RL Circuit: Difference between revisions
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<math>I = \frac{emf}{R}(1 - e^{-\frac{R}{L}t})</math> | <math>I = \frac{emf}{R}(1 - e^{-\frac{R}{L}t})</math> | ||
Therefore, | Therefore, when the circuit is connected, the current in the RL circuit increases exponentially over time to the maximum value of <math>\frac{emf}{R}</math>. | ||
===A Computational Model=== | ===A Computational Model=== | ||
From the above section we were able to see that | |||
From the above section we were able to see that the current in the RL circuit grows exponentially over time. | |||
<math></math> | <math></math> | ||
<math></math> | <math></math> | ||
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===Difficult=== | ===Difficult=== | ||
==History== | ==History== | ||
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Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context? | Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context? | ||
==References== | ==References== |
Revision as of 06:29, 5 December 2015
Written by Jiwon Yom
The RL circuit is one of the simple circuit applications and is composed of a power source, a resistor and an inductor. Figure 1 illustrates a symbolic representation of a simple RL circuit. Typically, an inductor in the RL circuit is a solenoid. The RL circuit can be frequently seen in fluorescent light choke, also known as electrical ballast (Figure 2), where the RL circuit limits the current that flows through the fluorescent light tube in order to prevent destruction of the tube.3 Also, the RL circuit can act as a high-pass or low-pass filter for voltage supply of varying frequencies.4
The Main Idea
A Mathematical Model
Let's take a look at the simple RL circuit illustrated in Figure 1. Before we examine the RL circuit, we must examine the effects of an inductor. An inductor is a coiled current-carrying wire. Due to its coiled structure, it surrounds a certain area where the magnetic field is varying over time. When the inductor is connected to a power source, current flows through the coil and such change in current leads to an additional emf in the coil. As Faraday’s law in motional emf shows that the magnitude of emf is equal to the magnitude of rate of change in magnetic flux, we can calculate the magnitude of emf produced by the inductor.
[math]\displaystyle{ |{emf}| = N|\frac{d\Phi}{dt}| }[/math], where [math]\displaystyle{ \Phi = N\frac{\mu_0I}{l}\times\pi r^2 }[/math], [math]\displaystyle{ N = }[/math] the number of coils in an inductor, [math]\displaystyle{ r = }[/math] radius of the coil, and [math]\displaystyle{ l = }[/math] the length of the inductor.
[math]\displaystyle{ emf = N \frac{d}{dt}[N\frac{\mu_0I}{l}\pi r^2] }[/math]
Since N, [math]\displaystyle{ \mu_0 }[/math], [math]\displaystyle{ \pi }[/math], and [math]\displaystyle{ r }[/math] are constants,
[math]\displaystyle{ emf = \frac{\mu_0N^2}{l}\pi r^2\frac{dI}{dt} }[/math]
Here, the terms [math]\displaystyle{ \frac{\mu_0N^2}{l}\pi r^2 }[/math] is a proportionality constant called "inductance" and can be summed up by letter L. Therefore, we can get the magnitude of emf of the inductor, which is given by
[math]\displaystyle{ emf_{inductor} = L\frac{dI}{dt} }[/math]
Going back to the RL circuit, now we can apply conservation of energy in circuit (loop rule) to the RL circuit. By the law of conservation of energy, we can obtain the following result.
[math]\displaystyle{ \Delta V_{battery} + \Delta V_{resistor} + \Delta V_{inductor} = 0 }[/math]
[math]\displaystyle{ emf_{battery} - IR - L\frac{dI}{dt} = 0 }[/math]
From this point, [math]\displaystyle{ emf_{battery} }[/math] will be shortened to [math]\displaystyle{ emf }[/math]. Rearrange the above result to solve for [math]\displaystyle{ I }[/math].
[math]\displaystyle{ \frac{dI}{dt} = \frac{emf-IR}{L} }[/math]
Using separation of variables,
[math]\displaystyle{ \frac{dI}{emf-IR} = \frac{dt}{L} }[/math]
[math]\displaystyle{ \int\frac{dI}{emf-IR} = \int\frac{dt}{L} }[/math]
Solving the integrals, we obtain
[math]\displaystyle{ -\frac{ln(emf-IR)}{R} = \frac{1}{L}t + C }[/math]
Since I = 0 when t = 0, [math]\displaystyle{ C = -\frac{ln(emf)}{R} }[/math], thus
[math]\displaystyle{ -\frac{ln(emf-IR)}{R} = \frac{1}{L}t - \frac{ln(emf)}{R} }[/math]
[math]\displaystyle{ ln(emf-IR) - ln(emf) = -\frac{R}{L}t }[/math]
[math]\displaystyle{ ln(\frac{emf-IR}{emf}) = -\frac{R}{L}t }[/math]
[math]\displaystyle{ \frac{emf-IR}{emf} = e^{-\frac{R}{L}t} }[/math]
[math]\displaystyle{ IR = emf - emf\times e^{-\frac{R}{L}t} }[/math]
[math]\displaystyle{ I = \frac{emf}{R}(1 - e^{-\frac{R}{L}t}) }[/math]
Therefore, when the circuit is connected, the current in the RL circuit increases exponentially over time to the maximum value of [math]\displaystyle{ \frac{emf}{R} }[/math].
A Computational Model
From the above section we were able to see that the current in the RL circuit grows exponentially over time. [math]\displaystyle{ }[/math] [math]\displaystyle{ }[/math]
Examples
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