From wikipedia:
"In physics, resonance is the tendency of a system to oscillate with greater amplitude at some frequencies than at others. Frequencies at which the response amplitude is a relative maximum are known as the system's resonant frequencies, or resonance frequencies. At these frequencies, even small periodic driving forces can produce large amplitude oscillations, because the system stores vibrational energy.
Resonance occurs when a system is able to store and easily transfer energy between two or more different storage modes (such as kinetic energy and potential energy in the case of a pendulum). However, there are some losses from cycle to cycle, called damping. When damping is small, the resonant frequency is approximately equal to the natural frequency of the system, which is a frequency of unforced vibrations. Some systems have multiple, distinct, resonant frequencies.
Resonance phenomena occur with all types of vibrations or waves: there is mechanical resonance, acoustic resonance, electromagnetic resonance, nuclear magnetic resonance (NMR), electron spin resonance (ESR) and resonance of quantum wave functions. Resonant systems can be used to generate vibrations of a specific frequency (e.g., musical instruments), or pick out specific frequencies from a complex vibration containing many frequencies (e.g., filters)."
Reference: http://en.wikipedia.org/wiki/Resonance
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Resonance: In an electrical circuit, the condition that exists when the inductive reactance and the capacitive reactance are of equal magnitude, causing electrical energy to oscillate between the magnetic field of the inductor and the electric field of the capacitor.
Note 1: Resonance occurs because the collapsing magnetic field of the inductor generates an electric current in its windings that charges the capacitor and the discharging capacitor provides an electric current that builds the magnetic field in the inductor, and the process is repeated.
Note 2: At resonance, the series impedance of the two elements is at a minimum and the parallel impedance is a maximum. Resonance is used for tuning and filtering, because resonance occurs at a particular frequency for given values of inductance and capacitance. Resonance can be detrimental to the operation of communications circuits by causing unwanted sustained and transient oscillations that may cause noise, signal distortion, and damage to circuit elements.
Note 3: At resonance the inductive reactance and the capacitive reactance are of equal magnitude.
Source: http://www.its.bldrdoc.gov/fs-1037/dir-031/_4576.htm
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In a resonant circuit, the inductive and capacitive reactance are equal and opposite, thus cancelling each other. This leaves the fundamental resistance of the circuit as the only impedance.
If either the inductance or capacitance is greater than the other resulting in non-resonance, the remaining uncancelled impedance adds to the overall impedance of the circuit. The key words here are "at resonance".
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When a parallel RLC circuit is at resonance, the reactive impedance is at a maximum. At that point, we have two impedances, call them R_{1} (R) and R_{2} (~Z) in parallel. The resistance of two parallel resistors is
\(R_1R_2 \over R_1+R_2\) where \(R_2 \gg R_1\)
this is \(\approx R_1\) (the circuit resistance).
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Here's a good explanation of Q:
https://en.wikipedia.org/wiki/Q_factor
Basically, Q measures how much a circuit or antenna prefers to resonate at a particular frequency. High Q means narrow bandwidth.
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The important word here is "circulating." While the input current is minimum at resonance, the current circulating between the inductor and the capacitor can be very large, hence the answer to this question is "maximum."
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The inductor acts as a short circuit at low frequencies while the capacitor acts as a short circuit at high frequencies. Therefore, input current will be high when driven at either very low or very high frequencies.
The definition of resonance in a parallel RLC (resistance (R), inductance (L) and capacitance (C)) circuit is when the impedance of the circuit is at a maximum. If Z is at a maximum, then the input current I must be at a mininum: \(I=\frac{E}{Z}\); as \(Z\) goes to infinity, I goes to zero.
Likewise, the definition for resonance in a series RLC circuit is when \(Z\) is at a minimum, \(I\) is at a maximum.
Memory tip: The question involved two "i" letters -- input and current (I). Choose the answer with two "i" letters -- minimum.
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When the circuit is in resonance, the inductive reactance and capacitive reactance are equal, so, they effectively cancel each other out with a resulting 0 degree phase angle. This in turn makes the circuit resistive with the voltage and current in phase.
