Generators initially produce AC voltage with V(rms)=10.0 kV. Then transformers r

Question

Generators initially produce AC voltage with V(rms)=10.0 kV. Then transformers raise it to 138 kV for transmission on very tall power poles. Next, "substation" transformers lower it to 12.0 kV to run along the tops of telephone poles along our streets. Finally, the gray trash-can sized canisters on the telephone poles transform it to 120 kV for use in the our homes.

A. For each of the three transformerations in voltage, is it done by a "step-up" or "step-down" transformer?

Fill in the three blanks: 10.0 kV_________ 138 kV ___________ 12.0 kV _________ 120.V(you)

B. Which one of the three transformers above hasa ration of 11.5:1 (or 1:11.5) for the number of windings of its two sides? Show your work.

C. Suppose that you are drawing a total of 50.A in your entire house or apartment right now. If your home were the ONLY home serviced, and if you assume that all of the above transformers are exactly 100% efficient, how much current must flow in the 138 kV transmission wires in order to supply you with 50.A at 120. V in your home? Show your work.

Explanation / Answer

APPLY in a transforer Ns/Np = Vs/Vp = Ip/Is

A. 10 KV ---s tepped UP   -------138 KV ------ stepped down ----- 120 V---- stepped down ---- 12 V

B. Ns/NP = 138/10 = 13.8

Ns/Np = 12/138 = 0.0869 ( also 1: 11.5 = 1/11.5 = 0.0869)

so this is for transformer at substsation

c. at last step use,

Ip/Is = Vs/Vp

Ip = 120* 50/12000

Ip =0.5 is the out put of 12K volts

so Ip of las t = Is of 138 Kv( i,e its precedingtransforme)

so Ip = Vs*Is/VP = I2000* 0.5/138000

Ip =0.0434 Amps is the that must flow in 138 KV transformer

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