In the given diagram, when the key is closed, what will be the readings of the ammeter and voltmeter?

Here, resistances are connected in series. 

So, the total resistance (R) will be, 
$\mathrm{R} =\mathrm{ }{\mathrm{R}}_1\mathrm{ }+\mathrm{ }{\mathrm{R}}_2\mathrm{ }+\mathrm{ }{\mathrm{R}}_3+...$
Where R₁ = 5_, R₂ = 8_, and R₃ = 12 are individual resistance of the resistors. 

In this case,
$\Rightarrow \mathrm{R} =\mathrm{ }5\mathrm{ }+\mathrm{ }8\mathrm{ }+\mathrm{ }12$
$\Rightarrow \mathrm{R}=\mathrm{ }25\mathrm{\Omega }$
By Ohm’s Law,
$\Rightarrow \mathrm{V} = \mathrm{IR}$, where V is the potential difference, I is the current flowing across the resistance and R is the value of the resistance.

$\Rightarrow \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$
$\Rightarrow \mathrm{I}=\frac{6}{25}$
$\Rightarrow \mathrm{I}=0.24 \mathrm{A}$
Hence, the reading in the ammeter will be 0.24A.
The voltage across a resistor of $12 \mathrm{\Omega }$ will be 
    $\mathrm{V} = \mathrm{IR}$
$\Rightarrow \mathrm{V}=0.24\times 12\mathrm{ }$
$\Rightarrow \mathrm{V}=\mathrm{ }2.88V$.