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Fundamentally, regardless of circuit configuration, \(Q\) factor is defined as the ratio of reactive power, \(P_X\), over dissipative (resistive) power, \(P_R\):
\[Q = \frac{P_X}{P_R}\]
In a parallel RLC circuit, the input voltage is applied equally across all three branches:
\[V_{In} = V_R = V_L = V_C = V\]
Using the version of the power equation with voltage and impedance terms, and considering that at resonance, reactive impedance of the inductance equals that of the capacitance, solve for reactive and resistive power:
\[P_X = \frac{V_L^2}{X_L} = \frac{V_C^2}{X_C} = \frac{V^2}{X}\]
\[P_R = \frac{V_R^2}{R} = \frac{V^2}{R}\]
Finally, substituting the power equations into the \(Q\) equation and simplifying:
\[Q = \frac{V^2 \over X}{V^2 \over R} = \frac{1 \over X}{1 \over R} = \frac{R}{X}\]
Therefore, \(Q\) of a parallel RLC circuit is calculated as the ratio of resistance divided by the reactance of either the capacitance or inductance.
Test Hint: Resistance is the first word in only one of the answers for the "parallel" version of this question.
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Fundamentally, regardless of series or parallel configuration, \(Q\) factor is defined as the ratio of reactive power, \(P_X\), over dissipative (resistive) power, \(P_R\):
\[Q = {P_X \over P_R}\]
In a series network, current is equal through each component and through the entire network:
\[I_{in} = I_R = I_L = I_C = I\]
Substituting the version of the power equation with current and resistance terms into the \(Q\) equation:
\[Q = \frac{I_X^2 X}{I_R^2 R} = {I^2X \over I^2 R} = {X \over R}\]
So the Q factor of a series RLC network in resonance equals the ratio of the reactance of either the inductance or capacitance over resistance.
The \(Q\) of an RLC series circuit is defined as
\(Q = {\sqrt{L \over C} \over R}\), and using a little algebra,
\(Q = \frac{X_L}{R}\) and
\(Q = {X_C \over R}\)
so the answer "Reactance of either the inductance or capacitance divided by the resistance" is correct. (- nojiratz)
Memory tip: Q is a ratio of resistance and reactance. If circuit is series (note the 2 "s"), the denominator has 2 "s", namely resistance. If the circuit is parallel (note the 2 "a"), the denominator has 2 "a", namely reactance. Also the only answer ending with "divided by resistance" WP4AES
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\(Q\) is defined as the ratio of the center or resonant frequency \(f\) to the \(3 \text{ dB}\) (half-power) bandwidth:
\[Q = {f \over \text{BW}}\]
therefore
\[\text{BW} = \frac{f}{Q}\]
and in our case
\begin{align} \text{BW} &= {7.1 \text{ MHz} \over 150}\\ &= 0.04733 \text{ MHz} \\ &=47.333 \text{ kHz} \end{align}
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Given:
\(f_{\text{resonant}} = 3.7 \text{ MHz}\)
\(Q = 118\)
The half-power bandwidth may be calculated as the quotient of the frequency and the \(Q\):
\[BW_{\text{half-power}} \text{ (in MHz)} = \frac{f \text{ (in MHz)}}{Q}\]
So,
\[BW_{\text{half-power}} = \frac{3.7\text{ MHz}}{118} = 0.0314 \text{ MHz}\]
Converted to kHz:
\[ 0.0314 \text{ MHz} \times \frac{1000 \text{ kHz}}{1 \text{ MHz}} = 31.4 \text{ kHz} \]
Quick Clue: Interestingly enough, either way you divide, the first number is "3". Only one answer starts with a "3".
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Remember \(Q\) represents the frequency peak of the resonance. Thus deviating in any direction changes the capacitive/inductive reactance causing components comprising the resonant circuit to begin increasing their currents and voltages.
The inverse is also the case when \(Q\) is decreased.
