In the provided circuit diagram, ammeter A_1 shows a reading of 2.4...

Let the current through the $20 \Omega$ resistor be $I_1$ and through the $30 \Omega$ resistor be $I_2$.

Given that the reading of ammeter $A_1$ is $I_1 = 2.4 \text{ A}$.

Since the $20 \Omega$ and $30 \Omega$ resistors are connected in parallel, the potential difference across them is the same.

$V = I_1 \times 20 = 2.4 \times 20 = 48 \text{ V}$

The reading of ammeter $A_2$ is the current $I_2$ through the $30 \Omega$ resistor:

$I_2 = \dfrac{V}{30} = \dfrac{48}{30} = 1.6 \text{ A}$

The reading of ammeter $A_3$ is the total current in the circuit, which is the sum of the currents in the parallel branches:

$I_3 = I_1 + I_2 = 2.4 + 1.6 = 4.0 \text{ A}$

Therefore, the readings of ammeters $A_2$ and $A_3$ are $1.6 \text{ A}$ and $4.0 \text{ A}$ respectively.