Three resistors `2Omega, 4Omega` and `5Omega` are connected in parallel.If the combination is connected to a battery of emf `20 V` and negligible internal resistance, determine the current through each resistor and the total current drawn from the battery.

Explanation

Main current
`I=E/(R_(p)+r)`

i.e., `I=(20)/(20/19+10)=19A`
Main current I = 19 A
Branch current `I_(1)=(20V)/(R_(1))=20/2=10A`
`I_(2)=(20V)/(R_(2))=20/4=5A`
and `I_(3)=(20V)/(R_(3))=20/5=4A`