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Resonant circuits and filters questions
The reactance of an inductor rises with frequency and that of a capacitor falls, so at one frequency they are equal and cancel. That is resonance, and it is how a tuned circuit selects one radio station from many. A resistor and capacitor alone make a high-pass or low-pass filter, keeping one end of the spectrum and discarding the other.
17 original questions · 51 marks · the resonant circuits and filters notes · Electronics
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State how the reactance of an inductor and of a capacitor each change as the frequency rises, and state what happens at resonance in a series LC circuit.
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An inductor's reactance 2πfL rises with frequency; a capacitor's reactance 1/2πfC falls (1). At resonance the two are equal in size and opposite in sense, so they cancel and only the circuit's resistance is left to limit the current, which is therefore a maximum (1). Saying the reactances 'become zero' is the common error: each is still large, they simply cancel.Define the Q factor of a resonant circuit, and state the unit in which it is quoted.
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Q = f0/fB, the resonant frequency divided by the bandwidth, the bandwidth being the width of the peak between the two frequencies at which the current has fallen to 1/√2 of its peak value (1). It is a pure number with no unit (1). Quoting a Q factor as a percentage is the common error, and it earns nothing: a Q of 10 is not 10%, it is a peak ten times narrower than its own centre frequency.A series LC circuit with resistance is driven by a variable-frequency source. State where the output must be taken to give a band-pass response and where to give a band-stop response.
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Band-pass: take the output across the resistor, because the current, and so the pd across R, is largest at resonance (1). Band-stop: take the output across the series inductor and capacitor, whose combined reactance vanishes at resonance, so the output there falls to zero while everything far from resonance gets through (1). Naming the component without saying why is half an answer; a complete one names the output terminals and gives the reason.A resistor and a capacitor are connected in series across a signal source. State which component the output is taken across for a low-pass filter and which for a high-pass filter.
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Output across the capacitor gives a low-pass filter, since the capacitor's reactance is large at low frequency and small at high, so high frequencies are shorted away (1). Output across the resistor gives a high-pass filter (1). Swapping the two is the common error; anchor it on the capacitor, which passes fast signals and blocks slow ones.Define the cut-off frequency of an RC filter, and state the output there as a fraction of the full output.
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It is the frequency at which the capacitor's reactance equals the resistance, fc = 1/2πRC (1), and there the output has fallen to 1/√2, about 0.71, of its full value, which is the half-power point (1). Describing the cut-off as the frequency where the output 'stops' is the common error: the response slides away over a decade or more and there is no sharp edge.State the effect of adding resistance to a tuned circuit on its resonant frequency, on the height of its resonance peak and on its bandwidth.
Mark scheme
The resonant frequency does not move at all, since f0 depends on L and C alone (1). The peak is lower and broader, because Q has fallen and the bandwidth f0/Q has widened (1). Claiming that resistance shifts the peak sideways is the common error, and it costs the mark every time.A tuned circuit is built from an inductor of 1.0 mH and a capacitor of 100 nF. Calculate its resonant frequency.
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LC = 1.0 × 10−3 × 100 × 10−9 = 1.0 × 10−10 (1), whose square root is 1.0 × 10−5 s (1). f0 = 1/(2π × 1.0 × 10−5) = 1.59 × 104 Hz, that is 15.9 kHz (1). Clear the millihenries and nanofarads first, multiply, then take the root: leaving the prefixes in, or taking the root before multiplying, is where almost every lost mark in this calculation goes.A tuned circuit resonates at 15.9 kHz and has a Q factor of 10. Calculate its bandwidth and give the range of frequencies it passes.
Mark scheme
fB = f0/Q = 15.9/10 = 1.59 kHz (1). The passband runs half a bandwidth either side of resonance (1), from 15.1 kHz to 16.7 kHz (1). Running the band a whole bandwidth either side of f0, which doubles it, is the common error: fB is the total width, not the half-width.A series circuit of 1.0 mH and 100 nF, resonating at 15.9 kHz, has a total resistance of 25 Ω. Using Q = 2πf0L/R, calculate the Q factor and the bandwidth.
A filter is made from a 1.6 kΩ resistor in series with a 0.10 μF capacitor. Calculate its cut-off frequency.
A low-pass filter is to have a cut-off frequency of 2.5 kHz using a 4.7 kΩ resistor. Calculate the capacitance required, and state which component the output is taken across.
A radio's tuned circuit uses a fixed 1.0 mH inductor with a variable capacitor adjustable from 40 pF to 360 pF. Calculate the resonant frequency at each end of the range, and state the ratio of highest to lowest frequency.
A measured resonance curve peaks at 40.0 kHz, and the current has fallen to 1/√2 of its peak value at 38.4 kHz and at 41.6 kHz. Calculate the bandwidth and the Q factor.
A receiver's tuned circuit is centred on 200 kHz and must pass a signal 8.0 kHz wide, so its bandwidth may not be less than 8.0 kHz. The only inductor available is 250 μH. Calculate the capacitance needed, the largest Q factor the circuit may have, and the smallest total circuit resistance that is acceptable, using Q = 2πf0L/R. Comment on what would go wrong if the resistance were made very much smaller.
Speech occupying 300 Hz upwards shares a line with 50 Hz mains hum. A filter built from a 22 kΩ resistor is to have a cut-off frequency of 120 Hz. State which type of filter is needed, calculate the capacitance, and estimate the fraction of the hum that survives, given that a high-pass filter's output fraction is (f/fc)/√(1 + (f/fc)2).
An audio line is to have 50 Hz mains hum removed by a band-stop filter built from an inductor in series with a 1.0 μF capacitor. Calculate the inductance required, comment on whether the component is practical, and state where the output of the filter must be taken.
Medium-wave stations are spaced 9 kHz apart. A receiver is tuned to a station on 909 kHz and its designer has wound the tuned circuit to a Q factor of 300. Calculate the bandwidth this gives, state the Q the receiver should have instead, and explain what the listener would hear with the Q of 300.
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