Question

A simple transformer has a 100-turn primary coil and a 1000-turn secondary coil. The primary is connected to 120V AC source and the secondary is connected to an electrical device with a resistance of 1,000ohms.

-What will the voltage output of the secondary be?
-What current flows in the secondary circuit?
-What is the power in the secondary coil?

Answer 1

The voltage ratio is equal to the turn ratio so where N = number of turns and V = votlage:
\[\frac{Ns}{Np}=\frac{Vs}{Vp}=>\frac{1000}{100}=\frac{1200V}{120V}=>V_{secondary}=1200V\]For the second part you have a resistance of 1,000 ohms so the current in the secondary is \[I=\frac{V}{R}=\frac{1200V}{1000\Omega}=12A\]Lastly for the power: \[P=IV=12A*1200V=14,400W=14.4 kW\]

Answer 2

How did you get 1200V over 120V?

Answer 3

For simple transformers:\[\frac{Ns}{Np}=\frac{Vs}{Vp}\] so
Using this you can see that the secondary has 10x the number of turns so it must have 10x the voltage.

Answer 4

and you were told that the source voltage (primary side of the transformer) was 120V.

Answer 5

OH!!!! Thank you thank you!

Answer 6

no problem :)

Answer 7

Wait then how would you find the power in the "primary" coil?

Answer 8

The the extended equation of this would also relate the current:
\[\frac{N_s}{N_p}=\frac{V_s}{V_p}=\frac{I_p}{I_s}\]So you can solve for the primary current as follows:
\[\frac{1000}{100}=\frac{1200V}{120V}=\frac{120A}{12A}\]So the primary current will be 120A.