Chapter 31 Alternating Current

Transcription

1 Chapter 31 Alternating Current In this chapter we will learn how resistors, inductors, and capacitors behave in circuits with sinusoidally vary voltages and currents. We will define the relationship between the voltage and the current in AC circuits, as well as the concept of resonance in AC circuits. 1 Phasors and Alternating Currents To supply an alternating current to a circuit, a source of alternating emf or voltage is required. The sinusoidal voltage might be described by a function such as v = V cos ωt where V is the peak voltage, and ω = 2πf, and f is the frequency in Hz. Sinusoidal Alternating Current Figure 1: i(t) = I cos(ωt) 1

2 1.1 Phasor Diagrams Figure 2: 2

3 1.2 Rectified Alternating Current Rectified Average Current Figure 3: I rav = 2 π I = I 3

4 1.3 Root-Mean-Square (rms) Values Figure 4: What do we mean when we talk about the root-mean-square current? Why is this important? i 2 = I 2 cos 2 ωt i 2 is never less than zero as a function of time. The time-average of i 2 is: T ) < i 2 > T = 1 T 0 I 2 cos 2 (ωt) dt = I 2 ( 1 2 I rms = < i 2 > T = I 2 (root-mean-square current) (1) where I is the maximum current in the ac circuit 4

5 The same sequence of calculations occur for defining the root-mean-square voltage. V rms = V 2 (root-mean-square voltage) (2) where V is the maximum voltage in the ac circuit. Example: 5

6 2 Resistance and Reactance Figure 5: The current i and v are in-phase. v R = ir = (IR) cos ωt v R = V R cos ωt 6

7 2.1 Inductor in an ac Circuit Figure 6: v L = L di dt = ILω sin ωt = I(ωL) sin ωt = I X L sin ωt X L = ωl (inductive reactance) (3) V L = I ωl V L = I X L (Ohm s Law) (4) Ex. 11 A H inductor is connected in series with a 84.5 Ω resistor and an ac source. The voltage across the inductor is v L = (13.0V ) sin [(489 rad/s) t] (a) Derive an expression for the voltage v R across the resistor. Express your answer in terms of the variables L, R, V L (amplitude of the voltage across the inductor), ω, and t. (b) What is v R at 2.13 ms? 7

8 2.2 Capacitor in an ac Circuit Figure 7: i = dq dt = I cos ωt q = I ω sin ωt v CC = I sin ωt ω v C = I ωc sin ωt X C = 1 (capacitive reactance) ωc V C = I X C (Ohm s Law) (5) 8

9 2.3 Comparing ac Circuit Elements Figure 8: 9

10 3 The L-R-C Series Circuit Figure 9: V R = IR V L = I X L V C = I X C V = V 2 R + (V L V C ) 2 = I R 2 + (X L X C ) 2 = I Z 10

11 Z = R 2 + (X L X C ) 2 (the impedance) (6) Figure 10: This figure shows the voltages across the various circuit elements v R, v L, and v C. The resultant voltage (shown in blue) is not in phase with the current (synonymous with v R ) and is slightly leading the current. Ex. 27 You have a 210 Ω resistor, a H inductor, a 4.97 µf capacitor, and a variable-frequency ac source with an amplitude of 2.95 V. You connect all four elements together to form a series circuit. (a) At what frequency will the current in the circuit be greatest? (b) What will be the current amplitude at this frequency? (c) What will be the current amplitude at an angular frequency of 396 rad/s? (d) At this frequency, will the source voltage lead or lag the current? 11

12 3.1 The Meaning of Impedance and Phase Angle Z = R 2 + (X L X C ) 2 (the impedance) tan φ = V L V C V R = X L X C R tan φ = ωl 1/ωC R (the phase angle) (7) 4 Power in Alternating-Current Circuits p = v i (the instantaneous power) Figure 11: 4.1 Power in a General ac Circuit P av = 1 2 V I cos φ = V rms I rms cos φ (8) where φ is the phase angle and cos φ is called the power-factor. 12

13 Figure 12: Ex. 43 A series circuit has an impedance of 65.0 Ω and a power factor of at a frequency of 55.0 Hz. The source voltage lags the current. (a) What circuit element, an inductor or a capacitor, should be placed in series with the circuit to raise its power factor? (b) What size element will raise the power factor to unity? 5 Resonance in Alternating-Current Circuits The resonance angular frequency occurs when the impedance Z is at its minimum, thus causing the current I to be at its maximum. This occurs when X L = X C, or when: ω o = 1 LC (Resonance Angular Frequency) (9) 13

14 Figure 13: 14

15 Figure 14: Ex. 63 In an L-R-C series circuit, the source has a voltage amplitude of 124 V, R=81.0 Ω, and the reactance ofo the capacitor is 457 Ω. The voltage amplitude across the capacitor is 358 V. (a) What is the current amplitude in the circuit? (b) What is the impedance? (c) What two values can the reactance of the inductor have? (d) For which of the two values found in part (c) is the angular frequency less than the resonance angular frequency? 15

16 6 Transformers Figure 15: 16