91Ó°ÊÓ

Theroot mean square (RMS) is the square root of the mean square, which is the arithmetic mean of the squares of a group of values.

$l_{rms}=\frac{1}{\sqrt{2}}$, where I is the maximum current amplitude.

Z is defined as the impedance of the circuit which is the effective resistance of an electric circuit or component to alternating current, arising from the combined effects of ohmic resistance and reactance.

$Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}$, where Z is the impedance

The equivalent Ohm’s law relation to get the amplitude voltage V in the circuit.

$$\begin{gathered} \mathit{V}\mathbf{=}\mathit{I}\mathit{Z} \\ {\mathit{V}}_{\mathbf{r}\mathbf{m}\mathbf{s}}\mathbf{=}{\mathit{I}}_{\mathbf{r}\mathbf{m}\mathbf{s}}\mathit{Z} \end{gathered}$$

Similarly, the amplitude voltage across the resistor, capacitor and inductor is found by the relation

$$\begin{gathered} \mathit{V}\mathbf{=}\mathit{I}\mathit{X} \\ {\mathit{V}}_{\mathbf{r}\mathbf{m}\mathbf{s}}\mathbf{=}{\mathit{l}}_{\mathbf{r}\mathbf{m}\mathbf{s}}\mathit{X} \end{gathered}$$

Where X is the reactance which is equal to${\mathbf{X}}_{\mathbf{L}}\mathbf{=}\mathbf{Ó\neg L}$ for inductor and${\mathbf{X}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{Ó\neg C}}$ for capacitor andRfor resistor.

$$\begin{gathered} \mathrm{Ó\neg }=200\frac{\mathrm{rad}}{\mathrm{s}} \\ L=0.400\mathrm{H} \\ \mathrm{C}=6.00\mathrm{μ¹ó} \\ \mathrm{V}=30\mathrm{V} \\ {\mathrm{X}}_{\mathrm{L}}=200×0.400 \\ =80\mathrm{Î\copyright } \\ {\mathrm{X}}_{\mathrm{c}}=\frac{1}{\left(200\right)\left(6×{10}^{−6}\right)} \\ =\frac{1}{0.0012} \\ =833.3\mathrm{Î\copyright } \\ \mathrm{Z}=\sqrt{{200}^2+(80−833.33{)}^2} \\ =779.4\mathrm{Î\copyright } \end{gathered}$$

Now,

$$\begin{gathered} \mathrm{I}=\frac{\mathrm{V}}{Z} \\ {\mathrm{I}}_{\mathrm{rms}}=\frac{\frac{30\mathrm{V}}{779.4\mathrm{Î\copyright }}}{\sqrt{2}} \\ =0.0272\mathrm{A} \end{gathered}$$

The current$l_{rms}$ in the circuit is 0.0272A.

Now plug the values for I and Z to get V

$$\begin{gathered} V_1=\left(0.0272\mathrm{A}\right)\left(200\mathrm{Î\copyright }\right) \\ =5.44\mathrm{V} \end{gathered}$$

$$\begin{gathered} {\mathrm{V}}_2=\left(0.0272\mathrm{A}\right)\left(80\mathrm{Î\copyright }\right) \\ =2.18\mathrm{V} \\ {\mathrm{V}}_3=\left(0.0272\mathrm{A}\right)\left(833.33\mathrm{Î\copyright }\right) \\ =22.7\mathrm{V} \\ {\mathrm{V}}_4={\mathrm{V}}_3−{\mathrm{V}}_2 \\ =22.7−2.18 \\ =20.5\mathrm{V} \\ {\mathrm{V}}_5=\frac{\mathrm{V}}{\sqrt{2}} \\ =\frac{30}{\sqrt{2}} \\ =21.2\mathrm{V} \end{gathered}$$

Therefore, The voltmeter readings are

a)$V_1=5.44V,V_2=2.18V,V_3=22.7V,V_4=20.5V,V_5=21.2V$

$$\begin{gathered} \mathrm{Ó\neg }=1000\frac{\mathrm{rad}}{\mathrm{s}} \\ \mathrm{L}=0.400\mathrm{H} \\ \mathrm{C}=6.00\mathrm{μ¹ó} \\ \mathrm{V}=30\mathrm{V} \\ {\mathrm{X}}_{\mathrm{L}}=1000×0.400 \\ =400\mathrm{Î\copyright } \\ {\mathrm{X}}_{\mathrm{C}}=\frac{1}{\left(1000\right)\left(6×{10}^{−6}\right)} \\ =167\mathrm{Î\copyright } \\ \mathrm{Z}=\sqrt{{200}^2+{\left(400-167\right)}^2} \\ =307\mathrm{Î\copyright } \end{gathered}$$

Now,

$$\begin{gathered} \mathrm{I}=\frac{\mathrm{V}}{Z} \\ {\mathrm{I}}_{\mathrm{ms}}=\frac{\frac{30\mathrm{V}}{307\mathrm{Î\copyright }}}{\sqrt{2}} \\ =0.0691\mathrm{A} \end{gathered}$$

The current$l_{rms}$in the circuit is 0.0691A.

Now plug the values for I and Z to get V

$$\begin{gathered} {\mathrm{V}}_1=\left(0.0691\mathrm{A}\right)\left(200\mathrm{Î\copyright }\right) \\ =13.8\mathrm{V} \\ {\mathrm{V}}_2=\left(0.0691\mathrm{A}\right)\left(400\mathrm{Î\copyright }\right) \\ =27.6\mathrm{V} \\ {\mathrm{V}}_3=\left(0.0691\mathrm{A}\right)\left(167\mathrm{Î\copyright }\right) \\ =11.5\mathrm{V} \end{gathered}$$

$$\begin{gathered} {\mathrm{V}}_4={\mathrm{V}}_3−{\mathrm{V}}_2 \\ =11.5−27.6\mathrm{V} \\ =−16.1\mathrm{V} \\ {\mathrm{V}}_5=\frac{\mathrm{V}}{\sqrt{2}} \\ =\frac{30}{\sqrt{2}} \\ =21.2\mathrm{V} \end{gathered}$$

Therefore, The voltmeter readings are

a)$V_1=5.44V,V_2=2.18V,V_3=22.7V,V_4=20.5V,V_5=21.2V$

b)$V_1=13.8V,V_2=27.6V,V_3=11.5V,V_4=-16.1V,V_5=21.2V$