Question:
A circuit has a current rating of 5 A. How many lamps of rating 40 W, 220 V can run simultaneously on it.
Answer:
Bulb rating as (p)= 40 w and (V)= 220 V
So, Resistance (R) = V2/p = (220)2/40 = 1210 Ω
Maximum current (I) =5 A
Therefore, Resistance (R1) = V/I
R1 =220/5 = 44 Ω
When n number of resistance are in parallel in line, then
R1 = R/n
44 = 1210/n
n = 1210/44 = 27
So, 27 number of bulb can be used simultaneously on it.
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