Question

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 = \( \frac{(C_2 – C_1 ) C_1 C_2}{Q_1 C_1 – Q_2 C_2}\)

I have been asked this question during an online interview.

The query is from Advanced Problems on Q Meter in section Electronic Instruments of Electrical Measurements

Answer 1

Correct choice is (a) Q = \(\frac{(C_1 – C_2 ) Q_1 Q_2}{Q_1 C_1 – Q_2 C_2}\)

The 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 = \(\frac{X_S}{R_S}  = \frac{(C_1- C_2) Q_1 Q_2}{Q_1 C_1-Q_2 C_2}\).