Parallel Circuits and power lost in leads?

Question

I have eight lights flowing in parallel to a 1.1E2-V source by two long leads of total resistance.

If the resistance of each equals 500ohms and the current equals 2.2E-1 A what percentage of the total power is wasted in the leads.

I'm lost...any help would be great! Thanks!

This is the complete question, its part of homework:
Eight lights are connected in parallel to a 1.1E2-V source by two long leads of total resistance 1.7 Ω.

Enter scientific notation as 1.23E4.

If 2.2E2 mA flows through each bulb, what is the resistance of each?


(I got 500ohms for this part)
[Num] [Units]



If 2.2E2 mA flows through each bulb, what percentage of the total power is wasted in the leads?

I can't get this part though!

Answer 1

Not quite enough information to tell you how much power usage is added by the extension cords.
Two items with 500ohm resistance in parallel would certainly have resulted in .22amps of current flowing in each string of lights since
V=I*R=110=.22*500
If the total amperage is really .22 then maybe the resistance of the cords is 500 and the resistance of the 4 lights at the end of each of the cords have an effective resistance of 500ohms.
The geometry of the problem needs to be known to provide a definitive answer. If the lights are equally spread along the long leads then the answer is different still.

Answer 2

The formula is E=IR
E= voltage
I = current
R=Resistance
Measure the Resistance with an ohm meter.
measure the applied voltage under load.
If u can measure the Current in each leg.