36 1 Model A phasor is a vector that rotates counterclockwise around the origin at angular frequency Solve a Referring to the phasor in Figure EX36 1 the phase angle is t 180 30 210 3 665 rad 2 2 4 10 rad s rad 180 3 665 rad 15 10 s 3 b The instantaneous value of the emf is E E 0 cos t 12 V cos 3 665 rad 10 4 V Assess Be careful to change your calculator to the radian mode to work with the trigonometric functions 36 2 Model A phasor is a vector that rotates counterclockwise around the origin at angular frequency Solve a Referring to the phasor in Figure EX36 2 the phase angle is t 135 135 rad 1178 rad s rad 180 3 4 3 4 2 0 ms E E 0 cos t E 0 E cos t 50 V cos 3 4 rad 71 V b From Figure EX36 2 36 3 Model A phasor is a vector that rotates counterclockwise around the origin at angular velocity Solve The emf is E 0E cos t 50 V cos 2 110 rad s 3 0 10 3 s 50 V cos 2 074 rad 50 V cos119 36 4 Model A phasor is a vector that rotates counterclockwise around the origin at angular velocity Solve The instantaneous emf is given by the equation E 170 V cos 2 60 Hz t At t 60 ms E 170 V cos 22 619 rad An angle of 22 619 rad corresponds to 3 60 periods which implies that the phasor makes an angle of 0 60 2 rad or 0 60 360 216 in its fourth cycle 36 5 Visualize Please refer to Figure EX36 4 for an AC resistor circuit a For a circuit with a single resistor the peak current is Solve I R E 0 R 10 V 200 0 050 A 50 mA b The peak current is the same as in part a because the current is independent of frequency 36 6 Model Current and voltage phasors are vectors that rotate counterclockwise around the origin at angular frequency Visualize Please refer to Figure EX36 6 Solve a The frequency is b From the figure we note that VR 10 V and IR 0 50 A Using Ohm s law f 2 1 T 1 0 04 s 25 Hz R V R I R 10 V 0 50 A 20 c The voltage and current are v R V R cos t 10 V cos 2 25 Hz t i R I R cos t 0 50 A cos 2 25 Hz t For both the voltage and the current at t 15 ms the phase angle is t 2 25 Hz 15 ms 2 0 375 rad 135 That is the current and voltage phasors will make an angle of 135 with the starting t 0 s position Assess Ohm s law applies to both the instantaneous and peak currents and voltages For a resistor the current and voltage are in phase 36 7 Visualize Figure EX36 7 shows a simple one capacitor circuit Solve a The capacitive reactance at 2 f 2 100 Hz 628 3 rad s is b The capacitive reactance at 2 100 kHz 628 300 rad s is X C 1 C 1 628 3 rad s 0 30 10 F 6 5305 I C V C X C 10 V 3 5 305 10 1 88 10 A 1 88 mA 3 X C 1 C 1 6 283 10 rad s 0 30 10 F 6 5 5 305 I C V C X C 10 V 5 305 1 88 A Assess Using reactance is just like using resistance in Ohm s law Because CX 1 XC decreases with an increase in as observed above 36 8 Solve a For a simple one capacitor circuit I C V C X C V C 1 C CV C When the frequency is doubled the new current is 2 CV C C I CV C 2 CV C 2 I C 20 mA b Likewise when the voltage is doubled the current doubles to 20 mA c When the frequency is halved and the emf is doubled the current remains the same at 10 mA 36 9 Visualize Figure EX36 7 shows a simple one capacitor circuit Solve a From Equation 36 11 I C V C X C V C 1 C CV C 2 fCV C f 3 50 10 A 5 0 V 20 10 F 9 2 80 kHz b The AC current through a capacitor leads the capacitor voltage by 90 or 2 rad For a simple one capacitor circuit must be equal to 2n where n 1 2 This means cos For iC IC t t i C I C 1 2 1 2 At these values of t v C VC cos t t 11 2 0 V That is iC is maximum when vC 0 V 7 3 2 2 36 10 Solve From Equation 36 11 I C V C X C V C 1 C CV C C I C V C 3 65 10 A 2 15 000 Hz 2 6 0 V 9 81 10 F 81 nF 36 11 Solve a From Equation 36 11 I c V c X c V c C 1 CV c 2 fCV c C I 2 c fV c 6 250 10 Hz 2 2 V 330 10 A 3 2 9 5 10 11 C 95 pC b Doubling the frequency will halve the capacitive reactance Xc doubling the current so Ic 2 330 A 660 A 36 12 Model The current and voltage of a resistor are in phase but the capacitor current leads the capacitor voltage by 90 Solve For an RC circuit the peak voltages are related through Equation 36 12 We have 2 E 0 2 V R V R 2 V C 2 E 0 2 V C 10 0 V 2 6 0 V 2 8 0 V 36 13 Model The crossover frequency occurs for a series RC circuit when VR VC Visualize Please refer to Figure EX36 13a for the low pass filter circuit Solve From Equation 36 15 2 f c c C 1 RC 1 2 Rf c 100 2 1000 Hz 1 1 59 10 F 1 59 F 6 Assess The output for a low pass filter is across the capacitor 36 14 Model The crossover frequency occurs for a series RC circuit when VR VC Visualize Please refer to Figure EX36 13b for the high pass filter circuit Solve From Equation 36 15 2 f c c C 1 RC 1 2 Rf c 100 2 1000 Hz 1 1 59 10 F 1 59 F 6 Assess The output for a high pass filter is across the resistor 36 15 Model The current and voltage of a resistor are in phase but the capacitor current leads the capacitor voltage by 90 Visualize Please refer to Figure EX36 15 Solve From Equations 36 13 and …
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