Physics: Transmission of Electrical Power

iamapebble SG

Hello, I'm seriously very confused over this part of physics which I just learnt, so I really need your help. >.< Thank you very much!

1] A transformer is used to supply a voltage of 12V to the output circuit. The primary coil has 2000 turns and the secondary coil has 100 turns. With the secondary coil connected across a resistor, the power dissipated in the secondary circuit is 40W.

[a] What kind of transformer is it? (I know it's a step down transformer)

[b] Calculate the current flowing in the secondary coil.

[c] Calculate the voltage of the a.c. supply, if the efficiency of the transformer is 80%.

2] A consumer receives a power of 20kW at 400V at his end of transmission lines. If the resistance of the transmission lines is 0.2 ohms, calculate

[a] the current flowing in the lines

[b] the voltage drop in the lines

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You see, I know the formulas but I seriously don't know how to apply them, in case you guys need, these should be the forumulas I can used for my level as of now:

IpVp = IsVs , Vs/Vp = Ns/Np

If it's possible, can you guys try to explain to me why you do the steps you take? Because I'm really 100% clueless >.<. There are still a lot of questions, but I guess I'll see how these two questions are solved first and attempt the rest on my own afterwards, thank you in advance! =)

wishboy SG

[b] Calculate the current flowing in the secondary coil.

In secondary coil, voltage is 12V, power is 40W, so

P = VI
40W = 12V * I
I = 10/3 A = 3.33A (3 s.f.)

[c] Calculate the voltage of the a.c. supply, if the efficiency of the transformer is 80%.

Not sure but I know

80/100 * Pp = Ps
80/100 * Pp = 40W
Pp = 50W

after that i dunno liao

2] A consumer receives a power of 20kW at 400V at his end of transmission lines. If the resistance of the transmission lines is 0.2 ohms, calculate

[a] the current flowing in the lines

power is 20 000W, voltage is 400V

P = VI
20kW = 400V * I
I = 50A

[b] the voltage drop in the lines

dunno how to do

iamapebble SG

OHH, thanks so much, so it's basically just the old formulaes?

For part 1[c], can I do it this way?

[For 80% efficiency] Vs/Vp = Ns/Np

12V/Vp = 100/2000

Vp = 240V

Therefore voltage of a.c. supply (should be 100%) = 240/80% x 100%

= 300V

=X As for part 2[b], can I assume that voltage drop is same as power loss? No right?

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May I ask another question? >.<

3] Electricity is to be supplied from a power station to a factory by means of a transmission line of overall resistance 100 ohms. The power station produces 10000kW. This power is delivered to the line at a voltage of 50kV. Calculate

[a] the current that would flow in the transmission line.

[b] the power coverted to heat on the transmission line.

For part [a] I'm kind of confused whether to use P = IV or V = IR or P = I^2 R. As for part [b], my teacher mentioned that for power loss you must use P = I^2 R, but then isnt P = IV possible too? Sorry and thank you so much for replying! (:

wishboy SG

1c i dun think u can do it tat way

2b im not sure of wad is it asking lol

3a) i think should be use P = IV

without a diagram, very hard for mi to know wad is the qn trying to say

3b) use heat loss = I²R

eagle SG

Do you have the answers?

For part 1[c], can I do it this way?

[For 80% efficiency] Vs/Vp = Ns/Np

12V/Vp = 100/2000

Vp = 240V

Therefore voltage of a.c. supply (should be 100%) = 240/80% x 100%

= 300V

That depends on whether the 12V is a dc voltage or an ac voltage. Not really stated in question.

If it is a dc voltage, you need to multiply it by sqrt(2) to get the max ac voltage

iamapebble SG

Sigh. ): Hmmm, sorry but I don't have the answers. It's okay, I'll wait for my teacher to go through and post up the answers here for others' references. :D Thanks so much for trying to help.