(a) Three resistors `2 Omega, 4 Omega and 5 Omega` are combined in series. What is the total resistance of the combination ?
(b) 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

(a) Here, `R_(1) = 2 Omega , R_(2) 4Omega , R_(3) = 5 Omega , V = 20 V`
In parallel combination,total resistane `R_(P)` is given by
`1/R_(P) = 1/R_(1) + 1/R_(2) + 1/R_(3) = 1/2 + 1/4 + 1/5 = (10 + 5 + 4)/20 = 19/20 or R_(P) = 20//19 Omega`.
(b) Current through `R_(1) = V//R_(1) = 20//2 = 10A`, Current through `R_(2) = 20//4 = 5A`
Current through `R_(3) = 20//5 = 4A` , Total current `= 20//(20//19)=19A`