(a) Three resistors `2Omega ,4Omega` and `5Omega` 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

Here `R_1 = 2Omega , R_2 = 4Omega` and `R_3 = 5Omega`
In parallel combination resultant resistance R is given by
`1/R = (1)/(R_1) + (1)/(R_2) + (1)/(R_3) = 1/2 + 1/4 + 1/5= (10 + 5 + 4)/(20) = 19/20`
` rArr R = 20/19 Omega`
(b) Here `epsi = 20V` and r = 0
` therefore ` current through the three resistors are respectively
`I_1 = (epsi)/(R_1) = 20/2 = 10 A, I_2 = (epsi)/(R_2) = 20/4 = 5 A, and I_3 = (epsi)/(R_3) = 20/5 = 4A`
Total current drawn `I = I_1 + I_2 + I_3 = 10 + 5 + 4 = 19A`