MP_Assignment#3_soln

MP_Assignment#3_soln - MasteringPhysics: Course Home...

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Unformatted text preview: MasteringPhysics: Course Home Student View 2/22/11 9:05 AM Summary View Diagnostics View Print View withEdit Assignment Answers Settings per Student MP_Assignment#3 Due: 8:00am on Monday, February 21, 2011 [ Print ] Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy ± The R- L Circuit: Responding to Changes Description: ± Includes Math Remediation. Contains a derivation of the expressions for the current in a series R - L circuit and a number of related conceptual questions. It is a long problem, but it may be useful to tackle it in one sitting. Learning Goal: To understand the behavior of an inductor in a series R - L circuit. In a circuit containing only resistors, the basic (though not necessarily explicit) assumption is that the current reaches its steady - state value instantly. This is not the case for a circuit containing inductors. Due to a fundamental property of an inductor to mitigate any "externally imposed" change in current, the current in such a circuit changes gradually when a switch is closed or opened. Consider a series circuit containing a resistor of resistance and an inductor of inductance connected to a source of emf and negligible internal resistance. The wires (including the ones that make up the inductor) are also assumed to have negligible resistance. Let us start by analyzing the process that takes place after switch is closed (switch remains open). In our further analysis, lowercase letters will denote the instantaneous values of various quantities, whereas capital letters will denote the maximum values of the respective quantities. Note that at any time during the process, Kirchhoff's loop rule holds and is, indeed, helpful: . Part A Immediately after the switch is closed, what is the current in the circuit? Hint A.1 How to approach the problem Recall that inductors act to oppose changes in the current. So currents in a loop containing an inductor cannot be discontinuous (in time). ANSWER: http://session.masteringphysics.com/myct/courseHome Page 1 of 23 MasteringPhysics: Course Home 2/22/11 9:05 AM ANSWER: zero Part B Immediately after the switch is closed, what is the voltage across the resistor? Hint B.1 Ohm's law Recall that according to Ohm's law, the voltage current through the resistor by across a resistor of resistance is related to the . ANSWER: zero Part C Immediately after the switch is closed, what is the voltage across the inductor ? Hint C.1 A formula for voltage across an inductor The voltage drop given by across an inductor of inductance , where in the assumed direction of the current flow is is the current flowing through the inductor. Observe that the sign of the voltage drop is opposite to that of the induced EMF. ANSWER: zero Part D Shortly after the switch is closed, what is the direction of the current in the circuit? ANSWER: clockwise counterclockwise There is no current because the inductor does not allow the current to increase from its initial zero value. http://session.masteringphysics.com/myct/courseHome Page 2 of 23 MasteringPhysics: Course Home 2/22/11 9:05 AM Part E Shortly after the switch is closed, what is the direction of the induced EMF in the inductor ? ANSWER: clockwise counterclockwise There is no induced EMF because the initial value of the current is zero. The induced EMF does not oppose the current; it opposes the change in current. In this case, the current is directed counterclockwise and is increasing; in other words, the time derivative of the counterclockwise current is positive, which is why the induced EMF is directed clockwise. Part F Eventually, the process approaches a steady state. What is the current in the circuit in the steady state ? ANSWER: Part G What is the voltage Hint G.1 across the inductor in the steady state ? Induced EMF and current If the current is not changing, the derivative is equal to zero. Therefore, the EMF induced in the inductor is also zero. ANSWER: zero http://session.masteringphysics.com/myct/courseHome Page 3 of 23 MasteringPhysics: Course Home 2/22/11 9:05 AM Part H What is the voltage ANSWER: across the resistor in the steady state ? zero Part I Now that we have a feel for the state of the circuit in its steady state, let us obtain the expression for the current in the circuit as a function of time. Note that we can use the loop rule (going around counterclockwise): . Note as well that and . Using these equations, we can get, after some rearranging of the variables and making the subsitution , . Integrating both sides of this equation yields . Use this last expression to obtain an expression for . Remember that and that . Express your answer in terms of the notation exp(x) for , , and . You may or may not need all these variables. Use . ANSWER: = http://session.masteringphysics.com/myct/courseHome Page 4 of 23 MasteringPhysics: Course Home 