hwk_7_solutions

hwk_7_solutions - MasteringPhysics Assignment Print View

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Unformatted text preview: MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assigmnentPrint?assignmen... Assignment Display Mode: V PHYSICSGCF08 at: {he at mitigate {m tiréziay. Meantime 5:, aces View Gradng Details A Driven Series L-C Circuit Description: Apply current source to a series L—C circuit; find the total voltage; answer questions about the total voltage (zero!) at resonance. Learning Goal: To understand why a series L—C circuit acts like a short circuit at resonance. An AC source drives a sinusoidal current of amplitude Ia and fi'equency into an inductor having inductance I, and a capacitor having capacitance S? that are connected in series. The current as a fimction of time is given by E {it} Iggsiniwt’jl , Recall that the voltages Vg it} and Vg across the inductor and capacitor are not in phase with the respective currents 1;, {a} and 1C (5}. In particular, which of the following statements is true for a sinusoidal current driver? ANSWER: i V263 both lag irespective‘currents. V t} and Vng both lead their respective currents. i! ‘ lags 11:3} and leads a 'i-*},{t} leads {LG} and {shit} lags reg}. E The phase angle between voltage and current for inductors and capacitors is 90 degrees, or -2- radians. Among other things, this means that no power is dissipated in either the inductor or the capacitor, since the time average X of current times voltage, {6}?” {(3}, is zero. g Part B What is t” {it}, the voltage delivered by the current source? EHint 3.1 Current in a series circuit 3 Note that the current through the cmrent source, the capacitor, and the inductor are all equal at all times. Find the voltage across the capacitor I, : What is is}. {g}, the voltage across the capacitor as a fimction of time? Part 3.2.2 Find the charge on the capacitor What is the charge {I it} on the capacitor? Express your answer in terms of the capacitance £1? and the voltage V}; {1} across it. ANSWER: q {Q (fife; . {3) Part B.2.b What is the current in term of the charge on the capacitor? With the conventions in the circuit diagram, what is the current 19* {if} on the capacitor in tents of its voltage? Express your answer in terms of the capacitance {'5‘ and the voltage i? {3} across it and/or its derivative th};§§}jdi. ANSWER: ‘ , Ci“? 1‘ . {$83 = tilihii} Integrate both sides of the, equation {93} a: {git} av C M (once you substitute in the appropriate expression for {C it» The constant of integration is zero because there is no average (DC) current or voltage in a pure AC circuit. 1 ofll 12/5/2008 10:37 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignmen.. Express your answer interms of {5, 0, oz, and f. ANSWER: Part 33 What is the voltage across the inductor? What is h}, {I}, the voltage across the inductor as a firnction of time? H'mt B.3.a Voltage and current for an inductor Recall that in an inductor, Vfi?) Note that the voltage drop across the inductor has the opposite sign to that of the EMF generated by the inductor: E w L if: Express V; in terns of {9, u}, and 1s. ANSWER: . Vb {t} = {filmmfwii Hint 3.4 Total voltage You should have defined the voltages across the capacitor and inductor in such a way that the total voltage Vic”; is equal to legit} -§- Viffi, that is, so that total voltage is obtained by adding the voltages at time i, with careful attention paid to the signs. Express is” in terns of some or all of the variables {5, X; m g}, and Xe m , or C and in oz, and. of come, time I. 3 ANSWER: 's z i s rank: With if}, and V“; the amplitudes of the voltages across the inductor and capacitor, which of the following statements is true? i. ANSWER: ' At Very high frequencies if}: a V}, and at very low frequencies 1% at V5. ‘ At very high frequencies V3, as V: and at very low frequencies {[1, x; 1,15. V2: .fi 32‘} for all frequencies. 1”}, T1?