A battery of $15\text{ V}$ is connected to a parallel group of resistances of $4$, $10$ and $20\Omega$. The current in the circuit will be ($\text{A}=$ ampere)

1. Find the equivalent resistance ($R_{eq}$) of the three resistors connected in parallel:
$$\frac{1}{R_{eq}}=\frac{1}{4}+\frac{1}{10}+\frac{1}{20}$$
Find a common denominator, which is 20:
$$\frac{1}{R_{eq}}=\frac{5+2+1}{20}=\frac{8}{20}=\frac{2}{5}$$
$$R_{eq}=\frac{5}{2}=2.5\Omega$$
Using Ohm's Law ($V=I\times R_{eq}$) to calculate total circuit current ($I$):
$$I=\frac{V}{R_{eq}}=\frac{15}{2.5}=6\text{ A}$$