(a) Three resistors `2 Omega 4 Omega and 5 Omega` are combined in parallel. What is the total resistance of the combination ?
(b) If the combination is connected to a bettery of emf 20 V and neglibible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

Explanation

(i) Current passing through `R_(1)` is ,
`I_(1) = (epsilon)/(R_(1)) = (20)/(2) = 10 `A
(ii) Current passing through `R_(2)` is ,
`I_(2) = (epsilon)/(R_(2)) = (20)/(4) = 5 `A
(iii) Current passing through `R_(3)` is ,
`I_(3) = (epsilon)/(R_(3)) = (20)/(5) = 4 ` A
Applying Kirchhoff.s current law, (KCL also known as junction rule) at junction a,
`sum I = 0 `
`therefore I - I_(1) - I_(2) - I_(3) = 0`
`therefore I = I_(1) + I_(2) + I_(3) `
`= 10 + 5 + 4 `
`therefore I = 19 A ` .