Question

A generator has 534 circular loops with a diameter of 0.234 meters. The magnetic field applied through the region where our loops will spin is 0.456 T. It outputs an average of 66 MW. Assume the generator circuit is the source as described above in an AC-RL circuit where R is the equivalent resistance Reg and the inductor L is an ideal transformer. 8. If the wire loops spin at 9.33x104 rev/min, what is the Vmax produced by the generator? 9. What are the values of Req and L, knowing that the max voltage drop across the transformer must be 95% that of Vmax? 10.Power is transmitted to a town 33 km away through 2 wires with a linear resistance rating of 0.11 2/km. What must the current be in these wires in order to lose only 1.6% of average power output by the transformer due to heating the transmitting wires? 11. Let's assume our ideal transformer has its secondary current as the value above. What would the number of turns in the secondary coil be if the primary has 888?

Please answer all the questions!!!

Answer 1

8. The maximum voltage produce is, Vimax = 250 Ws - Here = B.A Area, A=TR? =TT*(0,234) = 0.043m2 $=0.456 x0.043m? =0.0196 Wb (1 min=cosed w = 9.33x10Yvew) mi'r = 1.55 geuls F10mo, Vman = 2x314 X0:0196 X 1.55 Vnox = 191.5V 9. cwovent , Jin = 66 xrobu P. Viaz = 344.64 k Amy 191.50 Voltoge dos op = I VR2X42 -0.95 Vnrar I VR3 x = 0.958 191.50 Jx2 . 7 0.95X191.50 x103 344.64 Amp Ba xº so,2 x 56 we know that, Inductane = Monza l L-415X1D7X(1063){x0.0430 n-1068 becawe m=2x circulos loop n = 2x 534 =1068 = 0.0616 HIM

then using voltige 2010P masimum R² = 0,29 4156 R=o.sim 10) Total gresistance of wide = 0.11'ex33 -3.632 - For a lines =2x3163 =7,26 total power loss is 1,6% 1.6 X 66 MW. 100 = 16x66x106w 100 we know that , loss equal to IRSO, 1.6 - X 66 X106 =1247,26 (00 I =381.38 Amp 1) Given that, loom above, I381.38 cmp. Is IP Np Ns NS-NO () -888*/ 344-64 amp) 381385 amp Ne = 802444