# Electrical Measurements Questions and Answers – Advanced Problems on Q Meter

This set of Electrical Measurements & Measuring Instruments Multiple Choice Questions & Answers (MCQs) focuses on “Advanced Problems on Q Meter”.

1. Q meter operator is the principle of __________
a) Series resonance
b) Current resonance
c) Self-inductance
d) Eddy currents

Explanation: We know that Q = $$\frac{ωL}{R}$$
From the above relation, we can say that it works on series resonance.

2. In a Q Meter, the values of tuning capacitor are C3 and C4 for resonant frequencies f3 and 2f4 respectively. The value of distributed capacitance is?
a) $$\frac{C_3-C_4}{2}$$
b) $$\frac{C_3-2C_4}{3}$$
c) $$\frac{C_3-4C_4}{3}$$
d) $$\frac{C_3-3C_4}{2}$$

Explanation: QX = $$\frac{R_P}{X_P} = \frac{(C_4 – C_3)Q_3 Q_4}{Q_3 C_3 – Q_4 C_4}$$
The main error in the measurement of Q is due to the distribution. To check for this, the Q value is measured at two frequencies f1 and 2f2. It should be same, if not then, $$\frac{C_3-4C_4}{3}$$.

3. A circuit tuned to a frequency of 1.5 MHz and having an effective capacitance of 150 pF. In this circuit, the current falls to 70.7 % of its resonant value. The frequency deviates from the resonant frequency by 5 kHz. Q factor is?
a) 50
b) 100
c) 150
d) 200

Explanation: Q = $$\frac{ω}{ω1 – ω2} = \frac{f}{f2-f1}$$
Here, f = 1.5 × 106 Hz
f1 = (1.5 × 106 – 5 × 103)
f2 = (1.5 × 106 + 5 × 103)
So, f2 – f1 = 10 × 103 Hz
Q = $$\frac{1.5 × 10^6}{10 × 10^3}$$ = 150.

4. A circuit tuned to a frequency of 1.5 MHz and having an effective capacitance of 150 pF. In this circuit, the current falls to 70.7 % of its resonant value. The deviates from the resonant frequency are 5 kHz. Effective resistance of the circuit is?
a) 2 Ω
b) 3 Ω
c) 5.5 Ω
d) 4.7 Ω

Explanation: R = $$\frac{f2-f1}{2πf^2 L}$$
Here, f = 1.5 × 106 Hz
f1 = (1.5 × 106 – 5 × 103)
f2 = (1.5 × 106 + 5 × 103)
So, f2 – f1 = 10 × 103 Hz
R = $$\frac{10 × 10^3}{2π(1.5 × 10^6)^2 L}$$
R = 4.7 Ω.

5. Q Meter is used to measure _________
a) Q factor of an inductive coil
b) Only the effective resistance
c) Only bandwidth
d) Q factor of an inductive coil, the effective resistance, and bandwidth

Explanation: Q meter can measure the Q factor of an inductive coil. It can also measure the effective resistance. Also, the bandwidth can be measured by the Q Meter. Therefore it can be used for all the above functions.

6. Q factor of a coil measured by the Q Meter is _________ the actual Q of the coil.
a) Equal to
b) Same but somewhat lesser than
c) Same but somewhat higher than
d) Not equal to

Explanation: The Q factor measured by the Q meter cannot be exactly equal to the actual Q of the coil because of the presence of errors. Also, it is not practically possible for the value to be higher than the actual one. But the value is somewhat lesser and almost equal to the actual value.

7. Consider a circuit consisting of two capacitors C1 and C2. Let R be the resistance and L be the inductance which are connected in series. Let Q1 and Q2 be the quality factor for the two capacitors. While measuring the Q value by the Series Connection method, the value of the Q factor is?
a) Q = $$\frac{(C_1 – C_2 ) Q_1 Q_2}{Q_1 C_1 – Q_2 C_2}$$
b) Q = $$\frac{(C_2 – C_1 ) Q_1 Q_2}{Q_1 C_1 – Q_2 C_2}$$
c) Q = $$\frac{(C_1 – C_2 ) Q_1 Q_2}{Q_2 C_2 – Q_1 C_1}$$
d) Q = $$\frac{(C_2 – C_1 ) C_1 C_2}{Q_1 C_1 – Q_2 C_2}$$

Explanation: ωL = $$\frac{1}{ωC}$$and Q1 = $$\frac{ωL}{R} = \frac{1}{ωC_1 R}$$
XS = $$\frac{C_1-C_2}{ωC_1 C_2 }$$, RS = $$\frac{Q_1 C_1 – Q_2 C_2}{ωC_1 C_2 Q_1 Q_2}$$
QX = $$\frac{X_S}{R_S} = \frac{(C_1- C_2) Q_1 Q_2}{Q_1 C_1-Q_2 C_2}$$.

8. Consider a circuit consisting of two capacitors C1 and C2. Let R be the resistance and L be the inductance which are connected in series. Let Q1 and Q2 be the quality factor for the two capacitors. While measuring the Q value by the Parallel Connection method, the value of the Q factor is?
a) Q = $$\frac{(C_1 – C_2 ) Q_1 Q_2}{Q_1 C_1 – Q_2 C_2}$$
b) Q = $$\frac{(C_2 – C_1 ) Q_1 Q_2}{Q_1 C_1 – Q_2 C_2}$$
c) Q = $$\frac{(C_1 – C_2 ) Q_1 Q_2}{Q_2 C_2 – Q_1 C_1}$$
d) Q = $$\frac{(C_2 – C_1 ) C_1 C_2}{Q_1 C_1 – Q_2 C_2}$$

Explanation: $$\frac{1}{R_P} = \frac{ωC_1}{Q_2} – \frac{1}{RQ_1^2}$$, XP = $$\frac{1}{ω(C_2 – C_1)}$$
Q = $$\frac{(C_2 – C_1 ) Q_1 Q_2}{Q_1 C_1 – Q_2 C_2}$$.

9. Consider the following statements regarding the sources of error in a Q Meter.

i) If a coil with a resistance R is connected in the direct measurement mode and
If the residual resistance of the Q Meter is 0.1 R,
Then the measured Q of the coil would be 1.1 times the actual Q.
ii) If the inductance to be measured is less than 0.1 μH.
The error due to the presence of residual inductance cannot be neglected.
iii) The presence of a distributed capacitance modifies the effective Q of the coil.


Which of the above statements are correct?
a) i, ii and iii
b) i and ii
c) ii and iii
d) i and iii

Explanation: We know that, Q = $$\frac{Lω}{R}$$
With Q Meter resistance considered, the measured or indicated Q is 1/1.11 times the actual Q. Thus, statement (i) is incorrect. Hence, statements (ii) and (iii) are correct.

10. The function of the Q- Meter is to _________
a) Measure capacitance
b) Measure inductance
c) Measure quality factor of capacitor and inductor
d) Measure form factor of capacitor and inductor

Explanation: Q-Meters are intended to measure the quality factor of a capacitor and inductor. Q = $$\frac{Lω}{R} = \frac{1}{ωCR} = \frac{V_C}{V_A}$$. They are not used for measuring capacitances and inductances, unlike AC Bridges.