AC Through Resistance and Inductance
A pure resistance R and a pure inductive coil of inductance L are shown connected in series in Figure (A) showing AC Through Resistance and Inductance.
Let V = r.m.s. value of the applied voltage,
I = r.m.s. value of the resultant current
VR = IR − voltage drop across R (in phase with I),
VL = I . XL –voltage drop across coil (ahead of I by 90°)
These voltage drops are shown in voltage triangle OAB in Figure (B). Vector OA represents ohmic drop VR and AB represents inductive drop VL. The applied voltage V is the vector sum of the two i.e. OB.
AdBlock-2
The quantity 
is known as the impedance (Z) of the circuit. As seen from the impedance triangle ABC (Figure (C)) Z2 = R2 + XL2.
i.e. (Impedance)2 = (resistance)2 + (reactance)2
From Figure (B), it is clear that the applied voltage V leads the current I by an angle φ such that
The same fact is illustrated graphically in Figure (D).
In other words, current I lags behind the applied voltage V by an angle φ.
Hence, if applied voltage is given by ν = Vm sin ω t, then current equation is
i = Im sin (ωt − φ) where Im = Vm / Z
In Figure (E), I has been resolved into its two mutually perpendicular components, I cos φ along the applied voltage V and I sin φ in quadrature (i.e. perpendicular) with V.
The mean power consumed by the circuit is given by the product of V and that component of the current I which is in phase with V.
So P = V × I cos φ = r.m.s. voltage × r.m.s. current × cos φ
The term ‘cos φ’ is called the power factor of the circuit.
Remember that in an a.c. circuit, the product of r.m.s. volts and r.m.s. amperes gives voltamperes (VA) and not true power in watts.
True power (W) = volt-amperes (VA) × power factor.
or
Watts = VA × cos φ
It should be noted that power consumed is due to ohmic resistance only because pure inductance does not consume any power.
Now P = VI cos φ = VI × (R/Z) = (V/Z) × I. R = I2R (ä cos φ = R/Z)
or
P = I2R watt
Let us calculate power in terms of instantaneous values.
Instantaneous power is = v i = Vm sinωt × Im sin (ωt − φ) = VmIm sin ω t sin (ωt − φ)
Obviously, this power consists of two parts (Figure (G)).
(i) a constant part 
which contributes to real power.
(ii) a pulsating component 
which has a frequency twice that of the voltage and current. It does not contribute to actual power since its average value over a complete cycle is zero.
Hence, average power consumed where V and I represent the r.m.s values. Symbolic Notation. Z = R + jXL
Impedance vector has numerical value of 
Its phase angle with the reference axis is φ = tan−1 (XL/R)
It may also be expressed in the polar form as Z = Z ∠ φ°
(i) Assuming V = V ∠ 0°;
It shows that current vector is lagging behind the voltage vector by φ°. The numerical value of current is V/Z.
(ii) However, if we assumed that I = I ∠ 0, then
V = IZ = I ∠ 0° × Z ∠ φ°
= IZ ∠ φ°
It shows that voltage vector is φ° ahead of current vector in ccw direction as shown in Figure (I).
Read article – AC Through Pure Capacitance Alone
Visit NCERTplanet.com for NCERT solutions and Textbook downloads
