What can cause the voltage across reactances in a series RLC circuit to be higher than the voltage applied to the entire circuit?
Why At series resonance the inductive and capacitive reactances are equal and cancel, so the circuit impedance drops to just the resistance and current reaches its maximum. That large current flows through XL and XC, producing voltages of Q times the applied voltage across each reactance. They are 180 degrees out of phase and cancel each other, so Kirchhoff's law still holds even though each individual reactive voltage can be many times the source voltage.
Watch out Low Q does the opposite: the voltage magnification factor is Q itself, so a high Q, not a low one, produces the big reactive voltages. Resistance limits the resonant current and therefore reduces the effect.
Series resonance = voltage magnification: V across L or C equals Q times the source voltage.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
What is the resonant frequency of an RLC circuit if R is 22 ohms, L is 50 microhenries, and C is 40 picofarads?
Why Resonance depends only on L and C: f = 1 / (2π√(LC)). Here LC = (50 × 10^-6)(40 × 10^-12) = 2 × 10^-15, whose square root is 4.472 × 10^-8. Multiplying by 2π gives 2.81 × 10^-7, and the reciprocal is about 3.56 MHz. The 22 ohms of resistance affects Q and bandwidth but has no effect on the resonant frequency.
Watch out The 22.36 MHz choice is what you get by computing 1/√(LC) and calling it hertz, which forgets the 2π factor; that number is actually the resonant frequency in radians per second.
Resonance ignores R. f = 1/(2π√LC), and never drop the 2π or you land on the tempting 22.36 MHz.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
What is the magnitude of the impedance of a series RLC circuit at resonance?
Why In a series RLC circuit the inductive and capacitive reactances are opposite in sign, so at the resonant frequency where XL = XC they cancel completely. What remains in the series path is only the resistance, so Z = R and the impedance reaches its minimum value. That is also why series resonance produces maximum current for a given applied voltage.
Watch out Saying the impedance is high compared to the resistance describes a parallel resonant circuit, where the tank looks like a large resistance at resonance. Matching XL or XC alone ignores the fact that those two reactances cancel each other out.
Series resonance: reactances cancel, only R is left, Z is minimum and current is maximum.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
What is the magnitude of the impedance of a parallel RLC circuit at resonance?
Why At resonance the inductive and capacitive susceptances cancel, so the L and C branches together draw no net current from the source. What is left is the resistive element, so the circuit's impedance is purely resistive and equal to that resistance. This is the maximum impedance a parallel RLC circuit can present, the opposite of a series resonant circuit, whose impedance drops to a minimum equal to its resistance.
Watch out Saying the impedance is high compared to the circuit resistance is tempting because parallel resonance does give a high impedance, but it is high compared to off-resonance values, not compared to R itself; it equals R.
Series or parallel, at resonance the reactances cancel and Z = R. Parallel peaks at R, series bottoms out at R.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
What is the result of increasing the Q of an impedance-matching circuit?
Why Q is the ratio of a resonant circuit's center frequency to its half-power bandwidth, BW = f0/Q. An impedance-matching network is a resonant circuit, so raising its Q sharpens the response and the range of frequencies over which a good match is maintained shrinks. A low-Q matching network is broadband and covers a whole band; a high-Q network may match only over a few tens of kHz and needs retuning as you move.
Watch out The idea that harmonics increase is backwards: a higher-Q network is more selective and actually attenuates harmonics better, which is why tank circuit Q matters in transmitter output stages.
BW = f0/Q. Q up, bandwidth down. High Q = sharp and narrow, low Q = broad.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
What is the magnitude of the circulating current within the components of a parallel LC circuit at resonance?
Why In a parallel LC tank at resonance, the inductive and capacitive branch currents are equal in magnitude and opposite in phase, so they largely cancel at the terminals. The energy sloshes back and forth between L and C, producing a circulating (tank) current that is roughly Q times the current drawn from the source, so it reaches its maximum at resonance. Meanwhile the impedance seen by the source is maximum and the line current is minimum.
Watch out The choice saying minimum describes the current drawn from the external source, not the current circulating inside the tank; the formula choices are versions of the resonant frequency equation, not a current.
Parallel resonance: line current minimum outside, circulating current maximum inside the tank (about Q times bigger).
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
What is the magnitude of the current at the input of a parallel RLC circuit at resonance?
Why In a parallel RLC circuit at resonance the inductive and capacitive branch currents are equal in magnitude but opposite in phase, so they cancel each other and circulate within the tank instead of being drawn from the source. What is left for the source to supply is only the current through the resistive element, so the circuit looks like a very high impedance and the input current drops to its lowest value. Parallel resonance means maximum impedance, therefore minimum line current.
