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 Select the correct answer from above options network theory questions and answers, network theory questions pdf, network theory question bank, network theory gate questions and answers pdf, mcq on network theory pdf, gate network theory questions and solutions, network theory mcq Test , ies network theory questions, Network theory Questions for GATE EC Exam, Network Theory MCQ (Multiple Choice Questions)

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