Chapter 36

# Chapter 36 - 36.1 Model A phasor is a vector that rotates...

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36.1. Model: A phasor is a vector that rotates counterclockwise around the origin at angular frequency ω . Solve: (a) Referring to the phasor in Figure EX36.1, the phase angle is 2 3 rad 3.665 rad 180 30 210 3.665 rad 2.4 10 rad/s 180 15 10 s t π ωω + ° × = = °× (b) The instantaneous value of the emf is ( ) ( ) 0 cos 12 V cos 3.665 rad 10.4 V t == = EE Assess: Be careful to change your calculator to the radian mode to work with the trigonometric functions.

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36.2. Model: A phasor is a vector that rotates counterclockwise around the origin at angular frequency ω . Solve: (a) Referring to the phasor in Figure EX36.2, the phase angle is rad 3 3 4 135 135 rad 1178 rad/s 180 4 2.0 ms t ππ π ωω × = = = ° (b) From Figure EX36.2, () 00 50 V cos 71 V cos cos 3 /4 rad t t ωπ =⇒ = = = E EE E
36.3. Model: A phasor is a vector that rotates counterclockwise around the origin at angular velocity ω . Solve: The emf is E = 0 E cos t = (50 V)cos(2 π × 110 rad/s × 3.0 × 10 –3 s) = (50 V)cos(2.074 rad) = (50 V)cos119 °

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36.4. Model: A phasor is a vector that rotates counterclockwise around the origin at angular velocity ω . Solve: The instantaneous emf is given by the equation E = (170 V) cos[(2 π × 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 π rad or (0.60)(360 ° ) = 216 ° in its fourth cycle.
36.5. Visualize: Please refer to Figure EX36.4 for an AC resistor circuit. Solve: (a) For a circuit with a single resistor, the peak current is 0 R 10 V 0.050 A 200 I R == = Ω E = 50 mA (b) The peak current is the same as in part (a) because the current is independent of frequency.

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36.6. Model: Current and voltage phasors are vectors that rotate counterclockwise around the origin at angular frequency ω . Visualize: Please refer to Figure EX36.6. Solve: (a) The frequency is 11 25 Hz 2 0.04 s f T π == = = (b) From the figure we note that V R = 10 V and I R = 0.50 A. Using Ohm’s law, R R 10 V 20 0.50 A V R I = Ω (c) The voltage and current are () ( ) RR cos 10 V cos 2 25 Hz vV t t ωπ ⎣⎦ ( )( ) cos 0.50 A cos 2 25 Hz iI t t For both the voltage and the current at t = 15 ms, the phase angle is t = 2 (25 Hz)(15 ms) = 2 π (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 ω = 2 π f = 2 π (100 Hz) = 628.3 rad/s is () C 6 11 5305 628.3 rad/s 0.30 10 F X C == = Ω × 3 C C 3 C 10 V 1.88 10 A 1.88 mA 5.305 10 V I X ⇒= = = × = ×Ω (b) The capacitive reactance at = 2 π (100 kHz) = 628,300 rad/s is ( ) C 56 5.305 6.283 10 rad/s 0.30 10 F X C = Ω ×× C C C 10 V 1.88 A 5.305 V I X = Ω Assess: Using reactance is just like using resistance in Ohm’s law. Because 1 C X , X C decreases with an increase in , as observed above.

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36.8.
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Chapter 36 - 36.1 Model A phasor is a vector that rotates...

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