Watch out Maximum current is what happens in a series RLC circuit at resonance, where the reactances cancel and leave only the small series resistance. R/L and L/R are time-constant style ratios and have nothing to do with resonant current.
Parallel = high Z = low current in; series = low Z = high current. Opposite twins.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
What is the phase relationship between the current through and the voltage across a series resonant circuit at resonance?
Why At resonance the inductive reactance and capacitive reactance are equal in magnitude and opposite in sign, so they cancel and leave only resistance. A purely resistive impedance produces no phase shift, so the current through the series circuit and the voltage across it rise and fall together. This zero phase angle is one of the defining tests for resonance, along with minimum impedance and maximum current in a series circuit.
Watch out The 90 degree answers describe a circuit dominated by a single reactance: voltage leads current by 90 degrees in a pure inductor, and current leads voltage by 90 degrees in a pure capacitor. At resonance neither dominates because they cancel each other.
Resonance = reactances cancel = looks like a plain resistor = zero phase angle, series or parallel.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
How is the Q of an RLC parallel resonant circuit calculated?
Why In a parallel RLC tank the resistance appears across the circuit, so a larger shunt resistance means less loss and a higher Q: Q = R / X, where X is the reactance of either the inductor or capacitor (they are equal at resonance). This is the reciprocal of the series case, where the resistance is in line with the reactance and Q = X / R. At resonance you can use either XL or XC since XL = XC.
Watch out The choice dividing reactance by resistance is the formula for a series resonant circuit; mixing the two up is the classic trap in this group.
Parallel: R is on top (R/X). Series: R is on the bottom (X/R). The R follows where it sits in the circuit.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
What is the resonant frequency of an RLC circuit if R is 33 ohms, L is 50 microhenries, and C is 10 picofarads?
Why Resonance depends only on L and C: f = 1 / (2π√(LC)). Here LC = (50 × 10^-6 H)(10 × 10^-12 F) = 5 × 10^-16, whose square root is 2.236 × 10^-8. Multiply by 2π to get 1.405 × 10^-7, and the reciprocal is about 7.12 × 10^6 Hz, or 7.12 MHz. The 33 ohm resistance affects Q and bandwidth but not the resonant frequency.
Watch out The 7.12 kHz choice has the right digits but is off by a factor of 1000, which is what you get from mishandling the micro and pico prefixes; watch that 50 µH is 50e-6 and 10 pF is 10e-12.
f = 1/(2π√LC), and R never enters it. Small L and C in the microhenry/picofarad range land you in the HF megahertz range.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
What is the half-power bandwidth of a resonant circuit that has a resonant frequency of 7.1 MHz and a Q of 150?
Why Half-power (3 dB) bandwidth of a resonant circuit is BW = f_resonant / Q. Here 7,100,000 Hz divided by 150 gives about 47,333 Hz, or 47.3 kHz. Higher Q means a narrower, sharper response; lower Q spreads the response out.
Watch out The 23.67 kHz choice is exactly half the right answer, what you get if you mistakenly divide by 2Q or take only one side of the curve. The answers in hertz come from dividing in the wrong direction or losing the MHz-to-Hz conversion.
BW = f / Q. Divide the resonant frequency by Q, and keep the frequency in hertz before you divide.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
What is the half-power bandwidth of a resonant circuit that has a resonant frequency of 3.7 MHz and a Q of 118?
Why Half-power (-3 dB) bandwidth of a resonant circuit is BW = f0 / Q. With f0 = 3.7 MHz and Q = 118, BW = 3,700,000 / 118 = 31,356 Hz, about 31.4 kHz. Higher Q means a narrower, sharper response for the same resonant frequency.
Watch out The 436.6 kHz figure comes from multiplying 3.7 MHz by 118 instead of dividing, and 15.7 kHz is that correct result cut in half, as if the formula used 2Q.
BW = f divided by Q. Q is quality, and high quality means narrow. Divide, never multiply.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.
What is an effect of increasing Q in a series resonant circuit?
Why In a series resonant circuit the current through L and C is maximized at resonance, and the reactive voltages across the inductor and capacitor are each Q times the applied voltage. These two voltages are equal and opposite so they cancel as seen by the source, but they are physically present across the components. So raising Q directly raises those internal voltages, which is why high-Q series circuits need components rated well above the supply voltage.
Watch out Minimizing parasitics is a way to get high Q, not a result of it; stray resistance and loss are what lower Q in the first place.
Series resonance: V across L and C = Q x source voltage. High Q means high internal volts.
HamSandwich explanation, first draft. The question and answers are the NCVEC text.