Encyclopedia of Electrical Engineering

For the series configuration of Fig. 1, the voltage $v_{element}$ of the
boxed-in element opposes the source e and thereby reduces the magnitude of the current i. The magnitude of the voltage across the element is determined by the opposition of the element to the flow of
charge, or current i. For a resistive element, we have found that the
opposition is its resistance and that $v_{element}$ and i are determined by
$v_{element} = iR$.
**Fig. 1: **Defining the opposition of an element to the
flow of charge through the element.
We found in the previous Chapter of inductor, that the voltage across an inductor is directly related to the rate of change of current through the coil. Consequently, the higher the frequency, the greater will be the rate of change
of current through the coil, and the greater the magnitude of the voltage. In addition, we found in the same chapter that the inductance of a coil will determine the rate of change of the flux linking a coil for a particular change in current through the coil. The higher the inductance, the greater the rate of change of the flux linkages, and the greater the
resulting voltage across the coil.
The inductive voltage, therefore, is directly related to the frequency
(or, more specifically, the angular velocity of the sinusoidal ac current
through the coil) and the inductance of the coil. For increasing values
of f and L in Fig. 2, the magnitude of vL will increase as described
above.
**Fig. 2: **Defining the parameters that determine the
opposition of an inductive element to the flow
of charge.
Utilizing the similarities between Figs. 1 and 2, we find that
increasing levels of vL are directly related to increasing levels of opposition in Fig. 1. Since vL will increase with both $w (= 2\pi f )$ and L,
the opposition of an inductive element is as defined in Fig. 2.
We will now verify some of the preceding conclusions using a more
mathematical approach and then define a few important quantities to be
employed in the sections and chapters to follow.
**Fig. 3: **Investigating the sinusoidal response of an
inductive element.
**Fig. 4: **For a pure inductor, the voltage across the coil leads the current through the
coil by 90.
For the inductor of Fig. 3,
$$ v_L = L{di_L \over dt}$$
and, applying differentiation
$$ v_L = L{di_L \over dt}\\
= L {d(I_m \sin wt) \over dt}\\
= w L I_m \cos wt\\
=V_m \sin(wt + 90)\\
$$
Where
$$\bbox[10px,border:1px solid grey]{V_m = wLI_m = 2 \pi fLI_m} \tag{1}$$
Note that the peak value of vL is directly related to $w (= 2\pi f )$ and L
as predicted in the discussion above.
A plot of vL and iL in Fig. 4 reveals that
If a phase angle is included in the sinusoidal expression for iL, such
as
$$ i_L = I_m \sin(wt + \theta)$$
then
$$ v_L = V_m \sin(wt + \theta + 90)$$
The opposition established by an inductor in a sinusoidal ac network
can now be found by:
$$effect = {cause \over opposition}$$
which, for our purposes, can be written
$$opposition = {cause \over effect}$$
Substituting values, we have
$$opposition = {V_m \over I_m} = {wLI_m \over I_m} = wL$$
revealing that the opposition established by an inductor in an ac sinusoidal network is directly related to the product of the angular velocity
($w = 2\pi f$ ) and the inductance, verifying our earlier conclusions.
The quantity $wL$, called the reactance (from the word reaction) of an
inductor, is symbolically represented by $X_L$ and is measured in ohms;
that is,
$$\bbox[10px,border:1px solid grey]{X_L = wL } (ohms) \tag{2}$$
In an Ohm's law format, its magnitude can be determined from
$$\bbox[10px,border:1px solid grey]{X_L = {V_m \over I_m} } (ohms) \tag{3}$$
Inductive reactance is the opposition to the flow of current, which
results in the continual interchange of energy between the source and
the magnetic field of the inductor. In other words, inductive reactance,
unlike resistance (which dissipates energy in the form of heat), does not
dissipate electrical energy (ignoring the effects of the internal resistance
of the inductor).
**Example 1: **The current through a 0.1-H coil is provided. Find
the sinusoidal expression for the voltage across the coil. Sketch the v
and i curves

a. $i = 10 \sin 377t$

b. $i = 7\sin (377t - 70)$

**Solution: **

a. Eq. (2): $X_L = wL = (377\, rad/s)(0.1\, H) = 37.7 Ω$

Eq. (3): $V_m = I_m /X_L = (10\, A)/(37.7 Ω) = 377 V$

and we know that for a coil v leads i by 90 degrees.

Therefore, $$v_L = 377 \sin(377t + 90)$$ The curves are sketched in Fig. 5. **Fig. 5: **Example 1(a).

b. $X_L$ remains at 37.7 Ω.

$V_m = I_m X_L = (7 A)(37.7 Ω) = 263.9 V$ and we know that for a coil v leads i by 90 degrees. Therefore $$v = 263.9 \sin(377t - 70 + 90)$$ and $$\bbox[10px,border:1px solid grey]{v = 263.9 \sin(377t + 20)}$$ The curves are sketched in Fig. 6. **Fig. 6: **Example 1(b).

a. $i = 10 \sin 377t$

b. $i = 7\sin (377t - 70)$

a. Eq. (2): $X_L = wL = (377\, rad/s)(0.1\, H) = 37.7 Ω$

Eq. (3): $V_m = I_m /X_L = (10\, A)/(37.7 Ω) = 377 V$

and we know that for a coil v leads i by 90 degrees.

Therefore, $$v_L = 377 \sin(377t + 90)$$ The curves are sketched in Fig. 5.

b. $X_L$ remains at 37.7 Ω.

$V_m = I_m X_L = (7 A)(37.7 Ω) = 263.9 V$ and we know that for a coil v leads i by 90 degrees. Therefore $$v = 263.9 \sin(377t - 70 + 90)$$ and $$\bbox[10px,border:1px solid grey]{v = 263.9 \sin(377t + 20)}$$ The curves are sketched in Fig. 6.