2/22/11 9:05 AM Part J Just as in the case of R - C circuits, the steady state here is never actually reached: The exponential functions approach their limits asymptotically as . However, it usually does not take very long for the value of to get very close to its presumed limiting value. The next several questions illustrate this point. Note that the quantity has dimensions of time and is called the time constant (you may recall similar terminology applied to R - C circuits). The time constant is often denoted by . Using , one can write the expression as . Find the ratio of the current at time to the maximum current . Express your answer numerically, using three significant figures. ANSWER: = Part K Find the time it takes the current to reach 99.999% of its maximum value. Express your answer numerically, in units of ANSWER: . Use three significant figures. = Part L Find the time and it takes the current to reach 99.999% of its maximum value. Assume that ohms millihenrys. Express your answer in seconds, using three significant figures. ANSWER: = seconds http://session.masteringphysics.com/myct/courseHome Page 5 of 23 MasteringPhysics: Course Home 2/22/11 9:05 AM It does not take long at all! However, the situation can be different if the inductance is large and the resistance is small. The next example illustrates this point. Part M Find the time it takes the current to reach 99.999% of its maximum value. Assume that ohms and henrys. Express your answer in seconds, using three significant figures. ANSWER: = seconds This is more than an hour and a half! Nobody would wait that long. Let us see what change would occur in such a circuit over a shorter period of time. Part N What fraction of the maximum value will be reached by the current one minute after the switch is closed ? Again, assume that ohms and henrys. Use three significant figures in your answer. ANSWER: = This is only 11.3% of the maximum value of the current! Now consider a different situation. After switch simultaneously, switch has been closed for a long time, it is opened; is closed, as shown in the figure. This effectively removes the battery from the circuit. The questions below refer to the time immediately after switch is opened and switch is closed. http://session.masteringphysics.com/myct/courseHome Page 6 of 23 MasteringPhysics: Course Home 2/22/11 9:05 AM Part O What is the direction of the current in the circuit? ANSWER: clockwise counterclockwise The current is zero because there is no EMF in the circuit. Part P What is happening to the magnitude of the current ? ANSWER: The current is increasing. The current is decreasing. The current remains constant. Part Q What is the direction of the EMF in the inductor ? ANSWER: clockwise counterclockwise The EMF is zero because the current is zero. The EMF is zero because the current is constant. Part R Which end of the inductor has higher voltage (i.e., to which end of the inductor should the positive terminal of a voltmeter be connected in order to yield a positive reading)? Hint R.1 How to approach the problem Recall that the induced electric field in the inductor deposits charge that produces another electric field that acts opposite to the induced electric field, since the net electric field in the inductor (assumed to be a perfect conductor) must be zero. The voltage drops in the direction of this second, oppositely directed, electric field that is due to static charges. ANSWER: left right The potentials of both ends are the same. The answer depends on the magnitude of the time constant. http://session.masteringphysics.com/myct/courseHome Page 7 of 23 MasteringPhysics: Course Home 2/22/11 9:05 AM Part S For this circuit, Kirchhoff's loop rule gives . Note that Hint S.1 , since the current is decreasing. Use this equation to obtain an expression for . Initial and final conditions Initially, the current in the circuit equals ; after a long time, the current drops down to zero (exponentially!). Hint S.2 Let us cheat a bit You may have guessed that the answer must contain the term Express your answer in terms of , , and . Use exp(x) for . It does! . ANSWER: = ± Transformers Description: ± Includes Math Remediation. Several conceptual and qualitative questions introducing the principles behind the transformer. Both ideal and nonideal transformers are discussed. Learning Goal: To understand the concepts explaining the operation of transformers. 