-KE0IPR
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We want to find the resonant frequency \((f_0)\) of a series RLC circuit, given:
\begin{align} R &= 22\ \Omega\\ L &= 50\ \mu\text{H} = 5.0 \times 10^{-5}\text{ H}\\ C &= 40\text{ pF} = 4.0 \times 10^{-11}\text{ F} \end{align}
The driven resonant frequency in a series or parallel resonant circuit is \[\omega_0 = \frac{1}{\sqrt{L C}}\]
The angular frequency is \[\omega_0 = 2\pi{f_0}\]
If we combine these two equations, we get \[2\pi{f_0} = \frac{1}{\sqrt{LC}}\]
Rearranging terms: \[f_0 = \frac{1}{2 \pi \sqrt{LC}}\]
Plugging in our values: \[f_0 = \frac{1}{2 \pi \sqrt{\left(5.0 \times 10^{-5}\text{ H}\right) \left(4.0 \times 10^{-11}\text{ F}\right)}}\]
If we plug this into a calculator: \begin{align} &= 3558815\text{ Hz}\\ &= 3.56\text{ MHz} \end{align}
If you forget your calculator on the test, this is going to be take a little work, but it is doable. Let's simplify:
Multiply fractional coefficients and add exponents: \begin{align} f_0 = \frac{1}{2 \pi \sqrt{20 \times 10^{-16}}} \end{align}
Simplify: \begin{align} f_0 &= \frac{1}{2 \pi \sqrt{(2 \times 2 \times 5) \times \left(10^{-8} \times 10^{-8}\right)}}\\ &= \frac{1}{2 \pi \times 2 \times 10^{-8} \times \sqrt{5}}\\ &= \frac{10^8}{4 \pi \sqrt{5}}\\ \end{align}
Estimating that \(4\pi \approx 12.5\) and \(\sqrt{5} \approx 2.2\), we can substitute those values in: \begin{align} f_0 \approx \frac{10^8}{12.5 \times 2.2} &= \frac{10^8}{27.5}\\ &= \frac{10^7}{2.75} \end{align}
\(\frac{10}{2.75}\) is between \(3\) and \(4\), so call it \(3.5\). That gives us: \begin{align} f_0 &\approx \frac{10^7}{2.75}\\ &\approx 3.5 \times 10^6\text{ Hz} = 3.5\text{ MHz}\\ \end{align}
Of the choices, \(3.56\text{ MHz}\) is the closest.
(Let this be a reminder to not forget your calculator for the test!)
https://en.wikipedia.org/wiki/RLC_circuit#Resonance
Multiply the inductance (\(L=50\)) by the capacitance (\(C=40\)) for a product of \(2000\). Take the square root of that product to find \(44.72\) and store that to memory (or write it down). Now multiply \(\pi\) by \(2\) for a product of \(6.283\). Multiply the stored value \(44.72\) by \(6.283\) for a product of \(280.98\). Now divide \(1\) by \(280.98\) for a value of \(0.0035589\). Multiply this by \(1000\) and round up for 3.56 MHz.
For the memory-limited folks like myself, here's a poor man's hint: given the first digits of the problem, the number 3 is missing. 3.56 is your answer.
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Higher \(Q\) indicates a lower rate of energy loss relative to the stored energy of the resonator; the oscillations die out more slowly. See Wiki
The formula for quality factor Q is: \[Q = \frac{2\pi f L}{R}\] where:
If we make \(R\) lower, \(Q\) will be larger, because we're dividing by \(R\). When we divide by a smaller number, the quotient is larger.
An example of raising the \(Q\) of an inductor (coil) would be to silver plate copper wire. This would raise the \(Q\), compared to copper wire without the silver plating, because silver has a lower resistance than copper.
Easy to remember: Q is quality. Lower losses = higher quality.
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The formula for resonant frequency is:
\[f=\frac{1}{2\pi\sqrt{LC}}\]
Where:
Notice that even though R is given in the question, it's irrelevant to the calculation. The resonant frequency is dependent on capacitance and inductance, but not on resistance.
Plugging in the values from the question yields:
\begin{align} f&=\frac{1}{2\pi\sqrt{\left(50\times 10^{-6}\right)\left(10\times 10^{-12}\right)}}\\&=7.12\;\text{MHz} \end{align}
Alternate for formula-weak folks like me: given the first digits in the question, 7 is the next number in the sequence.
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