In the shown circuit reading of ideal AC voltmeter V₁ and V₃ are 300 volt each then choose correct option/options if reading of ammeter is 10A:

${\mathrm{V}}_1=\mathrm{I}\sqrt{{\mathrm{X}}_{\mathrm{L}}^2+{\mathrm{R}}^2} \Rightarrow \sqrt{{{\mathrm{X}}_{\mathrm{L}}}^2+{\mathrm{R}}^2}=\frac{{\mathrm{V}}_1}{\mathrm{I}}=\frac{300}{10}=30 \mathrm{\Omega } \cdots \left(\mathrm{i}\right)$

Similarly, $\sqrt{{{\mathrm{X}}_{\mathrm{C}}}^2+{\mathrm{R}}^2}=\frac{{\mathrm{V}}_2}{\mathrm{I}}=\frac{300}{10}=30 \mathrm{\Omega } \cdots \left(\mathrm{ii}\right)$.

From (i) and (ii),  ${\mathrm{X}}_{\mathrm{L}}={\mathrm{X}}_{\mathrm{C}}$.

Hence, ${\mathrm{V}}_{\text{supply}}={\mathrm{V}}_{\mathrm{L}}+{\mathrm{V}}_{\mathrm{R}}+{\mathrm{V}}_{\mathrm{R}}+{\mathrm{V}}_{\mathrm{C}}=2{\mathrm{V}}_{\mathrm{R}} \left[\because {\mathrm{X}}_{\mathrm{L}}={\mathrm{X}}_{\mathrm{C}}, {\mathrm{V}}_{\mathrm{L}}+{\mathrm{V}}_{\mathrm{C}}=0\right]$.

Therefore, ${\mathrm{V}}_2=2{\mathrm{V}}_{\mathrm{R}}={\mathrm{V}}_{\text{supply}}=400 \mathrm{V}$.

Also, $2\mathrm{IR}={\mathrm{V}}_2 \Rightarrow \mathrm{R}=\frac{{\mathrm{V}}_2}{2\mathrm{I}}=\frac{400}{2\times 10}=20 \mathrm{\Omega }$.