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Problem 7.3 The voltage and current of an electric load can be expressed by the following equations v 340 sin (628.318 t 0.5236) i 100 sin (628.318 t 0.87266) Calculate the following: a. b. c. d. e. The rms voltage The frequency of the current The phase shift between current and voltage in degrees The average voltage The load impedance Solution: 340 240.41 V 2 a. V b. 2 f 628.318 rad/sec f c. 628.318 100 Hz 2 Phase shift 0.5236 0.87266 0.3491 rad 0.3491 180 20 o d. Average voltage is zero (the voltage waveform is sinusoidal) e. Load impedance Z V 340 / 2 20o 3.4 20o I 100 / 2 Problem 7.5 An ac source is feeding a load consists of a resistance and inductive reactance connected in series. The voltage and current of the source are v = 400 sin (377 t + 0.5236) i = 100 sin (377 t + 0.87266) Calculate the following: V A a. b. c. d. e. f. The resistance and inductive reactance of the load impedance The rms voltage across the resistance The frequency of the supply voltage Power factor at the source side, state leading or lagging The-real power delivered to the load The-reactive power delivered to the load Solution: 0.5236 *180 Z 4 20 o 0.87266 *180 100 100 4 cos 20 o 265.78 V VR I R 2 377 f 60Hz 2 2 pf cos(0.5236 0.87266) 0.939 leading 400 *100 P V I cos 0.939 18.78 kW 2 400 a. b. c. d. e. Q f. V I 2 2 400 * 100 2 P 18780 6.88 kVAr Capacitive 2 2 Problem 7.7 A load impedance consists of 25 resistance in series with 38 inductive reactance. The load is connected across a 240 V source. Compute the real and reactive powers consumed by the load. Solution: Z 252 382 45.5 ohm I V 240 5.28 A Z 45.5 pf cos( ) 25 R 0.55 45.5 Z P V I cos( ) 240 5.28 0.55 696.96 W sin( ) 38 XL 0.8352 45.5 Z Q V I sin( ) 240 5.28 0.8352 1.058 kVAr Problem 7.12 An electric load consists of 5 resistance, 20 capacitive reactance and 10 inductive reactance, all connected in parallel. A voltage source of 100V is applied across the load. Compute the following: a. b. c. The real power of the load The reactive power of the load The current of the load Solution: a. 100 2 2 kW 5 b. 100 2 100 2 500 VAr 10 20 c. S 2000 2 500 2 2.062 kVA I 2062 20.62 A 100