# Examples of Series Resonance Circuits   Whatsapp  Example 1:
a. For the series resonant circuit of Fig. 1, find $I$, $V_R$, $V_L$, and $V_C$ at resonance.
b. What is the $Q_s$ of the circuit?
c. If the resonant frequency is $5000 Hz$, find the bandwidth.
d. What is the power dissipated in the circuit at the half-power frequencies? Fig. 1: Example 1.
Solution:
a.
$$Z_{Ts} = R = 2Ω$$
$$I = {E \over Z_{Ts}} = {10V \angle 0^\circ \over 10 Ω \angle 0^\circ} = 5 A \angle 0^\circ$$
$$V_R = E = 5 V \angle 0^\circ$$
$$V_L = (I \angle 0^\circ)(X_L \angle 90^\circ)= 50 V \angle 90^\circ$$
$$V_C = (I \angle 0^\circ)(X_C \angle -90^\circ)= 50 V \angle -90^\circ$$
b.
$$Q_s = {X_L \over R} = { 10 Ω \over 2Ω} = 5$$
c.
$$BW = f_2 - f_1 ={ f_s \over Q_s} = 1000 Hz$$
d.
$$\begin{split} P_{HPF} &= {1 \over 2} P_{max} = {1 \over 2}I_{max}^2R \\ &={1 \over 2}(5A)^2(2 Ω) = 25 W \end{split}$$
Example 2: The bandwidth of a series resonant circuit is $400 Hz$.
a. If the resonant frequency is $4000 Hz$, what is the value of $Qs$?
b. If $R = 10 Ω$, what is the value of $X_L$ at resonance?
c. Find the inductance $L$ and capacitance $C$ of the circuit.
Solution:
a.
$$Q_s = {f_s \over BW} = {4000Hz \over 400Hz}=10$$
b.
$$X_L = Q_sR = (10)(10Ω)=100 Ω$$
c.
$$L = {X_L \over 2 \pi f_s} = { 100 Ω \over 2 \pi (4000Hz)} = 3.98 mH$$
$$C = {1 \over 2 \pi f_s X_C} = {1 \over 2 \pi (4000Hz)(100)} = 0.398 \mu H$$
Example 3: A series $RLC$ circuit is designed to resonant at $w_s=10^5 rad/s$, have a bandwidth of $0.15f_s$, and draw $16 W$ from a $120V$ source at resonance.
a. Determine the value of $R$.
b. Find the bandwidth in hertz.
c. Find the nameplate values of $L$ and $C$.
d. Determine the $Qs$ of the circuit.
e. Determine the fractional bandwidth.
Solution:
a.
$$P = {E^2 \over R}$$
$$R = {E^2 \over P} = { 120^2 \over 16} = 900Ω$$
b.
$$f_s = {w_s \over 2 \pi} = { 10^5 \over 2\pi} = 15915.5$$
$$BW = 0.15fs = 0.15(15915.5) = 2387.3Hz$$
c.
$$Q_s = {X_L \over R} \, and \, BW = {fs \over Qs}$$
$$Q_s = {fs \over BW}$$
$${X_L \over R} = {BW \over fs}$$
$$X_L = {fs\,R \over BW}$$
$$2 \pi fs L = {fs\,R \over BW}$$
$$L = {R \over 2 \pi BW} = { 900Ω \over 2 \pi (2387.3)} = 60mH$$
$$fs = {1 \over 2 \pi \sqrt{LC}}$$
and
$$C = { 1 \over 4 \pi^2 f_s^2 L} = 1.67nF$$
d.
$$Q_s = { X_L \over R} = { 2 \pi fs L \over R} = 6.67$$
e.
$${BW \over fs} = { 1 \over Qs} = { 1 \over 6.67} = 0.15$$

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