Find Impedance of LCR Series Circuit & Calculate Quality Factor

APPARATUS

A variable transformer as AC milliammeter of range 0-500 milliamp as AC voltage of range 0-250 volt, and AC voltmeter of range 0.25 volt, an inductance coil, a capacitor lamp resistance (60 watts) and connecting wires.

THEORY

In an Ac circuit containing only a resistance R. The Current is represented by I = Imax sin wt

= Imaxsin2πtn

Emax = ImaxR

E = Emax Sin W.t = Emassin2πnt

= RImax sin wt = RImaxsin2πnt

Imax = Emax/R as IV = Ev/R

Again, Emax = Emax/wl = Emax/2πnl = Emax/x₁

IV= EV/x₁

The quantity x₁ = wl = 2πnl is known as reactance due to inductance measured in ohm. And Imax = 1/wc Imax

= 1/2πnc Imax

Imax = Emaxwc = Emax2πnc

IV= EV/XC, The quantity Xc = 1/wc = 1/2πnc is known as the recutance due to capacitance and when AC circuits contain resistance R ohms. An inductance L and capacitance C ford is series the current is

Quality factor: The ratio of potential drop across the inductance or capacitance to potential drop across the resistance (or the applied voltage) at resonance is called quality factor.

Q = E₁/R = Lw/R

Similarly Q1 = Ec/ER = 1/RwRc

OBSERVATION:

In voltmeter 1 division = 2.5 volt

1 division = 0.25volt

Similarly , In ammeter 1 division = 1 amp

Mean (Z) = (0.2+0.3541+0.283+0.294)x10³ / 4

= 1.2184×10³ /4 = 0.2821×10³

= 282.1

Mean of quality factor (Q) = 1.25+1.33+ 2.153+1.9524 /4 = 1.6714

Mean of impedance = 282.1 +291.57 /2 = 286.83

Mean of quality factor (Q) = 1.6714 +9.03 /3 = 2.35

RESULT:

The impedance of the L.C.R series circuit is 286.83 and quality factor is 2.35.

PRECAUTION:

i) Only well insulated pieces of flexible wire should be used.

ii) Nakul part of the circuit should not be touched when the current is passing.

iii) Suitable values of R,L and C should be chosen.

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