RL Circuit

Written by Jiwon Yom

Figure 1. RL circuit representation¹

Figure 2. Fluorescent light choke²

Figure 3. Fluorescent light choke⁵ The RL circuit is shown by letter G

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] [math]\displaystyle{ }[/math]

A Computational Model

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