Chapter 33, Problem 33.76AP

A series RLC circuit in which R = l.00 Ω, L = 1.00 mH, and C = 1.00 nF is connected to an AC source delivering 1.0 V (rms). (a) Make a precise graph of the power delivered to the circuit as a function of the frequency and (b) verify that the full width of the resonance peak at half-maximum is R/2πL.

To determine

To draw: A precise graph of the power delivered to the circuit as a function of the frequency.

Answer to Problem 33.76AP

The precise graph of the power delivered to the circuit as a function of the frequency is shown below.

Explanation of Solution

Given info: The value of resistance is $1.00\Omega$, value of inductance is $1.00\text{mH}$, value of capacitance is $1.00\text{nF}$, and source with $1.00\text{V}$.

Formula to calculate the inductive reactance of the circuit is,

$X_{\text{L}}=\omega L$

Here,

$X_{\text{L}}$ is the inductive reactance of the circuit.

$\omega$ is the angular frequency of the source.

$L$ is the inductance of the inductor.

Formula to calculate the inductive reactance of the circuit is,

$X_{\text{C}}=\frac{1}{\omega C}$

Here,

$X_{\text{C}}$ is the inductive reactance of the circuit.

$C$ is the capacitance of the capacitor.

Formula to calculate the impedance of the circuit is,

$Z=\sqrt{R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2}$

Here,

$Z$ is the impedance in the circuit.

$R$ is the resistance in the circuit.

Substitute $\omega L$ for $X_{\text{L}}$ and $\frac{1}{\omega C}$ for $X_{\text{C}}$ to find $Z$.

$Z=\sqrt{R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2}$

Formula to calculate the rms current in the circuit is,

$I_{\text{rms}}=\frac{\Delta V_{\text{rms}}}{Z}$

Here,

$I_{\text{rms}}$ is the rms current in the circuit.

$\Delta V_{\text{rms}}$ is the rms source voltage.

Write the expression for the power deliver to the circuit.

$P={\left(I_{\text{rms}}\right)}^{2}R$

Substitute $\frac{\Delta V_{\text{rms}}}{Z}$ for $I_{\text{rms}}$ in above circuit.

$P={\left(\frac{\Delta V_{\text{rms}}}{Z}\right)}^{2}R$

Substitute $\sqrt{R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2}$ for $Z$.

$\begin{array}{c}P={\left(\frac{\Delta V_{\text{rms}}}{\sqrt{R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2}}\right)}^{2}R\\ =\frac{{\left(\Delta V_{\text{rms}}\right)}^2}{\left(R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2\right)}R\end{array}$

Substitute $1.00\text{V}$ for $\Delta V_{\text{rms}}$, $1.00\text{mH}$ for $L$ and $1.00\text{nF}$ for $C$ and $1.00\Omega$ for $R$.

$\begin{array}{c}=\frac{{\left(1.00\text{V}\right)}^2\left(1.00\Omega \right)}{\left({\left(1.00\Omega \right)}^2+{\left(\omega \left(1.00\text{mH}\times \frac{{10}^{-3}\text{H}}{1\text{mH}}\right)-\frac{1}{\omega \left(1.00\text{nC}\times \frac{{10}^{-9}\text{C}}{1\text{C}}\right)}\right)}^2\right)}\\ =\frac{1.00\text{VΩ}}{\left(1.00{\text{Ω}}^2+{\left(\left({10}^{-3}\text{H}\right)\omega -\frac{1}{\left({10}^{-9}\text{C}\right)\omega }\right)}^2\right)}\\ =\frac{1.00\text{VΩ}}{1.00{\text{Ω}}^2+\left({10}^{18}{\text{C}}^{-1}\right){\left(\left({10}^{-12}\text{HC}\right)\omega -\frac{1}{\omega }\right)}^2}\end{array}$

Draw the table for the power for different values of frequency.

$\frac{\omega }{{\omega }_0}\left({10}^6\text{rad}/\text{s}\right)$	$X_{\text{L}}=\omega L\left(\Omega \right)$	$X_{\text{C}}=\frac{1}{\omega C}\left(\Omega \right)$	$Z\left(\Omega \right)$	$P=\frac{1.00{\text{V}}^2\text{Ω}}{Z^2}\left(\text{W}\right)$
0.9991	999.1	1000.9	2.06	0.23569
0.9993	999.3	1000.7	1.72	0.33768
0.9995	999.5	1000.5	1.41	0.49987
0.9997	999.7	1000.3	1.17	0.73524
0.9999	999.9	1000.1	1.02	0.96153
1.0000	1000	1000.0	1.00	1.00000
1.0001	1000.1	999.9	1.02	0.96154
1.0003	1000.3	999.7	1.17	0.73535
1.0005	1000.5	999.5	1.41	0.50012
1.0007	1000.7	999.3	1.72	0.33799
1.0009	1000.9	999.1	2.06	0.23601

Draw precise graph of the power delivered to the circuit as a function of the frequency.

Figure (1)

To determine

To verify: The full width of the resonance peak at half maximum is $R/2\pi L$.

Answer to Problem 33.76AP

Hence, the full width of the resonance peak at half maximum is $R/2\pi L$.

Explanation of Solution

Given info: The value of resistance is $1.00\Omega$, value of inductance is $1.00\text{mH}$, value of capacitance is $1.00\text{nF}$, and source with $1.00\text{V}$.

Write the expression for the term $R/2\pi L$.

$\text{k}=\frac{R}{2\pi L}$

Substitute $1.00\Omega$ for $R$ and $1.00\text{mH}$ for $L$.

$\begin{array}{l}=\frac{1.00\Omega }{2\pi \left(1.00\text{mH}\times \frac{{10}^{-3}\text{mH}}{1\text{H}}\right)}\\ =159\end{array}$ (1)

From the graph the half maximum power occurs at two values of the angular frequencies that are ${\omega }_2=1.0005\times {10}^3\text{rad}/\text{s}$ and ${\omega }_1=0.9995\times {10}^3\text{rad}/\text{s}$.

Formula to calculate the angular bandwidth is,

$\Delta \omega ={\omega }_2-{\omega }_1$

Substitute $1.0005\times {10}^3\text{rad}/\text{s}$ for ${\omega }_2$ and $0.9995\times {10}^3\text{rad}/\text{s}$ for ${\omega }_1$ to find $\Delta \omega$

$\begin{array}{c}\Delta \omega =1.0005\times {10}^3\text{rad}/\text{s}-0.9995\times {10}^3\text{rad}/\text{s}\\ =1.00\times {10}^3\text{rad}/\text{s}\end{array}$

Write the expression for the frequency bandwidth.

$\Delta \omega =2\pi \Delta f$

Rearrange the equation for $\Delta f$.

$\Delta f=\frac{\Delta \omega }{2\pi }$

Substitute $1.00\times {10}^3\text{rad}/\text{s}$ for $\Delta \omega$ to find $\Delta f$.

$\begin{array}{c}\Delta f=\frac{1.00\times {10}^3\text{rad}/\text{s}}{2\pi }\\ =159.15\end{array}$ (2)

From equation (1) and equation (2), the RHS values are same that verify the full width of the resonance peak at the half power maximum is $R/2\pi L$.

Conclusion:

Therefore, the full width of the resonance peak at half maximum is $R/2\pi L$.