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We have n identical light bubls, which draw a power P when connected across supply of 220V. If all of them is series are connected to 220V supply then power drawn is P_(1) and when all of them are connected in parallel them power drawn is P_(2), Find P_(2)//P_(1) |
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Answer» Solution :Let R be the resistance of filament of a SINGLE BULB. USING `P= V^(2)//R`, we can write `P = (220)^(2)//R` …(i) When all of them are connected in series then equivalent resistance becomes nR, hence power drawn becomes `P_(1) = (220)^(2)//nR= P//n`. When all of them are connected in PARALLEL then the equivalent resistance becomes R/n. Hence, power drawn becomes `P_(2)= (220)^(2)//(R//n) = nP` Therefore, `P_(2)//P_(1) = n^(2)` |
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