Eight lights are connected in parallel to a 117 V source by two leads of total resistance 2.8 . If 240 mA flo?

Rtotal = (117/(8*0.24)) ohms = 60.9375 ohms

2.8 ohms of this is in the leads

So the resistance of all the 8 bulbs in parallel is (60.9375 - 2.8) ie 58.1375 ohms

The resistance of one bulb is 8 times this, ie, 465.1 ohms

Total power = (I^2)*Rtotal

Power in resistance of leads = (I^2)*Rleads

Pleads/Ptotal = Rleads/Rtotal = 2.8/60.9375 = .046 = 4.6%

What does the resistance 2.8 mean? I'll assume ohms. I'm also assuming that is one set of leads for each bulb, as opposed to one set for 8 bulbs. The question is vague.

R = E/I = 117/0.24 = 487.5 Ω

subtract the 2.8 Ω for the leads and you get 484.7 Ω for each bulb

power in the leads is I²R = 0.24²×2.8 = 0.161 watts

for 8 bulbs that is 1.290 watts

Power in a bulb is I²R = 0.24²×487.5 = 27.6 watts

percent in leads is 0.161 / (27.6+0.161) = 0.58%

answer will be different if it is one set of leads for 8 bulbs.

.

This is a great question