In the figure shown, hot wire voltmeter and hot wire ammeter are ideal. The reading of voltmeter is $5.0\times {10}^{\mathrm{P}}$. What is the value of P?

In the circuit,

$$\begin{gathered} \mathrm{Z}=\sqrt{{\mathrm{R}}^2+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^2}=\sqrt{5^2+{\left(7-2\right)}^2}=5\sqrt{2} \mathrm{\Omega } \\ {\mathrm{I}}_{\mathrm{rms}}=\frac{{\mathrm{V}}_{\mathrm{rms}}}{\mathrm{Z}}=\frac{\left(10/\sqrt{2}\right)}{\left(5\sqrt{2}\right)}=1 \mathrm{A} \\ {\mathrm{V}}_{\mathrm{L}}={\mathrm{IX}}_{\mathrm{L}}=1\times 7=7 \mathrm{V} \\ {\mathrm{V}}_{\mathrm{C}}={\mathrm{IX}}_{\mathrm{C}}=1\times 2=2 \mathrm{V} \end{gathered}$$

The phase difference between ${\mathrm{V}}_{\mathrm{L}}$ and ${\mathrm{V}}_{\mathrm{C}}$ is $\left[\frac{\mathrm{\pi }}{2}-\left(-\frac{\mathrm{\pi }}{2}\right)\right]=\mathrm{\pi }$.

$\therefore$ Reading of voltmeter $={\mathrm{V}}_{\mathrm{L}}-{\mathrm{V}}_{\mathrm{C}}=7-2=5 \mathrm{V}$.

Therefore, $\mathrm{P}=0$.