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