“ ii: for all frequencies. is}: and V; are about the same at all frequencies. ‘ Part1) um i The behavior of the L-C circuit provides one example of the phenomenon of resonance. The resonant fiequency is 3);, m 1 At this frequency, what is the amplitude 1:}, of the voltage supplied by the current sotn'ce? Express your answer using any or all of the constants given in the problem introduction. ANSWER: ' ti, = i; % g , g ’5? Part E Which of the following statements best explains this fact that at the resonant frequency, there is zero voltage across the capacitor and inductor? ANSWER: ‘ ‘ The voltage Mfg? is zero at all times because Vii: “3%{1}. The voltages V{£§; Vat); and Wag} are zero at all times. The voltage 13$} is zero only when the current is zero. The voltage 393‘} is zero only at times whenthe current is stationary (at a max or min). At the resonant frequency of the circuit, the current source can easily pushthe current through the series L—C circuit, became the circuit has no voltage drop across it at all! Of course, there is always a voltage across the inductor, and there is always a voltage across the capacitor, since they do love a current passing through them at all times; however, at resonance, these voltages are exactly out of phase, so that the net effect is a current passing through the capacitor and the inductor without any voltage drop at all. The L-C series circuit acts as a short circuit for AC currents exactly at the resonant fi'equency. For this reason, a series L-C circuit is used as a {k trap to conduct signals at the resonant frequency to ground. A Series L-R-C C' Description: Derive the impedance formula using phasor diagrarm and then analyze the formula to obtain the resonance conditions Learning Goal: To understand the use of phasor diagram in calculating the impedance and resonance conditions in a series L—R-C circuit 2 In this problem, you will consider a series L-R-C circuit, containing a resistor of resistance R, an inductor of inductance L, and a capacitor of capacitance if, all connected in series to an AC source providing an alternating voltage '8} :2 Varieties). You may have solved a number of problems in which you had to find the effective resistance of a circuit containing multiple resistors. Finding the overall resistance of a AC my“: ; circuit is ofien of practical interest In this problem, we will start our analysis of this IrR-C circuit by finding its effective overall resistance, or impedance. The i impedance 3 is defined by it“ 2:: $9, where :4; and [a are the amplitudes of the voltage across the entire circuit and the current, respectively E c ; 2 ofll 12/5/2008 10:37 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assigmnen... i Part A Find the impedance Z of the circuit using the phasor diagram shown Notice that in this series circuit, the current is same for all elements of the circuit: You may find the following reminders helpful: '2 3% 0 In a series circuit, the overall voltage is the sum of the individual voltages. 0 In an AC circuit, the voltage across a capacitor lags behind the current, whereas the voltage across an inductor leads the current. Vt I The reactance (efiecfive resistance) of an inductor in an AC circuit is given by 37;, «22.. I i V ii O The reactance (eEective resistance) of the capacitor in an AC circuit is given by Xg; . R M w won- } Hint A.1 Finding individual voltages : For each element of the circuit, "voltage amplitude" = "current amplitude" times "effective resistance." The value of the current amplitude, In, is, as we mentioned, the same for all three circuit elements. l-Iint A.2 Finding the overall voltage A phasor vector is defined in such a way that its "horizontal" component represents the instantaneous values of the corresponding quantities. Since the sum of the vector components is equal to the component of the sum of the ectors, the sum of the individual voltage vectors represents the hasor corres ndi to the "total" voltage, that is, the terminal voltage of the source. Hint .3 Combining the vectors Here is a diagram showing how to add the individual voltages. Since there are many right angles in this diagram, the Pythagorean theorem is useful here. To find the total voltage, first add the vectors representing V}, and VC. Since these vectors have opposite directions, the magnitude of their sum equals the difference of the vectors' rmgnitudes. We then add the vector representing t}: to get the resultant vector 33;. What is the magnitude it}, of this resultant vector? Express your answer in terms of 3%, is, (f and w. ANSWER: Part B Now find the tangent of the phase angle a: between the current and the overall voltage in this circuit. Express martinis} in terns of R, L, (7’, and ANSWER: . L w ( mute} = w R ote that if X; to Ag}, the overall voltage will lead the current; if 1th e: X , the overall voltage will lag the current. Inthe phasor diagram for this problem, the voltage leads, and the current lags. We may be iriierested in finding the resonance for the circuit, in other words, the conditions corresponding to the nmdnnnn current amplitude produced by a voltage source "of a gi‘ven‘amplihfie. Finding such conditions 3 has an immediate practical interest. For imtance, tuning a radio means, essentially, changing the parameters of the circuitry so that the signal of the desired frequency has the maximum possible amplitude. s E ‘ Part C Imagine that the parameters 3?, 1.