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**

X_{L} = 184 Ω; X_{C}
= 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/π^{2}
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^{-6}H;
C = 80/π^{2} ×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×10^{4} t) volt at i) 0 s
ii) 50 μs iii) 75 μs.

**Solution**

The given equation is υ
= 10sin (3 π×10^{4} 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)

X_{L} = ωL = 200
× 40 × 10^{-3} = 8 Ω

V_{m} = I_{m}
X_{L} = 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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