GATE 2020 IN Q.9: solution and answer

GATE 2020 · INQ.9 · 1 mark · MCQ

A Q meter is best suited for the measurement of the ____

(A)
Quality factor of a capacitance.
(B)
Distributed capacitance of a coil.
(C)
Quality factor of piezoelectric sensor.
(D)
Turns-ratio of a transformer

See the question in the paper

Answer and solution

ANSWER

(B) Distributed capacitance of a coil. The Q meter finds it from two resonance settings of its tuning capacitor, at $f_{1}$ and $2 f_{1}$ : $C_{d} = \frac{C _{1} - 4 C _{2}}{3}$ .

Official answer: BSource: Official answer key, GATE 2020 IN (IIT Delhi)

How a Q meter works (series resonance). The coil under test is connected in series with a calibrated variable capacitor $C$ , and a small, fixed RF voltage $E$ is injected into the loop. At resonance the voltage across the capacitor is $V_{C} = QE$ . With $E$ held constant, a voltmeter across $C$ that is calibrated in $Q$ reads the coil's quality factor directly. The instrument is designed to measure coils (inductors) at radio frequencies.
Distributed (self) capacitance is a standard Q-meter measurement. Resonate the coil at frequency $f_{1}$ with the tuning capacitor set to $C_{1}$ . Then resonate it at $f_{2} = 2 f_{1}$ with the capacitor set to $C_{2}$ . Each time, the coil's distributed capacitance $C_{d}$ is in parallel with the tuning capacitor.
Apply the resonance condition $f = \frac{1}{2 π L ( C + C _{d} )}$ at both frequencies. $L$ is the same in both cases, so it cancels: $\frac{f _{2}}{f _{1}} = 2 = \frac{C _{1} + C _{d}}{C _{2} + C _{d}} \Rightarrow C_{1} + C_{d} = 4 (C_{2} + C_{d})$
Solving gives $C_{d} = \frac{C _{1} - 4 C _{2}}{3}$ This needs only two readings of the capacitor dial, so the Q meter measures $C_{d}$ directly and simply. Option (B) fits.
Why not (A). A good capacitor has very low loss, so its $Q$ is much higher than a coil's, and the Q meter cannot read it directly. It has to be found indirectly: resonate a standard coil first without the capacitor and then with it, and calculate the capacitor's $Q$ from the small difference between two large readings. That makes the result much less accurate, so the Q meter is not best suited to this measurement.
Why not (C) and (D). A piezoelectric sensor is a capacitive, charge-generating transducer. Its quality factor describes an electromechanical resonance, not the resonance of a coil with a tuning capacitor, which is what the Q meter's circuit is built to measure. A transformer's turns ratio is found from a voltage-ratio test (a ratio meter or a ratio bridge), not from a resonance or a $Q$ reading.
So, of the four options, a Q meter is best suited to measuring the distributed capacitance of a coil. Answer: (B).

KEY IDEA

A Q meter is a series-resonant circuit made of the coil under test and a calibrated tuning capacitor. At resonance $V_{C} = QE$ , so it reads a coil's $Q$ directly. Resonating the same coil at $f_{1}$ and at $2 f_{1}$ gives its distributed capacitance from two dial readings: $C_{d} = (C_{1} - 4 C_{2}) /3$ .

WATCH OUT FOR

Picking (A) because the instrument is named after the quality factor. A Q meter reads a coil's Q directly. A low-loss capacitor has a much higher Q, which the meter can only find indirectly from the small difference between two readings, and that result is much less accurate. The standard "direct" uses of a Q meter are a coil's Q, its inductance and its distributed capacitance.

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