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 the one hand, the voltage supplied to the end users must be reasonably low for safety reasons (depending on the country, that voltage may be 110 volts, 220 volts, or some other value of that order). On the other hand, the voltage used in transmitting electric energy must be as high as possible to minimize losses in the transmission lines. A device that uses the principle of electromagnetic induction to increase or decrease the voltage by a certain factor is called a transformer . The main components of a transformer are two coils (windings) that are electrically insulated from each other. The coils are wrapped around the same core, which is typically made of a material with a very large relative permeability 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 alternating current in the primary coil induces the changing magnetic flux in the core that creates the emf in the secondary coil. The magnitude of the emf induced in the secondary coil can be controlled by the design of the transformer. The key factor is the number of turns in each coil. http://session.masteringphysics.com/myct/courseHome Page 8 of 23 MasteringPhysics: Course Home 2/22/11 9:05 AM Consider an ideal transformer, that is, one in which the coils have no ohmic resistance and the magnetic flux is the same for each turn of both the primary and secondary coils. If the number of turns in the primary coil is and that in the secondary coil is , then the emfs induced in the coils can be written as , and therefore, . Since both emfs oscillate with the same frequency as the ac source, the formula above can be applied to the instantaneous amplitude or the rms values of the emfs. Moreover, if the coils have zero resistance (as we assumed), then for each coil the terminal voltage will be equal to the induced emf. Therefore, we can write . Note that if , then . This is a case of a step - up transformer. Conversely, if , then . This is a case of a step - down transformer. Without energy losses, the power in the primary and secondary coils is the same: . If the secondary circuit is completed by a resistance , then . Combining this with the two equations above gives . Dividing the first and last expressions by and then inverting gives . In other words, the current in the primary coil is the same as if it were connected directly to a resistance equal to . In a way, transformers "transform" resistances as well as voltages and currents. In reality, no transformer is ideal. There are always some energy losses. However, modern transformers have very high efficiencies, usually well exceeding 90%. In answering the questions below, consider the transformer ideal unless otherwise noted. http://session.masteringphysics.com/myct/courseHome Page 9 of 23 MasteringPhysics: Course Home 2/22/11 9:05 AM Part A The primary coil of a transformer contains 100 turns; the secondary has 200 turns. The primary coil is connected to a size AA battery that supplies a constant voltage of 1.5 volts. What voltage would be measured across the secondary coil ? ANSWER: zero In order for an emf to be induced in the secondary coil, the flux through it must be changing; therefore, the current in the primary coil must also be changing. If a constant voltage is supplied to the primary coil, no emf would be induced in the secondary, and therefore, the secondary voltage would be zero. Part B A transformer is intended to decrease the rms value of the alternating voltage from 500 volts to 25 volts. The primary coil contains 200 turns. Find the necessary number of turns in the secondary coil. ANSWER: = This is a step - down transformer: The voltage decreases. 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 in the secondary coil. Hint C.1 How to approach this problem Recall that in an ideal transformer, the powers in the primary and secondary coils are equal. Therefore, if the current decreases by a factor of 20, the voltage must increase by a factor of 20. ANSWER: = This is a step - up transformer: The voltage increases by the same factor by which the current decreases. http://session.masteringphysics.com/myct/courseHome Page 10 of 23 MasteringPhysics: Course Home 2/22/11 9:05 AM Part D In a transformer, the primary coil contains 400 turns, and the secondary coil contains 80 turns. If the primary current is 2.5 amperes, what is the secondary current ? Express your answer numerically in amperes. ANSWER: = Part E The primary 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 in the secondary coil if it is terminated by a 20 - ohm resistor? Hint E.1 In the ideal world... This is an ideal transformer: There is no energy loss. ANSWER: In case of an ideal transformer, the power in the primary circuit is the same as that in the secondary circuit. Part F The primary coil of a transformer has 200 turns, and the secondary coil has 800 turns. The transformer is connected to a 120- volt (rms) ac source. What is the (rms) current in the primary coil if the secondary coil is terminated by a 20 - ohm resistor? Hint F.1 How to approach the problem Recall that transformers "transform" resistances as well as voltages and currents. Express your answer numerically in amperes. http://session.masteringphysics.com/myct/courseHome Page 11 of 23 MasteringPhysics: Course Home ANSWER: 2/22/11 9:05 AM = Part G 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- volt (rms) ac source. What is the current in the primary coil ? Hint G.1 How to approach the problem 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: = Part H The voltage and the current in the primary coil of a nonideal transformer are 120 volts and 2.0 amperes. The voltage and the current in the secondary coil are 19.4 volts and 11.8 amperes. What is the efficiency of the transformer ? The efficiency of a transformer is defined as the ratio of the output power to the input power, expressed as a percentage: . Express your answer as a percentage. ANSWER: = % Decay of Current in an L- R Circuit Description: To understand the decay of current in an L - R circuit, derive the mathematical form of I(t), and find the time constant. Learning Goal: To understand the mathematics of current decay in an L - R circuit A DC voltage source is connected to a resistor of resistance and an inductor with inductance , forming the circuit shown in the figure. For a long time before , the switch has been in the position shown, so that a current has been built up in the circuit by the voltage source. At the switch is thrown to remove the voltage source from the circuit. http://session.masteringphysics.com/myct/courseHome Page 12 of 23 MasteringPhysics: Course Home 2/22/11 9:05 AM This problem concerns the behavior of the current through the inductor and the voltage across the inductor at time after . Part A After , what happens to the voltage inductor relative to their values prior to Hint A.1 across the inductor and the current through the ? What is the relation between current and voltage for the inductor? The current through an inductor cannot change discontinuously unless the voltage across the inductor is infinite. This follows from the fundamental equation relating voltage and current for an inductor. Find the voltage drop across the inductor in the direction of the current shown (clockwise). Express your answer in terms of , , , and any necessary constants. ANSWER: = ANSWER: changes slowly and changes slowly and changes abruptly. changes abruptly. Both change slowly. Both change abruptly. After , the battery no longer provides a voltage that drives current around the circuit. If the circuit did not contain an inductor, then the current would drop to zero immediately. However, inductors act to keep the current flowing. If the current starts to change, this causes an electromotive force (EMF) to form across the inductor that (by Lenz's law) opposes the tendency for the current to change. Here, this causes the current through the inductor to persist for a while as it decays toward zero. Part B What is the differential equation satisfied by the current Hint B.1 after time ? Kirchhoff's loop law Kirchhoff's loop law tells us that the EMFs and potential differences around a circuit will sum to zero. If we follow the drops and increases in potential completely around a circuit, then we will return to the potential at the point at which we started. Use Kirchhoff's loop law to write down a differential equation http://session.masteringphysics.com/myct/courseHome Page 13 of 23 MasteringPhysics: Course Home in 2/22/11 9:05 AM for the sum of the voltage drops that arise from traversing the circuit in the direction of the current arrow (clockwise). The equation will have a term provided by the inductor and a term provided by the resistor, from Ohm's law. Hint B.1.1 Sign of the voltage drop across the inductor The voltage drop across the inductor along the direction of current flow is Express your answer in terms of , its derivative , , and . . ANSWER: = Express in terms of , , and . ANSWER: = The minus sign in this equation tells us that the current is decreasing with time. The current is decaying. This is the case because the DC voltage source no longer acts to sustain the current. Part C What is the expression for Hint C.1 obtained by solving the differential equation that satisfies after ? Separation of variables To solve the differential equation we may separate the variables and integrate. The variables in this equation are and . To separate the variables, rearrange the equation so that all the and terms are on one side, and all the and terms are on the other side. Complete the equation below in this format. ANSWER: = Hint C.2 Integrating Now integrate both sides of the differential equation. We know the limits of integration: http://session.masteringphysics.com/myct/courseHome runs from to Page 14 of 23 MasteringPhysics: Course Home , while 2/22/11 9:05 AM runs from 0 to . Recall also that Express your answer in terms of the initial current . , as well as , , and . ANSWER: = Part D What is the time constant Hint D.1 of this circuit? Definition of time constant The time constant is the time, commonly represented by or , that it takes for a variable to fall to (or about 0.37) of its initial value. The time constant is, as its name suggests, a constant. Hence a quantity that falls to 0.37 of its initial value after time will fall to of its initial value after time . Writing this relation mathematically, for a quantity , where Hint D.2 is the initial value (at time ) of that decays