91影视

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

• 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,$\text{tan}{\mathrm{蠒}}_0=\frac{{\mathrm{蝇}}_{\mathrm{d}}\mathrm{L}-\frac{\text{1}}{{\mathrm{蝇}}_{\mathrm{d}}\mathrm{C}}}{\mathrm{R}}=\frac{{\mathrm{Z}}_{\mathrm{LC}}}{\mathrm{R}}$.
• Current in circuit, ${\mathrm{I}}_{\mathrm{z}}=\frac{{\mathrm{蔚}}_{\mathrm{m}}}{{\mathrm{Z}}_{\mathrm{LC}}}$.
• The impedance of the circuit, $$\begin{gathered} {\mathrm{Z}}_{\mathrm{LC}}={\mathrm{蝇}}_{\mathrm{d}}\mathrm{L}-\frac{\text{1}}{{\mathrm{蝇}}_{\mathrm{d}}\mathrm{C}} \\ \mathrm{蝇}=\text{2}\mathrm{蟺蹿} \end{gathered}$$

.

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

$$\begin{gathered} \text{tan}{\mathrm{蠒}}_0=\frac{{\mathrm{蝇}}_{\mathrm{d}}\mathrm{L}-\frac{\text{1}}{{\mathrm{蝇}}_{\mathrm{d}}\mathrm{C}}}{\mathrm{R}} \\ \tan \left(-20.0^{掳}\right)=\frac{{\mathrm{蝇}}_{\mathrm{d}}\mathrm{L}-\frac{\text{1}}{{\mathrm{蝇}}_{\mathrm{d}}\mathrm{C}}}{\mathrm{R}} \\ \tan \left(-20.0^{掳}\right)=\frac{{\mathrm{蝇}}_{\mathrm{d}}\mathrm{L}-\frac{\text{1}}{{\mathrm{蝇}}_{\mathrm{d}}\mathrm{C}}}{\mathrm{R}}路路路路路路\left(1\right) \end{gathered}$$

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

In this case,

$$\begin{gathered} \text{tan}{\mathrm{蠒}}_1=\frac{{\mathrm{蝇}}_{\mathrm{d}}\mathrm{L}-\frac{\text{1}}{\text{2}{\mathrm{蝇}}_{\mathrm{d}}\mathrm{C}}}{\mathrm{R}} \\ \tan \left({10}^{掳}\right)=\frac{{\mathrm{蝇}}_{\mathrm{d}}\mathrm{L}-\frac{\text{1}}{\text{2}{\mathrm{蝇}}_{\mathrm{d}}\mathrm{C}}}{\mathrm{R}}路路路路路路\left(2\right) \end{gathered}$$

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

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

From equation 1,

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

Using the equation (3) in the above equation,

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

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

From the above three equations,

$$\begin{gathered} \mathrm{L}=\frac{{\mathrm{蔚}}_{\mathrm{m}}}{{\mathrm{蝇}}_{\mathrm{d}}{\mathrm{I}}_2}\left(\text{1 -}\frac{\text{2tan}{\mathrm{蠒}}_1}{\text{tan}{\mathrm{蠒}}_0}\right) \\ \mathrm{L}=\frac{\text{120}}{\text{2}\mathrm{蟺}\left(\text{60.0}\right)\left(\text{2.0}\right)}\left(\text{1 -}\frac{2\tan \left({10}^{掳}\right)}{2\tan \left(-{20}^{掳}\right)}\right) \\ \mathrm{L}=0.313\mathrm{H} \\ =313\mathrm{mH} \end{gathered}$$

Hence, the value of inductance is 313 mH.

From the above three equations,

$$\begin{gathered} \mathrm{C}=\frac{{\mathrm{I}}_2}{\text{2}{\mathrm{蝇}}_{\mathrm{d}}{\mathrm{蔚}}_{\mathrm{m}}\left(\text{1 -}\frac{\text{tan}{\mathrm{蠒}}_{\text{1}}}{\text{tan}{\mathrm{蠒}}_{\text{0}}}\right)} \\ \mathrm{C}=\frac{2.00}{\text{2}\left(2\mathrm{蟺}\right)\left(\text{60.0}\right)\left(\text{120}\right)\left(1-{\displaystyle \frac{\tan \left({10}^{掳}\right)}{\tan \left({20}^{掳}\right)}}\right)} \\ {\text{颁=1.49脳10}}^{-5}\text{F} \\ =14.9\mathrm{渭}F \end{gathered}$$

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