$(a)$ 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.

(N/A) Given resistances are $R_{1} = 2 \; \Omega$,$R_{2} = 4 \; \Omega$,and $R_{3} = 5 \; \Omega$.
Since they are connected in parallel,the total resistance $R$ is given by the formula:
$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$
Substituting the values:
$\frac{1}{R} = \frac{1}{2} + \frac{1}{4} + \frac{1}{5} = \frac{10 + 5 + 4}{20} = \frac{19}{20} \; \Omega^{-1}$
Therefore,$R = \frac{20}{19} \; \Omega \approx 1.05 \; \Omega$.
$(b)$ The battery voltage is $V = 20 \; V$. In a parallel circuit,the voltage across each resistor is the same.
Current through $R_{1}$: $I_{1} = \frac{V}{R_{1}} = \frac{20}{2} = 10 \; A$.
Current through $R_{2}$: $I_{2} = \frac{V}{R_{2}} = \frac{20}{4} = 5 \; A$.
Current through $R_{3}$: $I_{3} = \frac{V}{R_{3}} = \frac{20}{5} = 4 \; A$.
Total current $I = I_{1} + I_{2} + I_{3} = 10 + 5 + 4 = 19 \; A$.