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 got this question in an online quiz.

The question is from Problems of Series Resonance Involving Quality Factor topic in division Resonance & Magnetically Coupled Circuit of Network Theory

Answer 1

Right answer is (a) Q = \(\frac{(C_1 – C_2) Q_1 Q_2}{Q_1 C_1-Q_2 C_2}\)

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