A transformer has 50 turns in its primary and 250 in its secoundary. 12 V AC is connected to the primary.?

Question

1. Find the volts available at the secoundary.
2. The current in a ten ohm device connected to the secoundary.
3. The power supplied to the primary.

I know the first one, 60V in the secoundary.

Not sure how to solve the 2nd. I know the formula is C=V/T....

the third the formula is Power=C*V

Anyone want to help me out. Not sure what to plug in to get the results....explain please

Answer 1

The current drops by the same ratio that the voltage increases.

By Ohm's law, V=IR so I = V/R = 60 volts / 10 ohms = 6 amperes

The power in the primary is the same as the power in the secondary. P=VI = 60V*6A = 360 W

Answer 2

Your question is unclear because it's not clear if you mean 12 V peak or 12 V rms; where, for 60 cycle AC, V rms = V peak/sqrt(2).

When doing power work, V rms is equivalent to V DC when doing P = I^2 R and V rms = I R. Assuming your 12 V is peak voltage, the answers are:

1. VN = vn; so v = VN/n where V = 12 volts peak, N = 250 turns, and n = 50 turns. v is the secondary peak voltage.

2. v rms/R = I; where v rms = v/sqrt(2) and R = 10 ohm. I, which you can now calculate, is the current equivalent to DC current and should be used for power calculations.

3. Assuming no heat loses (an unrealistic assumption), we can say the power in the primary = the power in the secondary. So the power in the secondary is P = I^2 R; where I was found in 2 above and R = 10 ohms.

In the future, please define your variables for clarity.