Chapter 25 Alternating Currents

Transcription

1 Chapter 25 Alternating Currents GOALS When you have mastered the contents of this chapter, you will be able to achieve the following goals: Definitions Define each of the following terms and use it in an operational definition: effective values of current and voltage power factor reactance resonance impedance Q-factor AC Circuits Solve alternating-current problems involving resistance, inductance, and capacitance in a series circuit. Phasor Diagrams Draw phasor diagrams for alternating current circuits. Transformer Explain the operation of the transformer. AC Measurements Describe the use of alternating currents in physiological measurements. PREREQUISITES Before you begin this chapter you should have achieved the goals of Chapter 22, Basic Electrical Measurements, and Chapter 24, Electromagnetic Induction. 211

2 Chapter 25 Alternating Currents OVERVIEW The electrical energy utilized in the homes of America is delivered by the use of AC circuits. In this chapter you will learn about the nature of the AC circuit and its energy equivalent relationship with the DC circuit. An in-depth study of the AC circuit: additional components such as inductors and capacitors give the series AC circuit many interesting phase relationships and the property of resonance as outlined in Section Then in Section 25.8, the Transformer and Its Application, you will study one of the major advantages in using AC circuits. SUGGESTED STUDY PROCEDURE The key Chapter Goals in this chapter are Definitions, AC Circuits, Transformer, and AC Measurement. When you begin your study of this chapter, be familiar with each of these goals. Remember that each of the terms listed under Definitions is discussed in some detail in the first section of this Study Guide chapter. Next, read Chapter Sections Answers to the questions you encounter in your reading are answered in the second section of this Study Guide chapter. In section 25.9, pay particular attention to Table Now turn to the end of the chapter and read the Chapter Summary and complete Summary Exercises 1-6 and Next, do Algorithmic Problems 1-4, 6, and 7, and Exercises and Problems 1-3, 5, 9, 11, 12, and 15. Check your answers to each of these questions against the answers given. If you have difficulty, refer back to the correct text section. For more practice with the concepts introduced in this chapter, see the Examples section of this Study Guide chapter. Now you should be prepared to attempt the Practice Test on Alternating Currents provided at the end of this chapter. Complete the entire test before you check your answers. If you have difficulty with any part of the test, refer to either the appropriate section of the Study Guide or the text. This Study Procedure is outlined below Chapter Goals Suggested Summary Algorithmic Exercises Text Readings Exercises Problems & Problems Definitions , 2, 3, 4 AC Circuits 25.2, 25.3, 5, 6 1, 2, 3, 4, 1, 2, 3, 5, 25.4, Transformers , 12 AC Measurements , Phasor Diagrams , 8 212

3 DEFINITIONS EFFECTIVE VALUES OF CURRENT & VOLTAGE - The values of AC current and AC voltage which will produce the identical heating in a resistor as a DC circuit. These values are referred to as RMS values meaning Root Mean Square. REACTANCE - The proportionality factor (measured in ohms) between the current and voltage for capacitors and inductors in AC circuits. Two typical kinds of reactance are capacitive reactance (X C ) and inductive reactance (X L ), IMPEDANCE - Just as ohms law for DC circuits defines resistance as the ratio of voltage (V) to the current (I), for AC circuits, the impedance of an AC circuit is the ratio of the effective voltage to the effective current. POWER FACTOR - Equals the cosine of the phase angle, f, between voltage and current in an AC circuit. When φ = 0 ø, the maximum power is dissipated in the AC circuit. RESONANCE - Occurs when the frequency of the external force equals a natural frequency of the system. The condition present in an RLC circuit when X L = X C. Q-FACTOR - A measure of the sharpness of resonance. ANSWERS TO QUESTIONS FOUND IN THE TEXT SECTION 25.1 Introduction In alternating current circuits the electrical voltage is continuously changing since the voltage is given by a periodic function of time such as sin wt where w is the frequency. In the United States of America 60Hz or 120π radians/second is the usual frequency of AC circuits. Since the sine function changes its sign from positive to negative and back, this kind of electrical circuit is called an alternating current circuit. Usually electrical devices with no moving parts will work on either AC or DC. However the transformer is a major exception. This device with no moving parts is used to connect two portions of a circuit by means of linking, fluctuating electromagnetic fields. They are highly efficient devices for changing one voltage into another. SECTION 25.2 Nomenclature Used for Alternating Currents Since we specify, in common usage, the effective, or rms, value of an AC current or voltage the heating effect of AC and DC circuits are the same for equal currents -- as long as we used rms values for the AC circuit parameters. Remember it means that the actual current and voltages in the AC circuit vary by plus or minus 140% of the stated AC values, e.g. for the usual 110V line in a house the voltage maximum varies from - 155V to + 155V. 213

