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Definition, Explanation, Formulas, Solved Example Problems | Alternating Current (AC) - Quality factor or Q–factor | 12th Physics : Electromagnetic Induction and Alternating Current

Chapter: 12th Physics : Electromagnetic Induction and Alternating Current

Quality factor or Q–factor

It is defined as the ratio of voltage across L or C to the applied voltage.

Quality factor or Q–factor

The current in the series RLC circuit becomes maximum at resonance. Due to the increase in current, the voltage across L and C are also increased. This magnification of voltages at series resonance is termed as Q–factor.

It is defined as the ratio of voltage across L or C to the applied voltage.


Q-factor = Voltage across LorC / Applied voltage

At resonance, the circuit is purely resistive. Therefore, the applied voltage is equal to the voltage across R.


The physical meaning is that Q–factor indicates the number of times the voltage across L or C is greater than the applied voltage at resonance.

Since the phase angle is positive, voltage leads current by 53.1 for this inductive circuit.

 

EXAMPLE 4.22

Find the impedance of a series RLC circuit if the inductive reactance, capacitive reactance and resistance are 184 Ω, 144 Ω and 30 Ω respectively. Also calculate the phase angle between voltage and current.

Solution

XL = 184 Ω; XC = 144 Ω

R = 30 Ω

(i ) The impedance is


Impedance, Z = 50 Ω

(ii) Phase angle is


φ = 53.1

 

EXAMPLE 4.23

A 500 μH inductor, 80/π2 pF capacitor and a 628 Ω resistor are connected to form a series RLC circuit. Calculate the resonant frequency and Q-factor of this circuit at resonance.

Solution

L=500×10-6H; C = 80/π2 ×10−12 F; R = 628Ω

(i) Resonant frequency is


 Q =12.5

 

EXAMPLE 4.24

Find the instantaneous value of alternating voltage υ = 10 sin(3 π×104 t) volt at i) 0 s ii) 50 μs iii) 75 μs.

Solution

The given equation is υ = 10sin (3 π×104 t)


 

EXAMPLE 4.25

The current in an inductive circuit is given by 0.3 sin (200t – 40°) A. Write the equation for the voltage across it if the inductance is 40 mH.

Solution

L = 40 × 10-3 H; i = 0.1 sin (200t – 40º)

XL = ωL = 200 × 40 × 10-3 = 8 Ω

Vm = Im XL = 0.3 × 8 = 2.4 V

In an inductive circuit, the voltage leads the current by 90o Therefore,

v = Vm sin ( ωt +90º)

 v = 2 . 4 sin(200t −40 + 90 º)

v = 2 . 4 sin(200t +50 º)volt

Tags : Definition, Explanation, Formulas, Solved Example Problems | Alternating Current (AC) , 12th Physics : Electromagnetic Induction and Alternating Current
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12th Physics : Electromagnetic Induction and Alternating Current : Quality factor or Q–factor | Definition, Explanation, Formulas, Solved Example Problems | Alternating Current (AC)


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