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physics problem please help!!


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Here it is and it has a few parts to it:

 

1.a. Power lines carry electricity at about 120,000 volts. If a power line were stepped down by a transformer to 120 volts, how many turns would there be on the secondary coil if there were 1000 turns on the primary?

 

b. if the secondary coil in part a is connected to a resistor with a resistance of 10 ohm's, what is the current in the secondary coil?

 

c. what is the power in the secondary coil?

d. what is the current drawn by the primary coil?

e. why is it useful for the power lines to transfer electricity with a current of the magnitude of your answer in part d?

 

Ok, this is a physics course but my son is studying electricity and magnetism. In the textbook it gives the following formula to use for part a, but he doesn't know what goes where.

 

primary voltage/number of primary turns= secondary voltage/number of secondary turns

 

Any help will be greatly appreciated!:001_smile:

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Here it is and it has a few parts to it:

 

I gave this solution in hurry, please wait for someone to point out if any errors.

 

1.a. Power lines carry electricity at about 120,000 volts. If a power line were stepped down by a transformer to 120 volts, how many turns would there be on the secondary coil if there were 1000 turns on the primary?

 

Your book formula is correct. Use V for volts, N for turns, subscript p for primary and subscript s for secondary. Rewrite the formula as below:

 

 

Vs/Vp = Ns/Np

 

Hence

 

Ns = Np (Vs/Vp) --------> (1)

 

Given:

 

Vp = 120,000 volts

 

Vs = 120 volts

 

Np = 1000 turns

 

 

 

Substituting above quantities in (1)

 

 

 

Ns = 1000 (120/120000)

 

= 1

 

b. if the secondary coil in part a is connected to a resistor with a resistance of 10 ohm's, what is the current in the secondary coil?

 

Current in secondary Is = Vs/Rs = 120/10 = 12 ampere

c. what is the power in the secondary coil?

Power in Watts Ps = Vs x Is = 120 volts x 12 ampere = 1440 watts

 

d. what is the current drawn by the primary coil?

 

Power in primary coil =
Power in secondary coil

 

Ip x Vp = Is x Vs = 1440

 

Ip = 1440/120000 = 0.012 ampere (which is comparatively low amperage)

 

e. why is it useful for the power lines to transfer electricity with a current of the magnitude of your answer in part d?

 

P
ower loss = (I^2) R.

 

Thus, power loss increases in square proportion of current I for a given resistance R. By reducing I to low level such as 0.012, the power companies avoid power loss during power delivery from power station which is far away fro home or place of power consumption.

 

 

 

Rigorous solution to similar question is in
. (
)

 

 

 

Best regards.

 

 

 

mpcTutor

 

www.mpclasses.com

 

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AP Calculus, AP Physics, Singapore Math Grades 7-12

 

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US Central Time: 2:43 PM 5/18/2010

 

Ok, this is a physics course but my son is studying electricity and magnetism. In the textbook it gives the following formula to use for part a, but he doesn't know what goes where.

 

primary voltage/number of primary turns= secondary voltage/number of secondary turns

 

Any help will be greatly appreciated!:001_smile:

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