In the circuits shown below, the readings of the voltmeters and the...

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 ${R}_{1} = 10Ω, {R}_{V} = \infty$ and both are in a parallel arrangement, so

$\dfrac{1}{{R}_{eq}} = \dfrac{1}{{R}_{1}} + \dfrac{1}{{R}_{V}} ⇒ \dfrac{1}{{R}_{eq}} = \dfrac{1}{10} + \dfrac{1}{\infty} ⇒ {R}_{eq} = 10Ω$

and current in the first circuit is, ${i}_{1} = \dfrac{{V}_{1}}{{R}_{eq}} = \dfrac{10}{10} = 1A$

So, for the first circuit, ammeter and voltmeter readings are ${V}_{1} = 10V, {i}_{1} = 1A$, respectively.

For the second figure voltmeter and resistant of $10Ω$ are in series and also connected parallel to another resistant of $10Ω$. So, equivalent resistant is

$\dfrac{1}{{R}^{'}eq} = \dfrac{1}{{R}_{1}} + \dfrac{1}{{R}_{2} + {R}_{V}} ⇒ \dfrac{1}{{R}^{'}eq} = \dfrac{1}{10} + \dfrac{1}{10 + \infty} = \dfrac{1}{10} + 0 = \dfrac{1}{10} ⇒ {R}^{'}eq = 10Ω$

$∴$ Current in the circuit is, ${i}_{2} = {V}_{2}{R}_{eq}^{'} = \dfrac{10}{10} = 1A$,

$∴$ Ammeter and voltmeter readings for the second circuit is ${V}_{2} = 10V, {i}_{2} = 1A$

$∴$ For both the circuit is
${V}_{1} = {V}_{2} = 10V, {i}_{1} = {i}_{2} = 1A$