Induced Voltage
Voltage or current is induced in a coil or conductor if there is a rate of change of magnetic flux linking that coil or conductor. This phenomenon is discussed in the "Faradays law of induction" which was developed by English physicist Michael Faraday in 1830.
If a coil of N turns is placed in the region of a changing flux, a voltage will be induced across the coil as determined by
Faraday's law:
If we write Eq. (1) as
Where L is the inductance of a coil is a measure of the change in flux linking
a coil due to a change in current through the coil; that is,
where N is the number of turns, $\phi$ is the flux in webers, and $i$ is the current through the coil. If a change in current through the coil fails to result in a significant change in the flux linking the coil through its center, the resulting inductance level will be relatively small. For this reason the inductance of a coil is sensitive to the point of operation on the
hysteresis curve shown in [Fig. 1].
Equation (3) also reveals that the larger the inductance of a coil
(with N fixed), the larger will be the instantaneous change in flux
linking the coil due to an instantaneous change in current through the
coil.
Eq.(2) revealing that the magnitude of the voltage across an inductor is directly
related to the inductance L and the instantaneous rate of change of current through the coil. Obviously, therefore, the greater the rate of change of current through the coil, the greater will be the induced voltage. This certainly agrees with our earlier discussion of Lenz's law.
When induced effects are employed in the generation of voltages
such as those available from dc or ac generators, the symbol e is appropriate for the induced voltage. However, in network analysis the voltage across an inductor will always have a polarity such as to oppose the source that produced it, and therefore the following notation will be used throughout the analysis to come:
If the current through the coil fails to change at a particular instant,
the induced voltage across the coil will be zero. For dc applications,
after the transient effect has passed, $di/dt = 0$, and the induced voltage
is
The average voltage across the coil is defined by the equation
where $\Delta$ signifies finite change (a measurable change).
$$\bbox[5px,border:1px solid grey] {{e_L= N { d\phi_B \over dt}}} \tag{1}$$
$$\begin{split}
e_L &= N { d\phi_B \over dt} = N {d\phi_B \over dt}{di \over di}\\
&= N {d\phi_B \over di}{di \over dt}\\
&= (N {d\phi_B \over di}){di \over dt}\\
&= (L){di \over dt}\\
e_L&= L{di \over dt}
\end{split}
\tag{2}
$$
$$\bbox[10px,border:1px solid grey]{ L = N {d\Phi \over di}} \, (H)\tag{3}$$
Fig. 1: Hysteresis curve.
$$\bbox[10px,border:1px solid grey]{V_L = L {di \over dt}} \tag{4}$$
$$v_L = L di/dt = L(0) = 0 V$$
$$\bbox[10px,border:1px solid grey]{v_L = L {\Delta_i \over \Delta_t}} \tag{5}$$
Example 1: Find the waveform for the average voltage across the
coil if the current through a $4mH$ coil is as shown in [Fig. 2].
Solution:
a. 0 to 2 ms: Since there is no change in current through the coil, there is no voltage induced across the coil; that is,
b. 2 ms to 4 ms:
c. 4 ms to 9 ms:
d. 9 ms to $\infty$:
The waveform for the average voltage across the coil is shown in
[Fig. 3]. Note from the curve that
A careful examination of [Fig. 3] will also reveal that the area
under the positive pulse from 2 ms to 4 ms equals the area under the
negative pulse from 4 ms to 9 ms. In the next sections, we will find that
the area under the curves represents the energy stored or released by
the inductor. From 2 ms to 4 ms, the inductor is storing energy,
whereas from 4 ms to 9 ms, the inductor is releasing the energy
stored. For the full period zero to 10 ms, energy has simply been
stored and released; there has been no dissipation as experienced for
the resistive elements. Over a full cycle, both the ideal capacitor and
inductor do not consume energy but simply store and release it in their
respective forms.
Fig. 2: For Example 1.
a. 0 to 2 ms: Since there is no change in current through the coil, there is no voltage induced across the coil; that is,
$$v_L = L {\Delta_i \over \Delta_t} = L { 0 \over dt} = 0 V$$
$$ \begin{split}
v_L &= L {\Delta_i \over \Delta_t} = (4 \times 10^{-3} H){ 10 \times 10^{-3} \over 2 \times 10^{-3}} \\
&=20 \times 10^{-3} = 20mV\\
\end{split}
$$
$$ \begin{split}
vL &= L {\Delta_i \over \Delta_t} = (-4 \times 10^{-3} H){ 10 \times 10^{-3} \over 5 \times 10^{-3}} \\
&=-8 \times 10^{-3} = -8mV\\
\end{split}
$$
$$v_L = L {\Delta_i \over \Delta_t} = L { 0 \over dt} = 0 V$$
Fig. 3: Voltage across a 4-mH coil due to the current of Fig. 2.
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