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 Req 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 S2/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?

Answer 1

* Answer: The given data ? has 534 ЭН generator circular loops - Diameter = 0.234 m → Magnetic field applied B: 0.456T -> Output P i 66 MW 8) Given that wire loops spin at 9.33 × 104 verlmin Vmax produced by generator is to be produced Vmax'a 2n ows 0 Where 0 = magnetic flux = B. A where Area (A) = TR? Diameter D = 0.23 um R. 0.234 2 R = 0.117m P = 1TR, T (0.1111 A=0.043 m 2 Q - BA 0.456 X 0.043 a = 0.0196 wb where given ws = 9.33 X 10" verlmin Ws 1555 revisec [lmin = Gosec)

* Substitute the value of 0; in equation 0 Vmax = 2T X 0.0196 x 1555 Vmax = 191.49 v 9 Given that max voltage drop across transformer must be 95%. that of Vman Given P=66 MW. T Р 66x106 Vmax 191.49 I 0.34466x 6x106 A - Votlage. drop drop = I VR2+*23 Given 95% of Vmax Then I R²+x2² I VR2+x23 = 0.95 .max I VR?+xm? = 0.95(191.49] R?+x2? 181.91 0.344 x 106 R²+x₂2 528.80 x 10 :10-6 where inductance Mon?A La im, n = ax tircular loops n = 2x 530 17 1068

La 477 X10 41x10.7 x 11068]? x0.043 1 TL 0.06163 Him Resistance R- Vmax Iin R2 191.49 0. 34466 X 106 R=555.59 x 106 - R = 0.55 m. 10) Power is transmitted to 33 km 1 through a wires resistance 0.11.7km lénear → → lose only 106% of average power output . So total resistance of wire is Oolla X 33 R total = 3,630 For two lines a 2x3.63 Rz lines) § 7.26 2 Given power loss is 1.6% of 1.6 average output = 66x106 X 100 2 P= Ih R (plines) I ?x 7.26 = 106 x 66x106 100

I? 1.056 xlob 7.26 I’ - 0.1454 x 106 H Vo 14568106 103 0.38131 x 10 I - 381.31 11 H 1) Given transformer has its Secondary current as solved above, Then Is = 381-31 888 then If primary turns hass secondary turns is to be calculated Number of primary tums Np=888 Number of secondary turns Ns Ns = ? From equation @ current 0.34466x 106 Ip is we know relation Is NP 3 Ip Ns

substitute the given values. in eq 888 381.31 lj 6 NS 0.34466 x 10 0. 3uu 66 * 104 Ns 888 = 381.31 NS 3 802648.91 Ns y 802609 Therefore the number of turns in secondary coil is 8026uq.
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