130604-0904 EDT
mark32:
You need to understand ohms law to use it.
It says if you have a resistor that has a constant resistance independent of the current thru the resistor that the voltage across the resistance can be represented by the equation V = I*R.
In your application you have a circuit analysis problem. This requires more than just ohms law. You need to know that the voltage around a closed loop is 0.
You have a series circuit in its simplest form that is an ideal voltage source (constant voltage), an internal transformer impedance (for simplicity assume resistance), a wire resistance from the transformer to the load, and a single load resistance (lump all your individual lamps at one point).
Thus, Vsource = Vinternal + Vwire + Vload.
Assume Vload = 12 V and as you calculated the current is 12.5 A.
Next assume a wire resistance of 0.2 ohms. The voltage drop along the wire is 0.2*12.5 = 2.5 V.
Assume another 1 V drop in the transformer, then the ideal source voltage is 12+2.5+1 = 15.5 V, and this is the open circuit voltage from the transformer. VA input to the transformer would be slightly greater than 15.5*12.5 = 194 VA. The some greater is to allow for transformer core losses.
If you have 12 V at the input to the string of lamps that I lumped all in one spot, but actually the lamps are distributed along additional wire, then the total current will be somewhat less than your calculated value when the 12 V is measured at the input end of the light string.
You really would need to analyze the light string with 12 V at the input (or some other value) to determine what the current is to obtain an accurate value. This is probably not necessary. Just the 12 V at 12.5 A is likely good enough.
As an example of how power varies for an incandescent bulb you can use the equation
P = V^1.6/21.217 for a 120 V 100 W bulb. This quite closely approximates the actual curve from 90 to 130 V input.
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