}, (.7, and the amplitude of the voltage 12}, are fixed, but the frequency of the voltage source is changeable. If the frequency of the source is changed from a very low one to a very high one, the curre amplitude lg will also change. The frequency at which .533 is at a maximum is called resonance. Find the frequency .333 at whichthe circuit reaches resonance. Hint C.l Analyzing the impedance The maximum current amplitude corresponds to the minimmn impedance value 2 . What is the minimum value of .3? At what value of as is it reached? Express your awer in terns of any or all of R, L, and 3 ofll 12/5/2008 10:37 AM MasteringPhysics: Assignment Print View http://sessionmasteringphysics.com/myct/assignrnentPrinflassignmerL.. % E E i ANSWER: ;' t At resonance, )5; ,1 _ Km. The volmge across the capacitor exactly cancels tint across the inductor, since the two are out of phase and have the same amplitude. Thus, the inductor and capacitor effectively cancel out inthe formula for the impedance of the circuit. It should not come as a 5 rise that the resonant fie uenc e uals the natural Ere uenc of the oscillations in a source-flee L-C circuit i What is the phase angle as betweenthe voltage and the current when resonance is reached? Look back at the original phasor diagram in Part A. How would it change when the impedance is at a minimum? Remember that when resonance is reached, KL M Xi» ANSWER: i ’ PaflE ~ Now imagine that the parameters 3, L, at, and the amplitude of the voltage ti; are fixed but that the value of {3‘ can be changed This is one of the easiest parameters to change when "tinting" such (radio frequency) circuits in orde to Hake them resonate. This is because the capacitance can be changed just by moving the capacitor plates closer or farther apart. Find the resonance value of capacitance (fa. < Express your answer in term of L and ANSWER: ’ 1 r: it When resonance is reached, XL xx X01. The average power dissipated in the circuit under this condition is at a maximum It” m Note that this expression does not depend on the values of L and (T’. A Resistor and a Capacitor in a Series AC Circuit i Description: Use reactance to find the current through and voltage across elements of simple RC circuit. 3 A resistor with resistance It and a capacitor with capacitance {2’ are connected in series to an AC voltage source. The time-dependent voltage across the capacitor is given by {git} m the Siiimf. Part A What is the amplitude a ofthe total current 1g; inthe circuit? Hint A.l How to approach the problem i For a single-loop circuit, the current flowing through the capacitor is equal to that flowing through the resistor, and is equal to the total current in the circuit. Hence you can work out the current in the circuit by using Ohm's law to find the current though the capacitor. ‘ ’ Hint A.2 Applying Ohm's law to a capacitor To apply Ohm's law to a capacitor, you must use the capacitor's reactance in place of resistance Ohrn's law for a capacitor is if a: 1.32;, Where 3 is the amplitllde ofthe Current through the CaPaCiiDI‘y V is the amplitude ofthe voltage, and X3 is the reactance of the capacitor. Hint A3 The reactance of a capacitor The reactance of a capacitor is given by X}; a: if, where a; is the angular frequency of the voltage across the capacitor. Substituting this expression for X5 into Ohm‘s law for the capacitor will enable you to calculate the g1 ,. ' amplitude of the current in the capacitor. 2 " Express your answer in terms of any or all of It, 0, 1/24,, and ANSWER: 1,6 = p;ka : V Part B What is the amplitude 1"}; of the voltage across the resistor? 3.1 ' aerating :0 to 9;, t The current thro the capacitor is the same as that thro the resistor, and you have already worked out that I V7 .363. Hence, you can apply Ohm's law to work out the amplitude of the voltage across the resistor. 3! f. n 'K Express your awerin term of V“. C, as, and R. ANSWER: : 3/8 = gywcw g {39 1);? Vet; mo RN, and u} w it? range. what is is}? Calculating the answer Use the equation obtained in Part B to work out the answer. Be careful of powers often in your calculation. Express your amwer numerically, in Inillivolts, to the nearest integer. 