1cat2ferrets Posted May 18, 2010 Share Posted May 18, 2010 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: Quote Link to comment Share on other sites More sharing options...
1cat2ferrets Posted May 18, 2010 Author Share Posted May 18, 2010 BUMP!! Quote Link to comment Share on other sites More sharing options...
mpcTutor Posted May 18, 2010 Share Posted May 18, 2010 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 math forum . ( See the link ) Best regards. mpcTutor www.mpclasses.com --------------------------------------------------------------------------- AP Calculus, AP Physics, Singapore Math Grades 7-12 --------------------------------------------------------------------------- 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: Quote Link to comment Share on other sites More sharing options...
1cat2ferrets Posted May 18, 2010 Author Share Posted May 18, 2010 Thanks so much for the help. My son got the answer for number 1 as being only 1 turn on the secondary coil, but he was unsure if it was right or not.:001_smile: Quote Link to comment Share on other sites More sharing options...
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