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How many 176Omega resistors (in parallel) are required to carry 5A on a 220 V line ? |
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Answer» Solution :I = 5A V = 220 V `Rp= (V)/(I) = (220)/(5) = 44 OMEGA` Now, if n resistors, each o.f resistance 176 `Omega` are connected in parallel to GIVE EQUIVALENT resistance of 44`Omega`, i.e. `R_(p)` `R_(p)= (R)/(n)` `44=(176)/(n)` `n=(176)/(44) = 4` |
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