For an ideal voltmeter, resistance is infinite $\left(R_V=\infty \right)$, so it will not draw any amount of current from the circuit and for ideal ammeter resistance is zero.

Then for the first figure $R_1=10 \mathrm{\Omega }, R_V=\infty$ and both are in a parallel arrangement, so

$$\begin{gathered} \frac{1}{R_{\mathrm{eq}}}=\frac{1}{R_1}+\frac{{\displaystyle 1}}{{\displaystyle R_V}} \\ \Rightarrow \frac{{\displaystyle 1}}{{\displaystyle R_{\mathrm{eq}}}}=\frac{{\displaystyle 1}}{{\displaystyle 10}}+\frac{{\displaystyle 1}}{{\displaystyle \infty }} \\ \Rightarrow R_{\mathrm{eq}} = 10 \mathrm{\Omega } \end{gathered}$$

and current in the first circuit is, $i_1=\frac{V_1}{R_{\mathrm{eq}}}=\frac{10}{10}= 1 \mathrm{A}$

So, for the first circuit, ammeter and voltmeter readings are $V_1=10 V, i_1=1 \mathrm{A}$, respectively.

For the second figure voltmeter and resistant of $10 \mathrm{\Omega }$ are in series and also connected parallel to another resistant of $10 \mathrm{\Omega }$. So, equivalent resistant is 

$$\begin{gathered} \frac{1}{{R^{\text{'}}}_{\mathrm{eq}}}=\frac{1}{R_1}+\frac{{\displaystyle 1}}{{\displaystyle R_2+R_V}} \\ \Rightarrow \frac{{\displaystyle 1}}{{\displaystyle {R^{\text{'}}}_{\mathrm{eq}}}}=\frac{{\displaystyle 1}}{{\displaystyle 10}}+\frac{{\displaystyle 1}}{{\displaystyle 10+\infty }}=\frac{{\displaystyle 1}}{{\displaystyle 10}}+0=\frac{{\displaystyle 1}}{{\displaystyle 10}} \\ \Rightarrow {R^{\text{'}}}_{\mathrm{eq}}=10 \mathrm{\Omega } \end{gathered}$$

$\therefore$ Current in the circuit is, $i_2=\frac{V_2}{R_{\mathrm{eq}}^{\text{'}}}=\frac{10}{10}=1 \mathrm{A}$, 

$\therefore$ Ammeter and voltmeter readings for the second circuit is $V_2=10 V, i_2=1 \mathrm{A}$

$\therefore$ For both the circuit is 
$$\begin{gathered} V_1=V_2=10 \mathrm{V}, \\ {i}_1=i_2=1 \mathrm{A} \end{gathered}$$