Three resistor 4 `Omega`, 6 `Omega` and 8 `Omega`, are combined in parallel.If the combination is connected to a battery of emf 25 V and negligible Internal resistance, then determine the current through each resistor and total current drawn from the battery.

Explanation

Given :
`R_(1)=4 Omega, R_(2) = 6 Omega,R_(3)= 8OmegaV= epsilon= 25 V, r=0`
a. the total resistance of the parallel combination
`(1)/(R_(rho))=(1)/(R_(1))+(1)/(R_(2))+(1)/(R_(3))`
`(1)/(R_(rho))=(1)/(4)+(1)/(6)+(1)/(8)`
`(1)/(R_(rho))=(6+4+3)/(24)=(13)/(24)`
`R_(rho)=(24)/(13)=1.846 Omega`
B. The current through `4 Omega` is `I_(1)=(V)/(R_(1))=(25)/(4)`
`I_(1)= 6.25 A`
The current through `6 Omega`is `I_(2) = (V)/(R_(2))=(25)/(6)=4.16` A
The current through `8 Omega` is `I_(3)=(V)/(R_(3))=(25)/(8) `
`I_(3)= 3.125`
Total current drawn from the batter is
`1=1_(1)+1_(2)+1_(3)=6.25+4.16+3.125`
1=13.535 A