The ac generator in Fig. 31-39 supplies $\mathbf{120}\mathbf{}\mathit{V}$at $\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{Hz}$. With the switch open as in the diagram, the current leads the generator emf by ${\mathbf{20}}^{\mathbf{\circ }}$. With the switch in position 1, the current lags the generator emf by $\mathbf{20}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}$. When the switch is in position 2, the current amplitude is $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{A}$. What are (a) R, (b) L, and (c) C?

• The emf in the circuit, ${\mathrm{\epsilon }}_{\mathrm{m}}=120\mathrm{V}$
• Frequency, $\mathrm{f}=60.0\text{Hz}$
• For open switch, currently leads the emf by ${\mathrm{\varphi }}_0=-20.0°$.
• In position 1, the current lags the emf by,${\mathrm{\varphi }}_1=10.0°$

• In position 2, the current amplitude${\mathrm{I}}_2=2.0\text{0}\text{A}$.

By using the quantities given in the problem and the formula of phase constant, current, and impedance in the circuit, find the values of resistance, inductance, and capacitance.

Formulae are as follows:

• Phase constant,$\text{tan}{\mathrm{\varphi }}_0=\frac{{\mathrm{\omega }}_{\mathrm{d}}\mathrm{L}-\frac{\text{1}}{{\mathrm{\omega }}_{\mathrm{d}}\mathrm{C}}}{\mathrm{R}}=\frac{{\mathrm{Z}}_{\mathrm{LC}}}{\mathrm{R}}$.
• Current in circuit, ${\mathrm{I}}_{\mathrm{z}}=\frac{{\mathrm{\epsilon }}_{\mathrm{m}}}{{\mathrm{Z}}_{\mathrm{LC}}}$.
• The impedance of the circuit, $$\begin{gathered} {\mathrm{Z}}_{\mathrm{LC}}={\mathrm{\omega }}_{\mathrm{d}}\mathrm{L}-\frac{\text{1}}{{\mathrm{\omega }}_{\mathrm{d}}\mathrm{C}} \\ \mathrm{\omega }=\text{2}\mathrm{\pi f} \end{gathered}$$

.

When the switch is open, a series RLCcircuit involving just one capacitor near the upper right corner, the phase constant is,

$$\begin{gathered} \text{tan}{\mathrm{\varphi }}_0=\frac{{\mathrm{\omega }}_{\mathrm{d}}\mathrm{L}-\frac{\text{1}}{{\mathrm{\omega }}_{\mathrm{d}}\mathrm{C}}}{\mathrm{R}} \\ \tan \left(-20.0^{°}\right)=\frac{{\mathrm{\omega }}_{\mathrm{d}}\mathrm{L}-\frac{\text{1}}{{\mathrm{\omega }}_{\mathrm{d}}\mathrm{C}}}{\mathrm{R}} \\ \tan \left(-20.0^{°}\right)=\frac{{\mathrm{\omega }}_{\mathrm{d}}\mathrm{L}-\frac{\text{1}}{{\mathrm{\omega }}_{\mathrm{d}}\mathrm{C}}}{\mathrm{R}}······\left(1\right) \end{gathered}$$

Now, when the switch is in position, the equivalent capacitance in the circuit is. In

In this case,

$$\begin{gathered} \text{tan}{\mathrm{\varphi }}_1=\frac{{\mathrm{\omega }}_{\mathrm{d}}\mathrm{L}-\frac{\text{1}}{\text{2}{\mathrm{\omega }}_{\mathrm{d}}\mathrm{C}}}{\mathrm{R}} \\ \tan \left({10}^{°}\right)=\frac{{\mathrm{\omega }}_{\mathrm{d}}\mathrm{L}-\frac{\text{1}}{\text{2}{\mathrm{\omega }}_{\mathrm{d}}\mathrm{C}}}{\mathrm{R}}······\left(2\right) \end{gathered}$$

Finally, with the switch in position, the circuit is simply an LCcircuit with a current amplitude

$$\begin{gathered} {\mathrm{I}}_2=\frac{{\mathrm{\epsilon }}_{\mathrm{m}}}{{\mathrm{Z}}_{\mathrm{LC}}}=\frac{{\mathrm{\epsilon }}_{\mathrm{m}}}{\sqrt{{\left({\mathrm{\omega }}_{\mathrm{d}}\mathrm{L}-\frac{\text{1}}{\text{2}{\mathrm{\omega }}_{\mathrm{d}}\mathrm{C}}\right)}^2}} \\ {\mathrm{I}}_2=\frac{{\mathrm{\epsilon }}_{\mathrm{m}}}{\frac{\text{1}}{{\mathrm{\omega }}_{\mathrm{d}}\mathrm{C}}-{\mathrm{\omega }}_{\mathrm{d}}\mathrm{L}}·····\left(3\right) \end{gathered}$$

From equation 1,

$$\begin{gathered} \mathrm{R}=\frac{{\mathrm{\omega }}_{\mathrm{d}}\mathrm{L}-\frac{\text{1}}{{\mathrm{\omega }}_{\mathrm{d}}\mathrm{C}}}{-\tan \left({20}^{°}\right)} \\ =\frac{\frac{\text{1}}{{\mathrm{\omega }}_{\mathrm{d}}\mathrm{C}}-{\mathrm{\omega }}_{\mathrm{d}}\mathrm{L}}{\tan \left({20}^{°}\right)} \end{gathered}$$

Using the equation (3) in the above equation,

$$\begin{gathered} \mathrm{R}=\frac{{\mathrm{\epsilon }}_{\mathrm{m}}}{{\mathrm{I}}_2\text{tan}\left({20}^{°}\right)} \\ \mathrm{R}=\text{165}\mathrm{\Omega } \end{gathered}$$

Hence, the value of resistance is $\text{165}\mathrm{\Omega }$.

From the above three equations,

$$\begin{gathered} \mathrm{L}=\frac{{\mathrm{\epsilon }}_{\mathrm{m}}}{{\mathrm{\omega }}_{\mathrm{d}}{\mathrm{I}}_2}\left(\text{1 -}\frac{\text{2tan}{\mathrm{\varphi }}_1}{\text{tan}{\mathrm{\varphi }}_0}\right) \\ \mathrm{L}=\frac{\text{120}}{\text{2}\mathrm{\pi }\left(\text{60.0}\right)\left(\text{2.0}\right)}\left(\text{1 -}\frac{2\tan \left({10}^{°}\right)}{2\tan \left(-{20}^{°}\right)}\right) \\ \mathrm{L}=0.313\mathrm{H} \\ =313\mathrm{mH} \end{gathered}$$

Hence, the value of inductance is 313 mH.

From the above three equations,

$$\begin{gathered} \mathrm{C}=\frac{{\mathrm{I}}_2}{\text{2}{\mathrm{\omega }}_{\mathrm{d}}{\mathrm{\epsilon }}_{\mathrm{m}}\left(\text{1 -}\frac{\text{tan}{\mathrm{\varphi }}_{\text{1}}}{\text{tan}{\mathrm{\varphi }}_{\text{0}}}\right)} \\ \mathrm{C}=\frac{2.00}{\text{2}\left(2\mathrm{\pi }\right)\left(\text{60.0}\right)\left(\text{120}\right)\left(1-{\displaystyle \frac{\tan \left({10}^{°}\right)}{\tan \left({20}^{°}\right)}}\right)} \\ {\text{C=1.49×10}}^{-5}\text{F} \\ =14.9\mathrm{\mu }F \end{gathered}$$

Hence, the value of capacitance is$14.9\mathrm{\mu F}$.