A.C. CIRCUIT CONTAINING RESISTANCE AND INDUCTANCE IN SERIES (L-R Circuit)
Consider a pure inductance L and pure resistance R connected in series to which an alternating voltage of r.m.s. value V is applied. Let the voltage across R be VR and that across L be VL. The current through L and R is the same and is represented along the positive X-axis. VR is in phase with / and is represented by the vector OA along X-axis. VL leads / by n/2 and hence it is represented by the vector OB along + K-axis. The resultant voltage V across L-R circ{uit} is given by the vector sum of VR and VL, which is the diagonal OC. Now OA = VR = IR and OB = VL= IXL = /L.co.
V2 = VR2 + VL2 = 12R2 + I2Xl2 V2 = 12 (R2 + X2)
j = ^R2 + Xl2 ...0)
By Ohm's law — = Resistance. Thus >lR2 + XL2 is the resistance offered by the LR series
circ{uit} to the flow of electric current through the circ{uit}. It is called the inductive impedance denoted by ZL.
Inductive impedance of an LR series circ{uit} is defined as the obstruction offered by the inductance (inductor) and resistance (resistance) to the flow of alternating current through them and is denoted by ZL.
ZL = ^R2+XL2 =^IR2 + (Im)2
The reciprocal of the impedance is called admittance. S.I. unit of impedance is ohm and admittance is (ohm-1). y.
Phase Angle
Let 0 be the phase difference between the current and voltage in an LR circ{uit}.
From the triangle OAC,
, AC VL IXL Loi
The voltage across the LR series circ{uit} leads the current (or current lags the voltage) by a phase angle (j).
i.e., If instantaneous value of the applied voltage is E and the instantaneous current
is I then,
E = E0 sin (Of ...(ii), I = Iq sin (wt - <J>) ...(iii)
where <t> = tan-1 —
K
The phase angle depends upon the value of the inductance L, the resistance R and the frequency of the a.c supply.
POWER ASSOCIATED WITH LR SERIES CIRCUIT
Instantaneous alternating e.m.f. = E= E0 sin cot Instantaneous alternating current = / = 70 sin ((0/ - <j>)
Instantaneous power = Eh E0 sin (Of x 70 sin ((Of - <|))
= E010 sin (Of [sin (Of cos (j) - cos (Of sin <j>] = E0 /0 sin2(Of cos <)> - E0Iq sin (Of cos (Of sin (J)
2 E0 Iq = E010 cos <|) sin (Of - ^ sin 2(0f sin <()
Eq Iq COS <f» fT . 2 Eq Iq ct .
Average power over one cycle of a.c. =--—- sin (Of dt - ~zzr sin d> sin 2(0f dt
2 1
Average value of sin (Of over a cycle = —
Average value of sin 2(0f over a cycle = 0
So average power over one cycle of a.c = E0 /0 cos 0 x — - 0 = ^ x cos <|)
P = Enns cos i|>
= Virtual volt x Virtual ampere x cos <|)
The product Ems x is called the virtual power or apparent power. The term cos <|> is called the power factor. I0 sin <|> is called the wattless component of the current.
Eq Iq Eq Iq
Instantaneous power = — — cos $ - cos (2(0f - <|))
The second term is a pulsating component which has a frequency twice that of the voltage and current. This term does not contribute the actual power because its average value over a complete cycle is zero. Graphical representation of power consumed is as shown in Fig. 17.37. The work done is positive during the parts of the cycle CDE and GHJ, because the current and voltage are in the same direction, i.e., the source of supply is doing work on the circ{uit}. During the portions of the power cycle ABC and EFG, the current and the voltage are in the opposite direction. The work done is negative i.e., the circ{uit} is doing work on the source of e.m.f. The
difference between the positive and negative areas gives the work done during the time T.
We can find the power consumed using vector diagram shown in Fig. 17.38. The average power consumed is equal to the product of potential difference E (or V) and that part of current in phase with the voltage.
True power = EI cos <t> The power consumed is due to ohmic resistance only
( Rs
True power = EI cos <|> = £7 —
ZlJ
(F\ i F
= IR = I2R where ^ = 7
[ZlJ
= P = I2R
Impedance Triangle
We can represent the resistance and inductive reactance of the L-R series circ{uit} by the adjacent side OA and opposite side AC of the triangle OAC respectively. The side OC will give the impedance of the triangle.
From the triangle OAC, OC2 = OA2 + AC2
Zl2 = R2+ X2
Inductive impedance of the LR circ{uit} = ZL
Also the power factor cos <|) = —. The phase difference <|) between current and voltage in
ZL
an L-R series can also be found out from the impedance triangle, tan .
R
Power Triangle
Let us multiply R, XL and Z by I2 where I is the r.m.s value of current. Then the side OA represents I2R, the side AC represents I2XL and the side OC represents I2ZL. Such a triangle is called power triangle (see Fig. 17.40). The power which is actually dissipated in the circ{uit} is called the active power.
Active power = I2R = EI cos <(>. The power developed in the inductive reactance of the circ{uit} is called the reactive power.
Reactive power = I2XL - I2ZL sin ^ - EI sin <|). The product of r.m.s values of applied voltage and circ{uit} current is called the apparent power.
Apparent power = EI = (IZL)I - 12ZL ^-factor of a Coil
^-factor is the quality factor of a coil. It represents figure of merit of the coil. Q-factor is defined as the reciprocal of the power factor
1 Zl
Q-factor =-- = —
cos 0 R
If the resistance R is small compared to reactance, then
2-factor = ZJR = Ua/R
Also 2 = 2K x ™"m energy sotred
energy dissipated per cycle