Is there anyone good at physics that can me answer this question. if your not able to answer could you at least show steps or try to explain

Here's what you need to know about a transformer:

-- The ratio of voltages is the same as the ratio of turns.

Say you put 10 volts across one side of the transformer.  If the other side has twice as many turns, there'll be (10 x 2) = 20 volts across that side.   If the other side has half as many turns, then there'll be (10 x 1/2) = 5 volts across that side.

-- The power is the same on both sides.  Whatever power goes in, the same power comes out.  (It's the amount of energy every second, so it can't be created or destroyed.)

-- Just like every other electrical situation ... Power = (voltage) x (current)

Oh, I should also mention that any time you're working with a transformer, you're working with AC (alternating current).  If you put DC into a transformer, the only thing you get out of it is smoke.  This question doesn't even mention AC or DC.  It's just something to remember about transformers.

= = = = =

So, now, let's see what we've got here:

-- 200 turns on the primary, 50 turns on the secondary.  ==> Whatever voltage we put across the primary, we'll get 1/4 of that voltage on the secondary.

-- Electric power in the primary = 80 KW. ==> SAME 80 KW in the secondary.

-- 10,000 volts (10 kV) across the primary. ==> 2,500 volts across the secondary.

(1).  Voltage across the secondary ?  1/4 of the primary voltage

Voltage = 2,500 V

(2).  Current through the secondary ?  Well, the power has to be the same in both windings = 80,000 watts (80 KW) = (voltage) x (current).

(2,500 V) x (current) = 80,000 watts

Current = (80,000 watts) / (2500 V)

Current = 32 Amperes

(3).  What is the resistance, connected to the secondary, that's eating all of this power ?

R = V / I

Resistance = (voltage) / (current)  

Resistance = (2500 V) / (32 A)

Resistance = 78.125 ohms

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete

i will answer your question if you answer mine