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

The question was posed to me in exam.

This is a very interesting question from Advanced Problems on Q Meter in division Electronic Instruments of Electrical Measurements

Answer 1

Right option is (b) Q = \(\frac{(C_2 – C_1 ) Q_1 Q_2}{Q_1 C_1 – Q_2 C_2}\)

Best explanation: \(\frac{1}{R_P}  = \frac{ωC_1}{Q_2} – \frac{1}{RQ_1^2}\), XP = \(\frac{1}{ω(C_2 – C_1)}\)

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