(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 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) Total resistance in parallel combination of three resistors of resistances `R_(1), R_(2) and R_(3)` is calculated by:
`(1)/(R_(p))= (1)/(R_(1)) + (1)/(R_(2)) + (1)/(R_(3))`
Here, `R_(1) = 2Omega, R_(2) = 4Omega, R_(3) = 5Omega`
`:. (1)/(R_(p)) = (1)/(2) + (1)/(4) + (1)/(5)`
`rArr (1)/(R_(p)) = (10 + 5+4)/(20)`
`rArr R_(p) = (20)/(19) Omega`
(b) Since the resistors are connected in parallel and the internal resistance of the battery is negligible, the potential difference across each of the resistors will be same and is equal to the emf of the battery. The current flowing through each reasistor is shown in the following figure:

`:. I_(1) = (E )/(R_(1)) = (20)/(2) = 10A`
`I_(2) = (E )/(R_(2)) = (20)/(4) = 5A`
`I_(3) = (E )/(R_(3)) = (20)/(5) = 4A`
Total current drawn from the battery, `I= I_(1) + I_(2) + I_(3) = 10 + 5+4= 19A`