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Download Chapter 32 - Alternating Current (AC)
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Chapter 31 - Alternating Current (AC) I. Look at the behavior of resistors, inductors, and capacitors in alternating current (AC) circuits. We will find that the relationship between current and voltage for the elements looks very much like Ohm's Law. II. Alternating current or AC means that there is a periodic variation in the current and voltage - any type of periodic variation whether it be sinusoidal or square or triangular or…. A. We will look at sinusoidal types of signals since any periodic signal can be analyzed as a combination of sinusoidal and/or cosinusoidal signals. For example, 1 2n sin t n 1,3,5,... n T 4 1. square wave with period of T and amplitude of 1: v(t ) 2. triangular wave with a period of T and amplitude of 1: vt 2 n 1 B. 1n1 sin 2n t n Mathematical representation of a sinusoidal voltage. v = Vm sin t = Vm sin 2ft where Vm = amplitude, maximum value, peak value f = frequency of the AC signal (hertz, cycles/sec, osc/sec, vib/sec) = 2f = angular frequency (radians/sec) T = 1/f = period of the oscillation The voltage can also be written in a more general way as v = Vm sin (t + ) , where = initial phase angle 31-1 T C. Graphical representations of a sinusoidal voltage. There are different ways to graphically represent a sinusoidal voltage. One way is to graph the voltage versus time on rectangular axes. A second way is to represent the oscillations with a rotating phasor (vector or rotor). The phasor is an arrow whose length equals the amplitude of the voltage and is rotating around the origin with an angular velocity . The projection of this phasor onto the y-axis gives the instantaneous values of voltage. Graph of voltage vs. time (v vs. t) 1. Corresponding phasor or rotor diagram = 0 v t 2. 2 v t 31-2 3. =- 2 v t 4. = 3 4 v t D. Another way to express the value of the voltage (or any sinusoidal signal) is give the root-meansquare (rms) value of the sinusoidal signal: Vrms v2 If v = Vm sin t , then v2 = Vm2 sin2 t . The average value of a sine squared function is ½ the V2 value of that function, therefore, v 2 m , and 2 Vm Vrms 0.707Vm . 2 31-3 III. Single Element circuits - We are interested in determining how the current and voltage are related to each other. In each of these cases we will be given a sinusoidal voltage source v = Vm sin t , and the resulting current in the circuit is to be found. Specifically, we want to find 1) the amplitude of the current relative to the amplitude of the voltage, and 2) the phase angle between the current and the voltage. A. Resistor, R v = Vm sin t B. v t i t Capacitor, C v = Vm sin t v t i t 31-4 C. Inductor, L v = Vm sin t v t i t C’. Inductor, L using a different approach i = Im sin t D. v t i t Mnemonic device to remember the relative phase between currents (i) and voltages (v) for the inductor (L) and capacitor (C): ELI the ICEman 31-5 IV. Combination of the three: the RLC series circuit for AC R L C i = Im sin t 31-6 Summary: XL L 2fL and Finally: R2 XL XC 2 Vm = ImZ , where Z = XC 1 1 C 2 fC If i = Im sin t , then v = Vm sin (t + ) . If v = Vm sin t , then i = Im sin (t - ) . In both cases: tan 1 XL XC R Example: Suppose we have an RLC series circuit where the resistor is 900 ohms, the inductor is 2 henries, and the capacitor is 1 microfarad. A signal generator supplies a sinusoidal voltage v = 100 sin 1000t where v is in volts and t is in seconds. Find the instantaneous value of current in the circuit. That is find Im , , and . Find also the amplitudes of the voltages across the resistor, inductor, and capacitor. 31-7 V. More on the RLC circuit. A. Look at the variation of impedance Z, current amplitude Im , and phase angle as a function of the frequency of the AC signal. The voltage amplitude Vm is fixed (constant). Im , Z, f, O B. fo Power delivered to the RLC circuit. Let i = Im sin t and v = Vm sin (t + ) , find the instantaneous power delivered to the circuit and the average power delivered to the circuit. 31-8 C. Graph of the average power delivered to an RLC series circuit as a function of frequency. 2 Pavg I rmsVrms cos Vrms R Z2 2 Vrms R 1 R 2 L C 2 Note that maximum power is delivered at the resonant frequency. Pavg f, fo o The sharpness of the peak is defined as the quality factor Q = o oL . R What is the meaning of the sharpness of the peak? That is, what is the difference between a high Q and low Q circuit? AM FM 31-9 KFI 740 kHz KNX 1070 KCSM 88.5 MHz KPCC 89.3 KCRW 89.9 KPFK 90.7 KZLA 93.9 VI. Filter Circuits A. High Pass Filter C R Vin Vout Vout Vin B. Low Pass Filter R Vin C Vout Vout Vin 31-10 VII. The parallel RLC circuit R C L VIII. The Transformer transformer V1 primary 31-11 N1 N2 V2 secondary