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

Consider a circuit consisting of two capacitors C1 and C2. Let R be th

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

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