Induced Voltage

Facebook
Whatsapp
Twitter
LinkedIn
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.
Faraday's law states The electromotive force around a closed path is equal of the time rate of change of the magnetic flux enclosed by the path.
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:
$$\bbox[5px,border:1px solid grey] {{e_L= N { d\phi_B \over dt}}} \tag{1}$$
If we write Eq. (1) as
$$\begin{split} e_L &= N { d\phi_B \over dt} = N {d\phi_B \over dt}{di \over di}\\ &= N {d\phi_B \over di}{di \over dt}\\ &= (N {d\phi_B \over di}){di \over dt}\\ &= (L){di \over dt}\\ e_L&= L{di \over dt} \end{split} \tag{2} $$
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,
$$\bbox[10px,border:1px solid grey]{ L = N {d\Phi \over di}} \, (H)\tag{3}$$
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.
Hysteresis curve
Fig. 1: Hysteresis curve.
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: $$\bbox[10px,border:1px solid grey]{V_L = L {di \over dt}} \tag{4}$$ 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
$$v_L = L di/dt = L(0) = 0 V$$
The average voltage across the coil is defined by the equation
$$\bbox[10px,border:1px solid grey]{v_L = L {\Delta_i \over \Delta_t}} \tag{5}$$
where $\Delta$ signifies finite change (a measurable change).
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].
Fig. 2: For Example 1.
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,
$$v_L = L {\Delta_i \over \Delta_t} = L { 0 \over dt} = 0 V$$
b. 2 ms to 4 ms:
$$ \begin{split} v_L &= L {\Delta_i \over \Delta_t} = (4 \times 10^{-3} H){ 10 \times 10^{-3} \over 2 \times 10^{-3}} \\ &=20 \times 10^{-3} = 20mV\\ \end{split} $$
c. 4 ms to 9 ms:
$$ \begin{split} vL &= L {\Delta_i \over \Delta_t} = (-4 \times 10^{-3} H){ 10 \times 10^{-3} \over 5 \times 10^{-3}} \\ &=-8 \times 10^{-3} = -8mV\\ \end{split} $$
d. 9 ms to $\infty$:
$$v_L = L {\Delta_i \over \Delta_t} = L { 0 \over dt} = 0 V$$
Fig. 3: Voltage across a 4-mH coil due to the current of Fig. 2.
The waveform for the average voltage across the coil is shown in [Fig. 3]. Note from the curve that
the voltage across the coil is not determined solely by the magnitude of the change in current through the coil ($\Delta_i$), but also by the rate of change of current through the coil ($\Delta_i/\Delta_t$).
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.

Do you want to say or ask something?

Only 250 characters are allowed. Remaining: 250
Please login to enter your comments. Login or Signup .
Be the first to comment here!
Terms and Condition
Copyright © 2011 - 2024 realnfo.com
Privacy Policy