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A battery of 12 V is connected to a series combination of resistors 3Omega , 4 Omega, 5 Omega and 12 Omega. How much current would flow through the 12 W resistor ? |
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Answer» Solution :Current REMAINS same in SERIES combination `R_("series ") =R_(1)+R_(2)+R_(3)+R_(4)` `(3+4+5+12)=24 Omega` V=12 V `I =(V)/(R)=(12V)/(24Omega)=(1)/(2) =0.5 A` |
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