$FIGURE\hspace{0.17em}CP32.68$shows voltage and current graphs for a series RLC circuit.

a. What is the resistance $R$?

b. If $L=200\mu H$, what is the resonance frequency in $Hz$?

We have been given that the voltage and current graphs for a series RLC circuit in $FIGURE\hspace{0.17em}CP32.68$.

We need to find the resistance$R$.

In RLC circuit, the total resistance represents the impedance $\left(Z\right)$in the circuit.

We can calculate the root mean square voltage $\left(V_{rms}\right)$by using impedance because the circuit draws root mean square current (localid="1650284073591" $I_{rms}$).

We know the formula that the emf (${\epsilon }_{rms}$) is given by:

${\epsilon }_{rms}=I_{rms}Z$

${\epsilon }_{rms}=I_{rms}\sqrt{R^2+{(X_L-X_C)}^2}$ (Let this equation be ($1$))

where $X_L$is the inductive reactance and $X_C$is the capacitive reactance for RLC circuit.

The phase angle ($\varphi$) between the current and the emf is given by the formula :

$\tan \left(\varphi \right)=\left(\frac{X_L-X_C}{R}\right)$

$R\tan \left(\varphi \right)=\left(X_L-X_C\right)$ (Let this equation be ($2$))

Substituting the equation $\left(2\right)$in equation $\left(1\right)$, we get:

localid="1650285214488" ${\epsilon }_{rms}=I_{rms}\sqrt{R^2+{\left(R\tan \left(\varphi \right)\right)}^2}$

Rearranging the terms, we get:

$\frac{{\epsilon }_{rms}^2}{I_{rms}^2}=\sqrt{R^2(1+{\tan }^2\varphi )}$

$R=\frac{{\epsilon }_{rms}}{I_{rms}\sqrt{(1+{\tan }^2\varphi )}}$ (Let this equation be ($3$))

Now substitute the values of ${\epsilon }_{rms},I_{rms},\varphi$in equation ($3$) from the graph given in $FIGURECP32.68$to find the value of $R$,

$R=\frac{{\epsilon }_{rms}}{I_{rms}\sqrt{(1+{\tan }^{2}60°)}}$

$R=\frac{10V}{2A\sqrt{(1+{\tan }^{2}60°)}}$

$R=2.5\Omega$

We have been given that the voltage and current graphs for a series RLC circuit in $FIGURECP32.68$.

We need to find the resonance frequency when$L=200\mu H$.

The time period in the given graph is $T=100\mu s$.

The end frequency is $f=\frac{1}{T}=\frac{1}{100\mu s}=10\times {10}^{3}Hz$.

We know that the resonance frequency occurs when $X_C=X_L$, where $X_C$is the capacitive reactance and $X_L$is the inductive reactance.

Now, find the Capacitance $\left(C\right)$by equating $X_C=X_L$:

localid="1650306547543" $X_C=X_L$

$\frac{1}{\omega C}=\omega L$, where $L$is inductance and $C$is capacitance.

Rearranging the terms, we get:

$C=\frac{1}{{\omega }^{2}L}$

The resonant angular frequency is given by :

${\omega }_o^2=\frac{1}{LC}$ (Let this equation be ($4$))

Now use the equation ($2$) obtained in Part(a) to get the resonance frequency ($f_o$) :

$R\tan \left(\varphi \right)=(X_L-X_C)$

$R\tan \left(\varphi \right)=\omega L-\frac{1}{\omega C}$

$R\tan \left(\varphi \right)=\omega L\left(1-\frac{1}{{\omega }^2\left(LC\right)}\right)$

$\frac{R\tan \left(\varphi \right)}{\omega L}=\left(1-\frac{1}{{\omega }^2\left(LC\right)}\right)$

$\frac{R\tan \left(\varphi \right)}{\omega L}=\left(1-\frac{1}{{\omega }^2}\left({\omega }_o^2\right)\right)$ (Using equation $\left(4\right)$)

${\omega }_o^2={\omega }^2\left(1-\frac{R\tan \left(\varphi \right)}{\omega L}\right)$

Substituting ${\omega }_o=2\pi f_o$and $\omega =2\pi f$, we get:

role="math" localid="1650308913749" ${(2\pi f_o)}^2={(2\pi f)}^2\left(1-\frac{R\tan \left(\varphi \right)}{2\pi fL}\right)$

Rearranging the terms, we get:

$f_o=f\sqrt{\left(1-\frac{R\tan \left(\varphi \right)}{2\pi fL}\right)}$

Substituting the values of $R,\varphi ,f,L$, we get:

$f_o=(10\times {10}^{3}Hz)\sqrt{\left(1-\frac{(2.5\Omega )\tan \left(60°\right)}{2\pi (10\times {10}^{3}Hz)(200\times {10}^{-6}H)}\right)}$

$f_o=8.1\times {10}^{3}Hz$

The resonance frequency is$f_o=8.1\times {10}^{3}Hz$when$L=200\mu H$.