Resonance
Resonance
Learning Objectives:
- Describe series and parallel resonances.
- Illustrate conditions for series and parallel resonances in terms of impedances and frequency.
- Electrical resonance occurs in an electric circuit at a particular resonant frequency when the impedances or admittances of the circuit elements cancel each other.
- In series RLC circuit, resonance occurs when the inductive and capacitive reactanecs are equal in magnitude but cancel each other beause they are 180$^\circ$ apart in phase. \begin{align*} X_L &= X_C\\ 2 \pi f L &= \dfrac{1}{2 \pi fC}\\ f^2 &= \dfrac{1}{4 \pi^2 LC}\\ f &= \sqrt{\dfrac{1}{4 \pi^2 LC}}\\ f &= \dfrac{1}{2 \pi \sqrt{LC}}\\ \omega &= \dfrac{1}{\sqrt{LC}} \end{align*}
- Quality factor, Q is a quantitative measure of sharpness of the peak, which relates to the maximum or peak energy stored in the circuit (the reactance) to the energy dissipated (the resistance) during each cycle of oscillation. In other words, it is the ratio of resonant frequency to bandwidth.
- Higher the quality factor, the smaller is the bandwidth. \[Q = \dfrac{f_r}{BW}\]
- In parallel resonance, the total admittance is given by, \begin{align*} Y_{total} &= Y_1 + Y_2 + Y_3\\ &= \dfrac{1}{R} + \dfrac{1}{j \omega L} + \dfrac{1}{\dfrac{-j}{\omega C}}\\ &= \dfrac{1}{R} + \dfrac{-j}{\omega L} + j\omega C\\ &= \dfrac{1}{R} + j(\omega C - \dfrac{1}{\omega L}) \end{align*} Resonance occurs when, \[\omega C = \dfrac{1}{\omega L}\]
Solved Example: 9961-01
In a RLC circuit Inductance is 20 mH and capacitance is 200 micro Farad. Find the resonance frequency of the circuit.
A. 1000 rad/sec
B. 250 rad/sec
C. 500 rad/sec
D. 50 rad/sec
\begin{align*} \omega &= \dfrac{1}{\sqrt{LC}}\\ &= \dfrac{1}{\sqrt{20 \times 10^{-3} \times 200 \times 10^{-6}}}\\ &= 500\ \mathrm{rad/s} \end{align*}
Correct Answer: C
Solved Example: 9961-02
For a series resonant circuit, what happens to the Q-factor when the capacitance of the circuit is increased three times and the frequency is slashed by four times?
A. It remains unchanged
B. It decreases by half
C. It doubles
D. It increases to 1.33 times
Correct Answer: D
Solved Example: 9961-03
A series RLC circuit resonates at 1.5 kHz and consumes 50 W from a 50 V AC source operating at the resonant frequency. If the bandwidth is 0.75 kHz, then what are the values of the circuit elements R and L?
A. 25 $\Omega$ and 5.31 mH
B. 50 $\Omega$ and 10.6 mH
C. 50 $\Omega$ and 66.6 mH
D. 2.5 $\Omega$ and 1.06 mH
At resonance, the circuit is purely resistive and Z = R. \begin{align*} P &= \dfrac{V^2}{R}\\ R &= \dfrac{V^2}{P}\\ &= \dfrac{50^2}{50}\\ &= 50 \Omega \end{align*} For series RLC: \begin{align*} BW &= \dfrac{R}{2 \pi L}\\ 0.75 \times 10^{3} &= \dfrac{50}{2 \pi L}\\ L &= \dfrac{50}{0.75 \times 10^{3} \times 2 \pi }\\ L &= 10.6 \times 10^{-3}\ H \end{align*}
Correct Answer: B
Solved Example: 9961-04
A coil of resistance 20 $\Omega$ and inductance 10 mH is in series with a capacitance and is supplied with a constant voltage, variable frequency source. The maximum current is 2 A at 1000 Hz. The Q-factor of the circuit is _______.
A. 31.8
B. 3.14
C. 314
D. 31.4
Correct Answer: B
Solved Example: 9961-05
A series RLC circuit has $\omega$= $10^5$, Q = 50, R = 400 $\Omega$. The value of C is:
A. 250 pF
B. 1000 pF
C. 500 pF
D. 1.25 pF
\begin{align*} Q &= \dfrac{1}{\omega_0 CR}\\ 50 &= \dfrac{1}{10^{5} \times C \times 400}\\ C &= \dfrac{1}{10^{5} \times 50 \times 400}\\ C &= 0.5\ nF = 500\ pF \end{align*}
Correct Answer: C