over time , we have . The time constant for this circuit Compare the mathematical description of the time constant with the expression for Express your answer in terms of and found earlier. ? ANSWER: = Growth of Current in an L- R Circuit Description: Works through the differential equation and its solution for growth of current in an L - R circuit connected to a battery at time t=0. Learning Goal: To review the procedure for setting up and solving equations for determining current growth in an L - R circuit connected to a battery Consider an L - R circuit as shown in the figure. The battery provides a voltage . The inductor has http://session.masteringphysics.com/myct/courseHome Page 15 of 23 MasteringPhysics: Course Home inductance 2/22/11 9:05 AM , and the resistor has resistance switch is initially open as shown. At time . The , the switch is closed. Part A What is the differential equation governing the growth of current in the circuit as a function of time after ? Hint A.1 Find the voltage change across the resistor What is the voltage change Express in terms of the resistance ANSWER: Hint A.2 across the resistor? and the current . = Find the voltage drop across the inductor What is the voltage change across the inductor ? Express your answer in terms of the inductance , the current , and ANSWER: = Hint A.3 Use Kirchhoff's loop law to sum the voltages Kirchhoff's loop law tells us that the EMFs and potential differences around a circuit will sum to zero: . This gives us the master equation for the circuit. This equation will connect the EMF provided by the battery to the potential across the resistor and the EMF in the inductor (treated here as a voltage drop). Give an expresion for the sum of the potential differences . Choose your loop in the direction of the current, and express your answer in terms of http://session.masteringphysics.com/myct/courseHome , , Page 16 of 23 MasteringPhysics: Course Home , 2/22/11 9:05 AM , and . ANSWER: Express the right- hand side of the differential equation for in terms of , , , and . ANSWER: = Part B We know that the current in the circuit is growing. It will approach a steady state after a long time (as tends to infinity), which implies that the differential term in our circuit equation will tend to zero, hence simplifying our equation. The current around the circuit will tend to an asymptotic value . What is ? Express your answer in terms of and and other constants and variables given in the introduction. ANSWER: = Part C Solve the differential equation obtained in Part A for the current is closed at Hint C.1 . Substituting into the differential equation The current in the equation will tend to the asymptotic value differential equation with that you found in Part B. Rewrite the substituted into the differential equation to eliminate constant, equation in as a function of time after the switch . Since is a . This mathematical substitution allows us to write a differential rather than and right side with http://session.masteringphysics.com/myct/courseHome . What is this differential equation with the left side eliminated by the substitution ? Page 17 of 23 MasteringPhysics: Course Home 2/22/11 9:05 AM Express your answer in terms of , , , and . ANSWER: Hint C.2 A helpful integral Recall that Hint C.3 . Solving the differential equation 1: a substitution The equation that we have obtained is similar to the equation for a decaying current in an L - R circuit. We need to use the technique of separating variables in order to solve this differential equation. What we have at this point is a differential equation in . We can simplify this equation by defining the function for , writing the simplified differential equation , and then replacing and by , solving . First rearrange the equation to place all the terms on one side of the equation and all the and terms on the other side. What is the resulting equation? Express your answer in terms of , , and . ANSWER: = Hint C.4 Solving the differential equation 2: new limits Now that we have simplified the differential equation and separated variables, we need to integrate it to obtain our answer (see one of the earlier hints for help with this). Also be careful with the limits of integration. We are integrating with respect to rather than . So, at , , and therefore . Therefore, the limits of integration are varies from 0 to . Next, substitute for , and then replace relation found in Part B. Finally, rearrange terms to obtain an equation for and and by as based on the in terms of ,, , . Express your answer in terms of , , and . Use the notation exp(x) for . ANSWER: http://session.masteringphysics.com/myct/courseHome Page 18 of 23 MasteringPhysics: Course Home 2/22/11 9:05 AM ANSWER: = The current approaches its asymptotic/steady - state value exponentially, i.e., usually very fast. However, the time constant depends on the actual values of the inductance and resistance. Energy within an L- C Circuit Description: This problem