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

Memory Aid: Resonance causes a value to be higher than the sum of its parts.

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Tags: arrl chapter 4 arrl module 4d

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}

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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

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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.

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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.

Another way to look at it is this is an amateur radio test. Only 3.56 is an authorized ham radio frequency.

Another HINT: This is twice the close contender of 1.78 & and is the correct answer 356 days a year. I know, but, now you will remember for sure.

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Tags: arrl chapter 4 arrl module 4d

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".

If you love formulas, here is the formula for the impedance of a circuit: \(Z=\sqrt{R^2+(X_L-X_C)^2}\) Thus, when the inductive and capacitive reactance cancel each other out, the formula simplifies to \(Z=\sqrt{R^2}\), or \(Z=R\). Thus, the impedance equals the resistance.

Memory tip: circuit at resonance is similar to equal to circuit in the correct answer

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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 R₁ (R) and R₂ (~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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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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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. Resonant circuits are often used as impedance-matching circuits. Because BW = f/Q, increasing the Q of a resonant circuit used for this application has the effect of decreasing the range of frequencies, or bandwidth, over which it can match the impedance between two circuits or between a transmitter and an antenna.

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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."

Simple memory trick: "magnitude" starts with 'ma' "maximum" starts with 'ma'

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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.

Rule at resonance:
RLC parallel: min current, max impedance
RLC series: max current, min impedance
Mnemonic: The word "series" has "se" twice, forward at the beginning and backwards at the end, the maximum number of ses. Se=C. C, current, is max.

Memory tip: The question involved two "i" letters -- input and current (I). Choose the answer with two "i" letters -- minimum.

Alternative tip: Min(imum) In(put)

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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.

Hint: There is no mention of inductance nor capacitance; therefore, no leading nor lagging of voltage or current.

Memory Aid: For those of us who aren’t as electrically savvy - the keyword here is resonance. This word should clue you in that answers referring to things that are approximately equal, or in this case, “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.

Study hints:

• Resistance is the first word in only one of the answers for the "parallel" version of this question.
• Q is a ratio therefore it'll be a division (not multiplication).
• RLC is Resistance divided by L or C (inductance or capacitance).
• The correct answer has the word "or" in it.
• The correct answer is the only one that starts with "Resistance."

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The formula for resonant frequency is:

\[f=\frac{1}{2\pi\sqrt{LC}}\]

Where:

• f - frequency in hertz (Hz)
• L - inductance in henrys (H)
• C - capacitance in farads (F)

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}

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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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\(Q\) 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} Just look at the numbers only from small to larger, the second is the answer, and this is true for other similar question on numbers.

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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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In a series resonant circuit, V_L = V_C = QV_S, in which V_S is the source voltage. Thus, for a series resonant circuit, as the Q increases, the voltages across the inductor and capacitor increase. (As a corollary, in a parallel resonant circuit, I_L = I_C = QI_S, in which I_S is the source current.)

KNØJI

Hint: Increasing, increase.

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