The Q of an unknown coil using parallel connection measurement method can be calculated using which of the following equations.

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UGC NET Paper 2: Electronic Science 29 Oct 2022 Shift 1
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  1. Qp = \(\rm\frac{C_1 \Delta Q}{\left(C_1-C_2\right)\left(Q_1 Q_2\right)}\)
  2. Qp = \(\rm\frac{\left(C_1-c_2\right)}{C_1 \Delta Q}\)
  3. Qp = \(\rm\frac{\left(C_1-C_2\right)\left(Q_1 Q_2\right)}{C_1 \Delta Q}\)
  4. Qp = \(\rm\frac{C_1 \Delta Q}{\left(C_1+C_2\right)\left(Q_1 Q_2\right)}\)

Answer (Detailed Solution Below)

Option 3 : Qp = \(\rm\frac{\left(C_1-C_2\right)\left(Q_1 Q_2\right)}{C_1 \Delta Q}\)
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Detailed Solution

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Concept:

Q-meter is an instrument used for the measurement of the Q-factor of the coil as well as for the measurement of self-inductance, self-capacitance and resistance. There are three methods to connect the unknown impedance to the testing terminals of the Q-meter:

  1. Direct Connection
  2. Series Connection
  3. Parallel Connection

 

Parallel Connection of Q-meter:

In a parallel connection, the unknown impedance/coil is connected in parallel with the tuning circuit as shown in the figure below.

F1 Madhuri Engineering 23.02.2023 D3

Measurement procedure:

  1. Before connecting the unknown impedance, the circuit is resonated and we get the values of the tuning capacitor and Q-meter as C1 and Q1 respectively.
  2. After connecting the unknown impedance, the circuit is resonated and we get the values of the tuning capacitor and Q-meter as C2 and Q2 respectively.

 

Results:

Solving using resonance equations, we get

\(R_p = {Q_1Q_2 \over \omega C_1(Q_1-Q_2)}\) and \(X_p = {1 \over \omega (C_1-C_2)}\)

Hence, Q of the unknown coil is obtained as

\(Q_p = {R_p \over X_p} = {{(C_1-C_2)Q_1Q_2} \over C_1(Q_1-Q_2)}\)

\(Q_p = {{(C_1-C_2)Q_1Q_2} \over C_1(\Delta Q)}\)

Additional Information

Series Connection of Q-meter:

F1 Madhuri Engineering 23.02.2023 D4


Using the resonance conditions before and after connecting the load impedance, we get

\(R_s = {C_1Q_1-C_2Q_2 \over \omega C_1C_2Q_1Q_2}\) and \(X_s = {C_1-C_2 \over \omega C_1C_2}\) 

Hence,

\(Q_s = {X_s \over R_s} = {{(C_1-C_2)Q_1Q_2} \over C_1Q_1-C_2Q_2}\)

 

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