Encyclopedia of Electrical Engineering
Examples of Series Resonance Circuits
Example 1:
a. For the series resonant circuit of Fig. 1, find I, VR, VL, and VC at resonance.
b. What is the Qs 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?
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. $$P_{HPF} = {1 \over 2} P_{max} = {1 \over 2}I_{max}^2R \\ {1 \over 2}(5A)^2(2 Ω) = 25 W$
a. For the series resonant circuit of Fig. 1, find I, VR, VL, and VC at resonance.
b. What is the Qs of the circuit?
c. If the resonant frequency is 5000 Hz, find the bandwidth.
d. What is the power dissipated in the circuit at the half-power frequencies?
Fig. 1: Example 1.
a. $ Z_{Ts} = R = 2Ω$
$ I = {E \over Z_{Ts}} = {10V \angle 0^\circ \over 10 Ω \angle 0^\circ} = 5 A \angle 0^\circ$
$V_R = E = 5 V \angle 0^\circ$
$V_L = (I \angle 0^\circ)(X_L \angle 90^\circ)= 50 V \angle 90^\circ$
$V_C = (I \angle 0^\circ)(X_C \angle -90^\circ)= 50 V \angle -90^\circ$
b. $Q_s = {X_L \over R} = { 10 Ω \over 2Ω} = 5$
c. $BW = f_2 - f_1 ={ f_s \over Q_s} = 1000 Hz $
d. $$P_{HPF} = {1 \over 2} P_{max} = {1 \over 2}I_{max}^2R \\ {1 \over 2}(5A)^2(2 Ω) = 25 W$
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. $$L = {X_L \over 2 \pi f_s} = { 100 Ω \over 2 \pi (4000Hz)} = 3.98 mH$$ $$C = {1 \over 2 \pi f_s X_C} = {1 \over 2 \pi (4000Hz)(100)} = 0.398 \mu H$$
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 1:
A series R-L-C 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 $120-V$ 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. $${BW \over fs} = { 1 \over Qs} = { 1 \over 6.67} = 0.15$$
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. $${BW \over fs} = { 1 \over Qs} = { 1 \over 6.67} = 0.15$$