91Ó°ÊÓ

We have been given that the voltage and current graphs for a series RLC circuit in $FIGURE CP32.68$.

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 (${\mathrm{Î\mu }}_{rms}$) is given by:

${\mathrm{Î\mu }}_{rms}=I_{rms}Z$

${\mathrm{Î\mu }}_{rms}=I_{rms}\sqrt{R^2+{(X_L-X_C)}^2}$ (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 ($\mathrm{Ï•}$) between the current and the emf is given by the formula :

$\tan \left(\mathrm{Ï•}\right)=\left(\frac{X_L-X_C}{R}\right)$

$R\tan \left(\mathrm{Ï•}\right)=\left(X_L-X_C\right)$ (Let this equation be ($2$))

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

localid="1650285214488" ${\mathrm{Î\mu }}_{rms}=I_{rms}\sqrt{R^2+{\left(R\tan \left(\mathrm{Ï•}\right)\right)}^2}$

Rearranging the terms, we get:

$\frac{{\mathrm{Î\mu }}_{rms}^2}{I_{rms}^2}=\sqrt{R^2(1+{\tan }^2\mathrm{Ï•})}$

$R=\frac{{\mathrm{Î\mu }}_{rms}}{I_{rms}\sqrt{(1+{\tan }^2\mathrm{Ï•})}}$ (Let this equation be ($3$))

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

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

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

$R=2.5\mathrm{Î\copyright }$

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$L=200\mathrm{μ}H$.

The time period in the given graph is $T=100\mathrm{μ}s$.

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

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" $X_C=X_L$

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

Rearranging the terms, we get:

$C=\frac{1}{{\mathrm{Ó\neg }}^{2}L}$

The resonant angular frequency is given by :

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

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

$R\tan \left(\mathrm{Ï•}\right)=(X_L-X_C)$

$R\tan \left(\mathrm{Ï•}\right)=\mathrm{Ó\neg }L-\frac{1}{\mathrm{Ó\neg }C}$

$R\tan \left(\mathrm{Ï•}\right)=\mathrm{Ó\neg }L\left(1-\frac{1}{{\mathrm{Ó\neg }}^2\left(LC\right)}\right)$

$\frac{R\tan \left(\mathrm{Ï•}\right)}{\mathrm{Ó\neg }L}=\left(1-\frac{1}{{\mathrm{Ó\neg }}^2\left(LC\right)}\right)$

$\frac{R\tan \left(\mathrm{Ï•}\right)}{\mathrm{Ó\neg }L}=\left(1-\frac{1}{{\mathrm{Ó\neg }}^2}\left({\mathrm{Ó\neg }}_o^2\right)\right)$ (Using equation $\left(4\right)$)

${\mathrm{Ó\neg }}_o^2={\mathrm{Ó\neg }}^2\left(1-\frac{R\tan \left(\mathrm{Ï•}\right)}{\mathrm{Ó\neg }L}\right)$

Substituting ${\mathrm{Ó\neg }}_o=2\mathrm{Ï€}{f}_o$and $\mathrm{Ó\neg }=2\mathrm{Ï€}f$, we get:

role="math" localid="1650308913749" ${(2\mathrm{Ï€}{f}_o)}^2={(2\mathrm{Ï€}f)}^2\left(1-\frac{R\tan \left(\mathrm{Ï•}\right)}{2\mathrm{Ï€}fL}\right)$

Rearranging the terms, we get:

$f_o=f\sqrt{\left(1-\frac{R\tan \left(\mathrm{Ï•}\right)}{2\mathrm{Ï€}fL}\right)}$

Substituting the values of $R,\mathrm{Ï•},f,L$, we get:

$f_o=(10×{10}^{3}Hz)\sqrt{\left(1-\frac{(2.5\mathrm{Î\copyright })\tan \left(60°\right)}{2\mathrm{Ï€}(10×{10}^{3}Hz)(200×{10}^{-6}H)}\right)}$

$f_o=8.1×{10}^{3}Hz$

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