4 ofll 12/5/2008 10:37 AM MasteringPhysics: Assignment Print View http://sessionmasteringphysics.com/myct/assigmnentPrint?assignmen... «mm «W.» Me, Description: Short quantitative problem on resonance in a series RLC circuit. Based on YoungGeller Quantitative Analysis 22.5. i In this problem we consider the resonance curve for a circuit with electrical components 3;, a, and (Z? and resonant frequency Part A t For given values of f; and f“, if you double the value of L, how does the new resonance curve differ fiomthe original one? Hint How to approach the problem i g In a series R—LvC circuit, the resonance frequency is determined by I. and whereas the peak value of the current in the circuit depends only on 22. Since the value of I? is being kept constant, the peak of the resonance curve is i 3 unchanged. To solve this problem, use proportional reasoning to find a relation between the resonance frequency jg and inductance L. I Find the simplest equation that contaim these variables and other known quantities from the problem 0 Write this equation twice, once to describe the original circuit and ayin for the circuit with greater inductance. I You then need to write each equation with all the constants on one side and the variables on the other. Since the variable is off} in this problem, you will write the equations in the form it, ~ ‘ ~. 0 To finish the problem you need to compare the two cases presented. For this question you should find the ratio of the resonance frequency of the original circuit to that of the circuit with greater inductance. Hint A.2 Resonance frequency In a series R—L—C circuit, the resonance frequency 3;; is given by Part A.3 Find an expression for the new resonance frequency Write an expression for the resonance frequency f, of the circuit when the indwtance 1, is doubled? Express your answer in terns of some or all of the variables 32, L, and (2. ANSWER: ANSWER ; Boththe peak heightandpeak aeqicmy double. The peak height will be half as great, and the peak frequency will double. Q The peak height won't change, and the peak frequency will be £565 times as great. The peak height won't change, and the peak frequency will be half as great. an R-"L-C circuit, theiresonance peak depends only on firwhile the resonant fi'equency is determined by both 3, and 0. In particular, for a given value of . , the resonance frequency is inversely proportional to the square root of 1,. Similarly, for a given value of L, the resonance fi-equency is inversely proportional to the square root of 1?. Part B For given values of {it and {,“f, if you double the value of L, how does the new rms current at resonance {gm difl‘er from is original value? Assume that the voltage amplitude of the ac source is the same. : Hint 3.1 How to approach the problem ’ Recall that in an ac circuit, for a given voltage, the rms current is always inversely proportional to the impedance of the circuit. This holds also at resonance. To solve this problem, use proportional reasoning and find a relation between the ms current 1% and the impedance Z . 0 Find the simplest equation that contains these variables and other known quantifies from the problem 0 Write this equation twice, once to describe the original circuit and again for the circuit with greater inductance. 0 You need to write each equation so that all the constants are on one side and the variables are on the other. Since the variable is {mm in this problem, you will write the equations in the form {m v a v 0 To finish the problem you need compare the two cases presented. For this question you should find the ratio of the rms current in the original circuit to that in the circuit with greater inductance. Hint 3.2 rm across an ac circuit '§ §. In a series R—L—C circuit1 the m5 voltage Km across the circuit is always proportional to the rim current {m in it If ,3 is the impedance of the circuit, then gins m Irma? Part 3.3 Find the impedance of the circuit : ' What is the impedance .5! of a series R-IIC circuit at resonance? mucosa H Impedance ti a'sciies R-L-C circuit ' The impedance Z of a series R—L—C circuit is a function of R, L, t", and the angular frequency ,9. It canbe expressed as are V3122 «r on - where X; and Xi»; are the respective inductive and capacitive reactances of the circuit. 