analyzes an L - C circuit with no voltage source. The question revises the equations for energy within the magnetic field of an inductor and the electric field of a capacitor. These are used to calculate the total energy within the circuit. The student is assumed to know some things about L C circuits. Consider an L - C circuit with capacitance , inductance , and no voltage source, as shown in the figure . As a function of time, the charge on the capacitor is and the current through the inductor is . Assume that the circuit has no resistance and that at one time the capacitor was charged. Part A As a function of time, what is the energy Express your answer in terms of and stored in the inductor ? . ANSWER: = An inductor stores energy in the magnetic field inside its coils. Part B As a function of time, what is the energy http://session.masteringphysics.com/myct/courseHome stored within the capacitor? Page 19 of 23 MasteringPhysics: Course Home 2/22/11 9:05 AM Express your answer in terms of and . ANSWER: = A capacitor stores energy within the electric field between its plates (conductors). Part C What is the total energy Hint C.1 stored in the circuit? Charge and current as functions of time In order to produce a solution that does not contain terms involving the capacitance and the charge, we need to express charge and current as functions of time. An L - C circuit will undergo resonance, with the current varying sinusoidally, where . Integrate this expression to obtain a similar expression for . Express your answer in terms of , , and . ANSWER: = Hint C.2 Find the resonance frequency We also require an expression for the angular frequency of oscillation of the circuit . What is the angular frequency of an L - C circuit? Express your answer in terms of and . ANSWER: = Hint C.3 Summing the contributions Combine the expressions we have obtained, equation for and to express the total energy in terms of the inductance You will need the trigonometric relation http://session.masteringphysics.com/myct/courseHome , and the and the maximum current . . Page 20 of 23 MasteringPhysics: Course Home 2/22/11 9:05 AM Express your answer in terms of the maximum current and inductance . ANSWER: = We could have eliminated the inductance from this expression instead of the capacitance to find that . This total energy remains constant; however, the location of the energy changes. We can tell from the and terms in the inductor's and the capacitor's energies that as time passes, the energy moves back and forth between the inductor and the capacitor. Problem 34.10 Description: The metal equilateral triangle in the figure, 20 cm on each side, is halfway into a 0.1 T magnetic field. (a) What is the magnetic flux through the triangle? (b) If the magnetic field strength decreases, what is the direction of the induced current... The metal equilateral triangle in the figure, 20 on each side, is halfway into a 0.1 magnetic field. Part A What is the magnetic flux through the triangle? ANSWER: Part B If the magnetic field strength decreases, what is the direction of the induced current in the triangle? ANSWER: http://session.masteringphysics.com/myct/courseHome Page 21 of 23 MasteringPhysics: Course Home ANSWER: 2/22/11 9:05 AM clockwise counterclockwise Problem 34.12 Description: The loop in the figure is being pushed into the 0.20 T magnetic field at 50 m/s. The resistance of the loop is 0.10 Omega. (a) What is the magnitude of the current in the loop ? (b) What is the direction of the current in the loop ? The loop in the figure is being pushed into the 0.20 is 0.10 magnetic field at 50 . The resistance of the loop . Part A What is the magnitude of the current in the loop ? ANSWER: Part B What is the direction of the current in the loop ? ANSWER: clockwise counterclockwise http://session.masteringphysics.com/myct/courseHome Page 22 of 23 MasteringPhysics: Course Home 2/22/11 9:05 AM Problem 34.43 Description: A loop antenna, such as is used on a television to pick up UHF broadcasts, is 25 cm in diameter. The plane of the loop is perpendicular to the oscillating magnetic field of a 150 MHz electromagnetic wave. The magnetic field through the loop is B ... A loop antenna, such as is used on a television to pick up UHF broadcasts, is 25 cm in diameter. The plane of the loop is perpendicular to the oscillating magnetic field of a 150 MHz electromagnetic wave. The magnetic field through the loop is . Part A What is the maximum emf induced in the antenna? ANSWER: V Part B What is the maximum emf if the loop is turned to be perpendicular to the oscillating electric field? ANSWER: V Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 12 points. http://session.masteringphysics.com/myct/courseHome Page 23 of 23 ...
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This note was uploaded on 12/28/2011 for the course PHYSICS 270 taught by Professor Drake during the Fall '08 term at Maryland.

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