4 SECTION 25.5 Resonance Questions - 1. The maximum current in a resonance circuit is determined by the effective resistance of the circuit. At resonance the current will be given by the voltage divided by the resistance. 2. At any instant in time the total energy of an AC circuit must add up to the input energy. But consider capacitors and inductors as devices that are able to store energy for some periods of time. Then at a time when energy has been stored in a capacitor it would be possible to take energy from the capacitor and from the emf itself to increase, for a short time, the energy stored in the inductor. SECTION 25.7 Q-Factor Questions - 3. The energy stored in an inductor is given by (½)LI 2 and the energy dissipated in a circuit is I2R. Clearly then the ratio of energy stored to energy dissipated is proportional to the ratio of L to R. But L/R is also proportional to the Q-factor. In fact, if you divide the ratio of the energy stored to energy dissipated by the period of a cycle then you have the following: [(1/2)LI 2 )/(I 2 R)]/T = (L/2R)/(1/f) = fl/2r = ωl/4πr = Q/4π so the ratio of energy stored to energy dissipated per cycle is proportional to Q, in fact, it is equal to Q divided by 4π. EXAMPLES AC CIRCUITS 1. Consider a series AC circuit which contains a resistor of 100 ohms, a capacitor of 16.7 microfarads, and an inductor of 240 millihenries connected to a 100 volt (rms) variable frequency power supply. (a) Calculate the resistance, capacitive reactance, inductive reactance, impedance, current, and phase angle for frequencies of 15.9 Hz, 47.7 Hz, 79.6 Hz, 111 Hz, and 143 Hz. (b) Draw the following graphs for this circuit (i) Capacitive reactance versus frequency (ii) Inductive reactance versus frequency (iii) Resistance versus frequency (iv) (X L - X C ) versus frequency (v) Z versus frequency (vi) Phase angle versus frequency (c) For what frequency is the phase angle equal to zero? (d) Draw the phasor diagram for each of the above frequencies. (e) Calculate the Q factor for this circuit and draw a graph of the current versus the frequency. Show the half-width on the graph and illustrate the Q factor. What data are given? The circuit as shown below is given: 214

5 What data are implied? All of the circuit components are assumed to be ideal. The power supply has zero effective impedance. The resistor has zero reactance and the reactances have zero resistance. What physics principles are involved? Ohm's Law and the conservation of energy and charge as applied to AC circuits as discussed in Section 25.4 of the text. What equations are to be used? Capacitive Reactance X C = 1 /2πfC ohms (25.11) Inductive Reactance X L = 2πfL ohms (25.9) Resistance R = R ohms; independent of f Total Reactance = X L - X C = 2πfL - 1/2πfC Impedance = SQR RT[R 2 + (X L - X C ) 2 ] (25.19) Phase Angle; tan θ = (X L - X C ) / R (25.20) Zero Phase Angle occurs when X L = X C ; ω o = 1 / SQR RT[LC] (25.23) Q-Factor Q = ω o L/R (25.24) Current = Applied Voltage/Z (25.12) Solutions To solve this problem we can now just enter the correct numbers from the problem into these various equations. Rather than show all of the arithmetic, which is easily accomplished with any hand calculator, let us just collect our numerical results in a Table. See Figs (c) Phase angle = zero at ω o = 500 rad/s; f = 79.6 Hz. (d) ω = 99.9 rad/s ω = 300 rad/s See Fig. (e) Q Factor = ω o L / R = (500)(0.240) / 100 = 1.2 ω 2 - ω 1 = ω o /Q = 500/1.2 = 417 rad/s See fig. Thinking about the answers Notice that in general the curves are not symmetrical about the resonance frequency. This occurs because the impedance does not depend in a symmetric way on the frequency when the reactance changes from being primarily capacitive to primarily inductive. The curves we have drawn are characteristic of all AC series circuits. 215

6 PRACTICE TEST 1. The apparatus shown below is used to show the properties of the transformer. The primary and secondary windings have a common core of heavy iron. The input is from a 120 volt - AC household line. a. If the voltmeter shows 6 volts, how many turns are in the secondary winding? I rms = b. If the resistor (R) in the external secondary circuit is 12 ohms, what I max = RMS current will be measured by the ammeter? What maximum "peak" current? (Show your calculations.) 2. Most electrical signals generated by the body are AC in nature. One common example is the voltage detected in a heart E.C.G. measurement. a. What is the approximate voltage of this body generated signal? b. Explain briefly what electrical hazards exists for a patient while this measurement is being made. 3. The patient is inadvertently connected in a series with a 50.0 volts of a 60.0 cycle AC line while being monitored by an ECG unit. Assume that the body in this situation acts like a series circuit of R = 2000 Ω and the ECG unit provides an inductance of.100 henry and a capacitance of µfd. a. At 60.0 cycles what is the equivalent impedance of this unfortunate circuit? b. What current will flow through the patient? c. Is this current dangerous? Explain your answer. ANSWERS: turns, Irms =.5 amp, IMAX =.71 amps 2. millivolts, most electrical hazards are associated with electrical leaks in equipment. If the patient touches a ground (wet floor, faucet, metal bed post, etc.) a complete circuit could result and allow current to flow through the body. Leakage currents higher than 10-3 amps are considered dangerous Ω,.02 amps. It is the current through the heart that is dangerous. At this level, 20 milliamps could be a problem, especially if the voltage is continuous, allowing the body's effective resistance to reduce as the current alters the body's resistance. 216