5 ofll 12/5/2008 10:37 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assigrnnentPrint?assigmneIL.. Hint B.3.b Condition for resonance 5 At resonance, the inductive and capacitive reactances must be equal. That is, Xx, as X“ Express your answer in terns of some or all ofthe variables Ii, 3;, 0', and use. ANSWER: 2 '=' f; At resonance, impedance circuit depend uni-iiv'lhus, for a given voltage, will in“, change when if, i In,» is twice as great. z Inn, is halfas great E i . {W is 35% times as great. 3 i e {m is unchanged. i p ‘ i In a series R—L—C circuit, for a given voltage, the me current is always inversely proportional to the circuit impedance. Since at resonance the impedance depends only on If, the me current in the circuit reneins constant is changed. Note that this is true only at resonance. § when either is or Leaning Goal: To understand the concepts explaining the operation of transfomters. One of the advantages of alternating current (ac) over direct current (dc) is the ease with which voltage levels can be increased or decreased. Such a need is always present due to the practical requirements of energy distribution. On g the one hand, the voltage supplied to the end 15ers must be reasonably low for safety reasons (depending on the country, that voltage my be 110 volts, 220 volts, or some other value of that order). On the other hand, the voltage used g in transmitting electric energy must be as high as possible to minimize {3}? losses in the transmission lines. A device that uses the principle of electmrmgnetic induction to increase or decrease the voltage by a certain factor is called Description: Several conceptual and qualitative questions introducing the principles behind the transforner. Both ideal and nonideal transformers are discussed. , i E a transformer. The main components of a transformer are two coils (winding) that are electrically insulated from each other. The coils are wrapped around the same core, which is typically made of a naterial with a very large relative ; pemeability to ensure maximum mutual inductance. One coil, called the primary coil, is connected to a voltage source; the other, the secondary coil, delivers the power. The altemating current in the primary coil induces the g changing magnetic flux in the core that creates the emf inthe secondary coil. The magnitude of the emf induced inthe secondary coil can be controlled by the design of the transformer. The key factor is the number of turns in each E coil. i ' Consider an ideal transformer, that is, one in whichthe coils have no ohmic resistance and the magnetic flux this is the same for eachtum of both the primary and secondary coils. If the number of turns in the primary coil is N, and and £2 3x3 , {fifths : i I Since both emfs oscillate with the sane frequency as the ac source, the formula above can be applied to the instantaneous amplitude or the m values of the emfs. Moreover, if the coils have zero resistance (as we assumed), then for ach coil the terminal voltage will be equal to the induced emf. Therefore, we can write = ; E Va}: 3 {dais- l )e If the secondary circuit is completed by a resistance R, then I2 15;. Combining this withthe two equations above gives 3 2 E i r i V3 .532 i V33 t; m as a; m A. ‘5 a {A} a Dividing the first and last expressions by V: and then inverting gives In other words, the current in the primary coil is the sane as if it were connected directly to a resistance equal to W . In a way, transformers "transform" resistances as well as voltages and currents. In reality, no transforner x? . is ideal. There are always some energy losses. However, modern transformers have very high efliciencies, usually well exceeding 90%. 3“ , _ In answering the questions below, consider the tramformer ideal unless otherwise noted. E E 6 ofll 12/5/2008 10:37 AM MasteringPhysics: Assi grimth Print View http://SeSSion.maSteringphysics.cont/myct/assignmentPrint7assignmen... Part A The primary coil of a transformer contains 100 turns; the secondary has 200 turns. The prinnry coil is connected to a size AA battery that supplies a constant voltage of l .5 volts. What voltage would be meastn'ed across the secondary coil? i ANSWER: ; Q zen; are V L5 V 3i} \3 n order for an emf to be induced inthe secondary coil, the flux through it must be changinggitlfiefore‘, the current in the prinar‘y‘c‘oil must also be changing If a constant voltage is supplied to the prirmry coil, no emf would be nduced in the secondary and therefore the