Question

help

Answer 1

@dumbcow

Answer 2

@e.mccormick @nincompoop @Compassionate

Answer 3

Uh - I don't know. I'm going to guess A.

Answer 4

for what reasons?

Answer 5

http://en.wikipedia.org/wiki/Parallel_circuits#Parallel_circuits

\[I =V (\frac{4}{R})\]

http://en.wikipedia.org/wiki/Electric_power#Resistive_circuits
\[P = \frac{V^2}{R} \rightarrow R = \frac{V^2}{P}\]
\[\rightarrow I = V(\frac{4P}{V^2})\]
\[I = \frac{4P}{V} = \frac{400}{220} = 1.8\]

Answer 6

Careful: there are only 3 bulbs fed by the designed current.

Answer 7

100/220 = 0.45, the current of one bulb. Since the ammeter is measuring the current used by ONLY three bulbs, the reading will be about 1.35