91Ó°ÊÓ

1. Amplitude of emf,${\mathrm{Î\mu }}_m=12.0\text{V}$
2. Driving frequency,$f_d=60.0\text{Hz}$
3. Phase angle between current and emf when both switches are open,${\mathrm{ϕ}}_1=30.9°$
4. Phase angle between current and emf when is closed and is open,${\mathrm{ϕ}}_2=15.0°$
5. Amplitude of current when both the switches are closed,$I=447\text{mA}$

We can find the values of resistance, capacitance, and inductance by considering the effective reactance and by using the formulae for phase angle and capacitive and inductive reactance.

The capacitive reactance of the capacitor,

${\mathit{X}}_{\mathit{C}}\mathbf{=}\frac{\mathbf{1}}{{\mathit{Ó\neg }}_{\mathit{d}}\mathit{C}}$ …(1)

The inductive reactance of the inductor,

${\mathit{X}}_{\mathit{L}}\mathbf{=}{\mathit{Ó\neg }}_{\mathit{d}}\mathit{L}$ …(2)

The current equation using Ohm’s law,

$\mathit{I}\mathbf{=}\frac{{\mathit{Î\mu }}_{\mathit{m}}}{\mathit{Z}}$ …(3)

Phase angle of the RLC circuit,

$\mathbf{}\mathit{t}\mathit{a}\mathit{n}\mathit{f}\mathbf{=}\left(\frac{{\mathit{X}}_{\mathit{L}}\mathbf{-}{\mathit{X}}_{\mathit{c}}}{\mathit{R}}\right)\mathbf{=}\frac{{\mathit{X}}_{\mathit{n}\mathit{e}\mathit{t}}}{\mathit{R}}$ …(4)

Here, R is the resistance of the resistor,Cis the capacitance of the capacitor, L is the inductance of the inductor and${\mathit{Ó\neg }}_{\mathit{d}}$is the driving angular frequency.

Whenthe switches$S_1$and$S_2$are closed,the current will not flow through R. This effectively removes the resistance from the circuit. Impedance will be equal to the net reactance.

Thus, the value of the net reactance is given using equation (3) as follows:

$$\begin{gathered} X_{net}=\frac{{\mathrm{Î\mu }}_m}{I} \\ =\frac{12.0\text{V}}{447×{10}^{-3}\text{A}} \\ =26.85\text{Ω} \end{gathered}$$

When$S_1$is closed and$S_2$is open, the resistorR is the part of circuit, so the reactance is the same as$X_{net}$.

So, using equation (4), we can get the resistance value of the circuit as follows:

$$\begin{gathered} R=\left(\frac{X_{net}}{tan{\mathrm{ϕ}}_2}\right) \\ =\frac{26.8\text{Ω}}{\tan 15.0°} \\ =100\text{Ω} \end{gathered}$$

Hence, the value of the resistance is $100\text{Ω}$.

Now consider the circuitwhen both the switches are open, and let$X_{net}^{\text{'}}$be the reactance offered by the circuit and${\mathrm{Ï•}}_1$be the phase angle.Then we have the net reactance using equation (4) as follows:

$$\begin{gathered} X_{net}\text{'}=100\text{Ω}×\tan 30.9° \\ =-59.96\text{Ω} \end{gathered}$$

Now let’s find the effect of closing$S_1$on the reactance.

We have the capacitive reactance as follows:

$$\begin{gathered} X_C=X_{net}-X_{net}\text{'} \\ =26.85\text{Ω}-\left(-59.96\right)\text{Ω} \\ =86.81\text{Ω} \end{gathered}$$

This is nothing but the capacitive reactance.

Thus, the capacitance value is given using equation (1) as follows:

$$\begin{gathered} C=\frac{1}{{\mathrm{Ó\neg }}_d{X}_c} \\ =\frac{1}{2\mathrm{Ï€}×60\text{Hz}×86.81\text{Ω}} \\ =30.6×{10}^{-6}\text{F} \\ =30.6\text{μ¹ó} \end{gathered}$$

Hence, the value of the capacitance is $30.6\text{μ¹ó}$.

We have$x_{net}=X_L-X_C$

Thus, the value of the inductive reactance is given by:

$$\begin{gathered} X_L=26.85\text{Ω}+86.81\text{Ω} \\ =113.66\text{Ω} \end{gathered}$$

Now, the value of the inductance is given using equation (2) as follows:

$$\begin{gathered} L=\frac{X_L}{2\mathrm{π}{f}_d} \\ =\frac{113.66\text{Ω}}{2\mathrm{π}×60\text{Hz}} \\ =0.301\text{H} \\ =301\text{mH} \end{gathered}$$

Hence, the inductance value is $301\text{mH}$.