Question

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An average power of 65 kW is delivered to a distant farm via a transmission line that has a resistance R = 3.00 Ohm. The generator has an rms voltage V_1 = 500 V. A step-up transformer raises the voltage to an rms value V_2 = 10,000 V. Calculate the following: The rms value I_2 of the current on the transmission line The average thermal power (P_thermal) lost on the transmission line. The rms voltage V'_2 just before the step-down transformer. The rms current I_1 provided by the generator to the step-up transformer. Assume that the step-up transformer does not have any losses. The ratio N_s/N_p of the number of turns in the secondary to the number of turns in the primary of the step-up transformer.

Answer 1

P3)P = 65 kW = 65000 W

I^2 R = 65000 => I = sqrt (65000/R)

I = sqrt(65000/3) = 147.2 A

So the current delivered is I = 147.2 A.

a)I2 = V2/R2

I2 = 10,000/3 = 3333.33 A

Hence, I2 = 3333.33 A

b)Power lost will be:

P' = I2^2R

P' = (3333.33)^2 3 = 33332666.67

Hence, P' = 33332666.67 = 3.33 x 10^7 Watts

c)RMS voltage V2' will be:

V2' = IR = 147.2 x 3 = 441.6 V (since V2 is yeilding I current to the farm house)

Hence, V2' = 441.6 Volts

d)Current provided by the generator will be:

I1 = V1/R = 500/3 = 166.67 A

Hence, I1 = 166.67 A

e)Ns/Np = V2'/V2

Ns/Np = 441.6/10,000 = 0.04416

Hence, Ns/Np = 0.0441