1.

How many 176 Omega resistor (in parallel) are requried to carry 5A on a 220V line?

Answer»

SOLUTION :Let n RESISTORS of `176 Omega` are joined in parallel. Then their combined resistances `R_p` is GIVEN by
`1/R_p=1/176+1/176….n time s=n/176 or R_p=176/n Omega`
It is given that V=220V and I=5A
`therefore R_p=V/I or 176/n=220/5=44`
`IMPLIES n=176/44=4 i.e., 4` resistors should be joined in parallel.


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