secondary voltage would be zero. Part B _ A transfonrer is intended to decrease the me value of the alternating voltage floor 500 volts to 25 volts. The primary coil contains 200 turns. Find the necessary number of turns 33,; in the secondary coil. 3 ANSWER: it»; = m This lS a step-down transformer: The voltage decreases. i § Part C A transformer is intended to decrease the rms value of the alternating current from 500 amperes to 25 amperes. The primary coil contains 200 turns. Find the necessary number of turns .35; inthe secondary coil. i i i » f Hint CJ How to approach this pmblem 3 ' Recall that in an ideal transformer, the powers in the prirmry and secondary coils are equal. Therefore, if the current decreases lay a factor of 20, the voltage must increase by a factor of 20. g This is a step-up transformer: The voltage increases by the same factor by which the current decreases. . Part 15 ' In a transformer, the primary coil contains 400 turns, and the secondary coil contains 80 turns. lfthe primary current is 2.5 amperes, what is the secondary current I2? Express your answer numerically in amperes. ANSWER: > 1% = 135 A Tart E : The prirmry coil of a transformer has 200 turns and the secondary coil has 800 turns. The power supplied to the primary coil is 400 watts. What is the power generated inthe secondary coil if it is terminated by a 20-ohm resistor? Hint EJ In the ideal world... This is an ideal transformer: There is no energy loss. ANSWER: ' w w i 3300 W‘ t cm w an) W Sim W M3009 WV 3% W i In case of an ideal transformer, the power in the primary circuit is the same as that in the secondary circuit. Part F The prirmry coil of a transformer has 200 turns, and the secondary coil has 800 turns. The transformer is connected to a lZO-volt (ms) ac source. What is the (rms) current A in the primary coil if the secondary coil is temiinated by a 20-ohm resistor? Hint F.1 How to approach the problem E t . Recall that transformers "transform" resistances as well as voltages and currents. Express your answer numerically in amperes. ANSWER: 3‘ 39% A _PartG ,3 § g z 7 ofll 12/5/2008 10:37 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assigmnentPrint?assignmen... A transformer supplies 60 watts of power to a device that is rated at 20 volts (rms). The primary coil is connected to a 120-v01t(rms) ac source What is the current I" in the primary coil? 5 Hint G.l How to approach the pmblem Find the current in the secondary coil and use the fact that the same power is generated in both coils. Express your answer in amperes. ANSWER: ; a = or; a i Express your amwer as a percentage. ANSWER: 6 = % 2 .3 § § Secondary Voltage and Current in a Transformer Ranking Task 1 Description: Rank the current and voltage inthe secondary coil of difi'erent transformers (ranking task) :PartA ’ Which of the transformers are step-up transformers? Which of the transformers are step-down transformers? Hint A.1 Step-up and step-down transformers A transformer is referred to as step-up if the secondary voltage is larger thanthe primary voltage. Similarly, a transformer is referre Hint A.2 Tum ratio The turns ratio, sift-’86}, is the factor that determines the characteristics of a transformer. If the ratio is greater than one, the transfomier "steps up" the voltage by this factor. If it is less than one, it "steps down" the voltage. 2 Place the appropriate tmmfonners into the two categories listed below. ANSWER: E View Part B Rank the transformers on the basis of their ms secondary voltage. t . Voltage and number of turns The re ortionality constant between voltage and number of turns is the same on both sides of the transformer. Hint 3.2 Since the proportionality constant between voltage and number of turns is the same on both sides of the transformer, we can set‘the ratios Vgat on both sides equal to each other: Secondary voltage 2v; Therefore, the secondary voltage is just the product of the primary voltage and the turns ratio. » I Rank from largest to smallest. To rank ite as equivalent, overlap them. ANSWER: 8 ofll 12/5/2008 10:37 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assigmnentPrint?assignmen.. i 100 A of ms current is incident on the prirmry side of each transfomier. Rank the transfomiers on the basis of their ms secondary current. Energy conservation ince energy flow (power) in a circuit is given by 1’ a It”. : nergy conservation requires that any increase in voltage be accompanied by a corresponding decrease