i) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?

ii) 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.

Given, three resistors 2$\mathrm{\Omega }, 4\mathrm{\Omega } \mathrm{and} 5\mathrm{\Omega }$ are combined in parallel. 
 
Therefore,
Total resistance of parallel combination,
                      $$\begin{gathered} \boxed{\frac{1}{\mathrm{R}}= \frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}+\frac{1}{{\mathrm{R}}_3}} \\ \mathrm{or} \frac{1}{\mathrm{R}} = \frac{1}{2}+\frac{1}{4}+\frac{1}{5} \\ = \frac{10+5+4}{20} \\ = \frac{19}{20}\mathrm{\Omega } \end{gathered}$$
                           
                             $\mathrm{R} = \frac{20}{19}\mathrm{\Omega }$ 

ii) The combination is connected to a battery of 20 V.
Let the current through resistances 2Ω, 4Ω and 5Ω are I₁, I₂ and I₃ respectively.
Now, using Ohm's law across each resistor we get,
                    $$\begin{gathered} {\mathrm{I}}_1 = \frac{\mathrm{V}}{{\mathrm{R}}_1} = \frac{20}{2} = 10 \mathrm{A} \\ {\mathrm{I}}_2 = \frac{\mathrm{V}}{{\mathrm{R}}_2} = \frac{20}{4} = 5 \mathrm{A} \\ {\mathrm{I}}_3 = \frac{\mathrm{V}}{{\mathrm{R}}_3} = \frac{20}{5} = 4 \mathrm{A} \end{gathered}$$  
Hence,
Total current is given by
             $$\begin{gathered} \mathrm{I} = {\mathrm{I}}_1+{\mathrm{I}}_2+{\mathrm{I}}_3 \\ = 10+5+4 \\ = 19\mathrm{A}. \end{gathered}$$