Three resistors of resistances `3Omega, 4Omega and 5Omega` are combined in parallel. This combination is connected to a battery of emf 12V and negligi
Three resistors of resistances `3Omega, 4Omega and 5Omega` are combined in parallel. This combination is connected to a battery of emf 12V and negligible internal resistance, current through each resistor in ampere is
A. 4,3,2.4
B. 8,7,3,4
C. 2,5,1,8
D. 5,5,8,2
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Correct Answer - A
Since, the voltage across the circuit is constant. Then current through `3 Omega` resistor
`I_(1)=(V)/(R_(1))=(12)/(3)=4A`
The current throught `4Omega` resistor `I_(2)=(V)/(R_(2))=(12)/(4)=3A`
and the current throught `5Omega` resistor, `I_(3)=(V)/(R_(3))=(12)/(5)=2.4A`
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