in current, and vice versa. Thus, a step-up transformer will step down current. Hint C.2 Stepping down current in a step-up transformer ‘ Since voltage is stepped up by the turns ratio, onservation of energy ensures that current is stepped down by the same ratio. This can be written as Rank from largest to smallest. To rank item as equivalent, overlap them. 2 z s i g i g i i i E z, § § E g $ Wi’roblem 28.28 Description: An LC circuit includes a 20(kem 1 pt) - (kem lpt) mu F capacitor and has a period of 5.0 ms The peak current is 25 mA. (a) Find the inductance. (b) Find the peak voltage. E An {(1}? circuit includes a 233 n: gt? capacitor and has a period of 5.0 any. The peak current is 25 mA. : 3 3 g § § § PARA . .. Findtheinductance. Express your answer using two significant figures. ANSWER: I i; = so in}! ; P.a « Find the peak voltage. I Express your answer using two significant figures. ANSWER: V = mm my Problem 28 29 WWW...“ \ Description: An FM radio transmitter requires an DC circuit that oscillates at the 89.5—MHZ transmitter fiequency. (a) What should the inductance be if the circuit capacitance is C? i An FM radio transmitter requires an Lei“ circuit that oscillates at the 89.5~ liin transmitter fiequency, yPartA ' What should the inductance be ifthe circuit capacitance is 41 33?? i ; xpress your answer using two significant figures. ANSWER: - ' ; % Problem 28.32 1 ‘Description: A series RLC circuit has R(kem lpt) = (kem 1 pt) 18( k)0mega , C (kem 1 pt) : (kem lpt) 14 mu F, and L = 0.20( H). (a) At wlmt frequency is its impedance lowest? (b) What is the value of the impedance at this frequency? Aseries RLC circuit has 3 m 18%}. C? m 3433?, and i} «a {kiwi}. . Part A : At what fi'equency is its impedance lowest? Express your answer using two significant figures. 2 3 : E E 9 ofll 12/5/2008 10:37 AM MasteringPhysics: Assigmnent Print View http://sessionmasteringphysics.com/myct/assignmentPrint?assignmen... i Part B ‘- What is the value of the impedance at, this frequency? ' Express your answer using two significant figures. g ANSWER: : st= ram 22 i’roblem 20.36 g Description: An electric water heater draws 20 A are at 240 V m and is purely resistive. A large AC motor has the same current and voltage, but inductance causes the voltage to lead the current. by 20 degree(s) . (a) Find the § power consumption in water heater. é An electric water heater draws 20 A nns at 240 \5 fins and is purely resistive. A large AC motor has the same current and voltage, but indwtance causes the voltage to lead the current by 20 “ . Part Find the power consumption in water heater. Express your amwer using two significant figures. ANSWER= f P= is»: w - Part B 7 Find the power consumption in AC motor. Express your answer using two significant frgmes. ANSWER: ' in 4500 w Description: A transformer is designed to step 230-V European power down to 120 V needed to operate North American devices. (a) If the primary has NO turns, how many should the secondary have? (b) If the primary current is s 10, what is the secondary current? A transformer is designed to step 230- V European power down to 120 V needed to operate North American devices. Problem 28.37 i i l Part A Ifthe primary has 900 turns, how tinny should the secondary have? Express your answer using two significant figures. ANSWER: Express your answer using two significant figures. E 5 i 3 ANSWER: in 9‘ s fig}. Problem 28.53 > Description: Figure below shows the phasor diagram for an RDC circuit. (a) Is the driving fi’equency above or below resonance? 2 i 3 Figure below shows the phasor diagram for an RLC circuit. 2 E i i g E i i-PartA E Is the driving fiequency above or below resonance? 10 of 11 7 12/5/2008 10:37 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/mchassigmnentPrint‘?assignmen... Summary 0 of 12 items complete (0% avg. scar a g o of 101 points a 11 0111 12/5/2008 10:37 AM i step-down transformers step-up transformers “ E _ § s; \ if \\\\\VM»WWWW\ 1;. 3 Rfi mm? : smallest : largest 3%} §~§§Y§§§ ¥3§§§§s smallest ’ <§§3€3§§ game KKK} Mme 1£§§§$ w mew: ‘ -\\ IL: The correct ranking cannot be determined. ‘ ...
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This note was uploaded on 03/08/2011 for the course PHYS 6c taught by Professor Smith,d during the Spring '08 term at UCSC.

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hwk_7_solutions - MasteringPhysics Assignment Print View

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