A 550V powersource provides energy to a load resistanceR L through a pair of cop

Question

A 550V powersource provides energy to a load resistanceRL through a pair of copper(=1.72x10-8m) wires that are20km long. The diameter of the wires is3.0mm.

A. 7.8km

B. 17.9km

C. 14.6km

D. 2.1km

QuestionDetails:

A 550V powersource provides energy to a load resistanceRL through a pair of copper(=1.72x10-8m) wires that are20km long. The diameter of the wires is3.0mm.

During a storm thecurrent, I, supplied by the source increases and the voltage acrossRL decreases. An engineer concludes that a fault has developedsomewhere along the line and that there is a shunt resistance RS atsome distance x from the source. If the load resistance isdisconnected, I is measured to be 3.8A. When the line terminals atthe load are shorted, I is 7.2A. What is x(2pts)?

A. 7.8km

B. 17.9km

C. 14.6km

D. 2.1km

Explanation / Answer

= 550 V, L = 20 km, I1 = 3.8 A, I2 = 7.2 A, d = 3.0 mm, areaA = d2/4 Let the resistance of the 2 segments of wires corresponding to xbe the resistance of the 2 segments of wires corresponding to L - xbe R2 = *2(L - x)/A = 8(L - x)/(d2) =8L/(d2) - R1 R1 + Rs =/I1                           (1) so R1 = /I1 - Rs R2 = 8L/(d2) - /I1 + Rs = Rs -47.405       (2) R1 + Rs*R2/(Rs + R2) = /I2     (3) (1) - (3): Rs*[1 - R2/(Rs + R2)] = /I1 - /I2 Rs2/(Rs + R2) = /I1 - /I2 = plug (2) into it, Rs2/(2Rs - 47.405) = 68.348 Rs2 - 68.348*(2Rs - 47.405) = 0 Rs2 - 136.696Rs + 3240.02 = 0 solve it and get Rs = 106.18 , or 30.514 R1 = /I1 - Rs = 38.557 , or 114.22 R1 = 8x/(d2) x = d2R1/(8) = 7.92 km or 23.47km 23.47 > 20, so 23.47 km should be discarded. answer: x = 7.92 km

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