A series resonant circuit has R = 2 , L = 1 mH and C = 0.1 μF, the value of quality factor Q is: 

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  1. 50
  2. 25
  3. 30
  4. 40

Answer (Detailed Solution Below)

Option 1 : 50
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CONCEPT:

The Quality factor: Quality factor of resonance is a dimensionless parameter that describes how underdamped an oscillator or resonator is.

Mathamaticaly, Q factor =  \( \frac{1}{R} \sqrt{ \frac{L}{C}}\)

Where, L, C and R are the inductance, capacitance and resistance respectively.

EXPLANATION:

We know,

⇒ Q =  \( \frac{1}{R} \sqrt{ \frac{L}{C}}\)

⇒ Q =\( \frac{1}{2} \sqrt{ \frac{1\times10^{-3}}{0.1\times10^{-6}}}\)

⇒Q=50

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