Question

Question-2: An industrial coil is modelled as a series combination of an inductance L and resistance R, as shown in Fig. 2. Since an ac voltmeter measures only the magnitude of a sinusoidal voltages, the following measurements are taken at 50 Hz when the circuit operates in the steady state: Vg1 = 300 V, V l = 60 V, IV.l = 120 V. Use these measurements to determine the values of L and R. (15 Marks) Coil 80 92 w V + R V Figure 2. An industrial coil of Question 2. Answer-2:

Answer 1

page, Answers Giren data sol 2 wme Vs = 300V (VL= Gov Vol = 1201 f=50 m2 69 = ant 27450 =10071 I 60 80 = 0.75 r 11=0.75

pager- & How , Voz 0.75 (R+jwc) we an F an (50) 1001 Vo = 0.75 (R+S(10072)) Hol= 0.75 VR²1 108 At (² dala vo =60 from given (6012 = 0.75 / R² H108A²2² 120 = 0.75 / 82100² A22 (120) / R²100²4222 0.75 Permia 160 Squaring on both sides (1601² R²100² A 22 - lus) 300V 0-75 (-2+R+j WL) =Vs » 0.75 (so tR+ 3 Clook 2) =Vs.

pgel- 0.75(80 +R)2 + C100ML)? = 300 300 (80tr) + C10074) 0.75 -400 (80+21² + (10074² Squaring both sides (80+R)2 + (100m2)* 40044 - © Now Solving ① and ② 2 400 180 +R} + 100ML R² + 1064²ť 160² 80 27R²-p² 4002 - 160² 80² + R² + 16or -pt 160000 - 25600 80+160R = 134400 160R a + 34400-6400 160R = 128000 R = 128000 - sooohms 16 R = 800 ohms

page! pul R valu 800 in Equat 100? A² (² 8002 160² 100 el ek connie - 1 = 6.995 L qua henry R=Soo ohms L = 2.ua herry