(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) There are the resistiors of resistances,
`R_(1)=2 Omega,R_(2) = 4 Omega, and R_(3) = 5 Omega`
They are connected in parallel. Hence, total resistance (R) of the combination is given by,
`(1)/(R)=(1)/(R_(1))+(1)/(R_(2))+(1)/(R_(3))`
`:. R = (20)/(19) Omega`
Therefore, total resistance of the combination is `(20)/(19) Omega`
(b) Emf of the battery, `V = 20 V`
Current `(I_(1))` flowing through resistor `R_(1)` is given by,
`I_(1)=(V)/(R_(1))`
`=(20)/(2)=10 A`
Current `(I_(2))` flowing through resistor `R_(2)` is given by,
`I_(2)=(V)/(R_(2))`
`= (20)/(4)=5A`
Current `(I_(2))` flowing through resistor `R_(3)` is given by,
`I_(3)=(V)/(R_(3))`
`=(20)/(5) = 4A`
The current, `I = I_(1)+I_(2)+I_(3)=10+5+4 = 19 A`
Therefore, the current through each resistor is 10 A, 5 A and 4 A respectively and the total current is 19 A.