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

Last edited by ctrstocks. Register to edit

Tags: none

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 capacitive reactance and the inductive reactance are of equal magnitude.

Source: http://www.its.bldrdoc.gov/fs-1037/dir-031/_4576.htm

Memory Trick: There are two answers with "equals" in them, but the one with reactance written twice is the correct answer.

Last edited by n9gmr. Register to edit

Tags: none

At resonance the impedance of the capacitance and inductance are equal in value but opposite in sign so they cancel each other completely leaving only the circuit resistance as magnitude of the circuit impedance.

Last edited by kd8pfe. Register to edit

Tags: none

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).

---

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".

Last edited by kd7bbc. Register to edit

Tags: none

Resonance in AC circuits implies a special frequency determined by the values of the resistance , capacitance , and inductance . For series resonance the condition of resonance is straightforward and it is characterized by minimum impedance and zero phase. This means that when a series RLC circuit is in resonance, the impedance is minimum. Impedance can be read as "resistance", so in a resonant series RLC circuit where the impedance or resistance is minimum, the current is maximum.

Last edited by mvs90. Register to edit

Tags: none

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'

Last edited by km4amj. Register to edit

Tags: none

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)

Last edited by dogshed. Register to edit

Tags: none

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.

Last edited by marvsherman419. Register to edit

Tags: none

When a circuit is in resonance the capacitive reactance (-90 degrees) and inductive reactance (+90 degrees) cancel because they are equal in magnitude (XC=XL). Therefore the voltage and current are in phase with each other:

phase angle is arctan((XL-XC)/R) = arctan (0/R) = arctan(0) = 0 degrees.

The voltage and current are in phase.

Last edited by ericthughes. Register to edit

Tags: none

For half power bandwidth use the formula:

B = Fr/Q Hz.

Where "Fr" is the resonant frequency and "Q" is the "quality" (or goodness) of the circuit.

1800 kHz / 95 = 18.9 kHz

Last edited by drichmond60. Register to edit

Tags: none

\(Q\) 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.

Last edited by davidportia. Register to edit

Tags: none

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".

Last edited by greyone. Register to edit

Tags: none

BW(kHz) = f(kHz) / Q, so

BW = 14.25*1000 / 187 = 76.2 kHz

Last edited by vaughanth. Register to edit

Tags: none

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.

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.

Last edited by yr7 - extra. Register to edit

Tags: none

The formula for a resonant frequency is 1/(2Pisqrt(L*C)) with L and C being henrys and farads.

1/(2Pisqrt(0.000040 * 0.000000000200)) which equals 1779406 hz

Or: 1/(2Pisqrt(40E-6 * 200E-12)) = 1,779,406

Divide by one million and round to find the answer of 1.78 Mhz.

On a calculator, first [Clear],
[4][0][x][2][0][0][=][√][x][6][.][2][8][=] 561.7
then, [Clear],
[1][0][0][0][0][0][0][÷][5][6][1][.][7][=] 1780

Last edited by kacela. Register to edit

Tags: none

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}

---

Alternate for formula-weak folks like me: given the first digits in the question, 7 is the next number in the sequence.

Last edited by nightdragon. Register to edit

Tags: none

The formula for a resonant frequency is 1/(2Pisqrt(L*C)) with L and C being henrys and farads.

1/(2Pisqrt(0.000025 * 0.000000000010)) which equals 10065842 hz

Divide this by one million and round to find the answer of 10.1 Mhz.

Another way:

= 1/(2 * 3.1416 * sqrt(2510^-6 * 1010^-12))

= 1/9.9345 10^-8 = 10,065,842 Hz ~= 10.1 MHz

Last edited by marjomercado. Register to edit

Tags: none