1.

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 ?

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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