An ac generator provides emf to a resistive load in a remote factory over a two-

Question

An ac generator provides emf to a resistive load in a remote factory over a two-cable transmission line. At the factory a step-down transformer reduces the voltage from its (rms) transmission value V_t to a much lower value that is safe and convenient for use in the factoiy. The transmission line resistance is 0.57 ohm/cable, and the power of the generator is 283 kW. If V_t = 100 kV, what are (a) the voltage decrease delta V along the transmission line and (b) the rate P_d at which energy is dissipated in the line as thermal energy? If V_t = 9.8 kV, what are (c) delta V and (d) P_d? If V_t = 0.92 kV, what are (e) delta V and (f) P_d?

Explanation / Answer

R = 2*0.57 ohm

part a )

rms current in cable = Irms = P/Vt = 283 * 10^3 /(100 x 10^3) = 2.83 A

voltage drop = dV = Irms*R = 3.2262 V

part b )

Pd = Irms^2*R = 11.87 W

part c )

Vt = 9.8 kV

Irms = 28.878 A

dV = Irms*R = 32.92 V

part d )

Pd = Irms^2*R = 950.66 W

part e )

Vt = 0.92 kV

Irms = 307.6 A

dV = 350.67 V

part f )

Pd = Irms^2*R

Pd = 107870.345 W

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.