constant amplitude V and variable angular frequency ω. a) Find expressions for the amplitudes IR , IL and IC of the currents through the resistor, inductor and capacitor as functions of ω. Solution: In the solution of d) we pointed out how these ampliudes was related to their reactances. We have IR = V , R IL = V , ωL 6 IC = ωCV. (20) Answer: IR = V , R IL = V , ωL IC = ωCV. (21) b) Graph IR , IL and IC as functions of ω for V = 200 V, R = 50 Ω, L = 1.0 H and C = 20 µF. Solution: A sketch of the behaviour of the different currents as a function of ω is shown in figure 5. Figure 4 Answer: c) Discuss the behaviour of IL and IC in the limits ω → 0 and ω → ∞. Explain why IL and IC behave why they do in these limits. Solution: In the limit where ω → 0, IL → ∞ and IC → 0. This happes because when the frequency becomes small, there almost no rate of change in the current which means the back emf of the inductor is really small, so a lot of current is let trough. The capacitor, on the other hand, ’fills up’ quickly and stops the current. In the limit ω → ∞, IC → ∞ and IL → 0 the rate of change of current trough the inductor is huge which causes it to set up a huge back emf which almost stops the current entirely. The capacitor has no time to ’fill up’ now so almost all the current can flow back and forth freely. d) Calculate the resonance frequency (in Hz) of the circuit, and sketch the phasor diagram at resonance. What is the current amplitude trough the source at this frequency? Solution: 7 Figure 5 f0 = ω0 1 1 √ √ = = Hz = 36.6 Hz 2π 2π 1 × 20 × 10−6 2π LC (22) Figure 6 We’ve calculated the impedance in the previous problem and found that s I= 2 1 1 + ωC − V. R2 ωL (23) At resonance, XL = XC which means that the term within the parentheses vanishes. The current amplitude trough the source is therefore 8 I= V = 4.0 A. R (24) The capacitor and inductor are out of phase by 180◦ as shown in figure 7. Answer: f0 = 36.6 Hz (25) Figure 7 I= V = 4.0 A. R (26) e) At the resonance frequency, what is the current amplitude trough the resistor, through the inductor and trough the capacitor? Solution: The current trough the inductor and capacitor is equal because their reactances are equal and they have the same voltage applied across them IL = IC = ω0 CV = 36.6 × 20 × 10−6 × 200 A = 146 mA (27) and the current amplitude trough the resistor is IR = V = 4 A. R (28) Because that the capacitor and inductor current are out of phase with eachother by 180◦ they exactly cancel eachother out, and all the current runs trough the resistor at resonance. 9 Answer: IL = IC = 146 mA IR = V = 4 A. R (29) (30) Figure 8 Exercise 12.5: A High-Pass Filter One application of of L-C-R series circuits is to high-pass or low-pass filters, which filter out either the low or the high frequencies in a signal. A high-pass filter is shown in figure 8, where the output voltage (filtered signal) is taken out between the a and b terminals. Derive an expression for Vout /V , the ratio of the output and source voltage amplitudes, as a function of the angular frequency ω of the source. Show that when ω is small, this ratio is proportional to ω and thus also is small. Show that the ratio approaches unity as the frequency gets large (The signal passes trough unaffected at large frequencies). Solution: Let’s first express the source voltage amplitude in terms of R, L, C, I and ω. We draw a phasor diagram as shown in figure 9 and from this we find that the voltage equals Figure 9 10 s V = p R2 + (XL − XC )2 I = R2 2 1 + ωL − I. ωC (31) By the exact same procedure we find the voltage amplidue across the a and b terminals which equals the voltage across the RL-combination Vout = q p R2 + XL2 I = R2 + ω 2 L2 I. (32) Thus their ratio is Vout = V s R2 + ω 2 L2 R2 + ωL − . 1 2 ωC (33) Now when ω is small the 1/ωC term domiates completely in the denominator, while in the numerator the R terms dominates. Therefore Vout → V s R2 = ωCR 1 2 (34) ωC showning that the output voltage indeed is proportional to the frequency of oscillation at low frequencies. At high frequencies however the 1/ωC term in the denominator becomes negigble and Vout → V r R2 + ω 2 L2 = 1. R2 + ω 2 L2 11 (35)