Question

Five infinite-impedance voltmeters, calibrated to read rms values, are connected as shown in Figure. Let R = 200 Ω, L = 0.400 H, C = 6.00 mF, and V = 30.0 V. What is the reading of each voltmeter if (a) $\mathit{\omega }$= 200 rad/s and (b) $\mathit{\omega }$= 1000 rad/s?

Short answer

The voltmeter readings are

a) ${\mathrm{V}}_1=5.44\mathrm{V},{\mathrm{V}}_2=2.18\mathrm{V},{\mathrm{V}}_3=22.7\mathrm{V},{\mathrm{V}}_4=20.5\mathrm{V},{\mathrm{V}}_5=21.2\mathrm{V}$

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

Step-by-step solution

Step-1: Formulas used  

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${\mathbf{X}}_{\mathbf{L}}\mathbf{=}\mathbf{\omega L}$ for inductor and${\mathbf{X}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\omega C}}$ for capacitor andRfor resistor.

Step-2: Calculations for Voltages Case 1

$$\begin{gathered} \omega =200\frac{\mathrm{rad}}{\mathrm{s}} \\ L=0.400\mathrm{H} \\ \mathrm{C}=6.00\mathrm{\mu F} \\ \mathrm{V}=30\mathrm{V} \\ {\mathrm{X}}_{\mathrm{L}}=200\times 0.400 \\ =80\mathrm{\Omega } \\ {\mathrm{X}}_{\mathrm{c}}=\frac{1}{\left(200\right)\left(6\times {10}^{-6}\right)} \\ =\frac{1}{0.0012} \\ =833.3\mathrm{\Omega } \\ \mathrm{Z}=\sqrt{{200}^2+(80-833.33{)}^2} \\ =779.4\mathrm{\Omega } \end{gathered}$$

Now,

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

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

Now plug the values for I and Z to get V

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

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

Therefore, The voltmeter readings are

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

Step-3: Calculations for Voltages Case 2

$$\begin{gathered} \omega =1000\frac{\mathrm{rad}}{\mathrm{s}} \\ \mathrm{L}=0.400\mathrm{H} \\ \mathrm{C}=6.00\mathrm{\mu F} \\ \mathrm{V}=30\mathrm{V} \\ {\mathrm{X}}_{\mathrm{L}}=1000\times 0.400 \\ =400\mathrm{\Omega } \\ {\mathrm{X}}_{\mathrm{C}}=\frac{1}{\left(1000\right)\left(6\times {10}^{-6}\right)} \\ =167\mathrm{\Omega } \\ \mathrm{Z}=\sqrt{{200}^2+{\left(400-167\right)}^2} \\ =307\mathrm{\Omega } \end{gathered}$$

Now,

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

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

Now plug the values for I and Z to get V

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

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

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