Download Calculating Current in an LRC Circuit using Undetermined

Austin Feldman
April 16, 2011
Brewer - Mat 275
Honors Project
Calculating Current in an LRC Circuit using
Undetermined Coefficients and Laplace Transforms
1. Current through an LRC circuit can be calculated in a wide variety of ways. Although
using undetermined coefficients is not the easiest method, it creates a conceptual
model of current that allows for a better understanding of the calculations behind the
formulas. Below is a simple LRC circuit with their corresponding expressions; R for
resistor, L for inductor, C for capacitor, and Vac for an AC source.
KVL: Sum of voltages in a closed loop is equal to zero.
−𝑉𝐴𝐶 + 𝑉𝑅 + 𝑉𝐿 + 𝑉𝐶 = 0
Note that the voltage source has a negative sign since the designated current will go
from the negative terminal to the positive terminal. Also, we know that the voltage
across a resistor is iR, the voltage across an inductor is Ldi/dt, the voltage across a
capacitor is Q/C, and the voltage across the given voltage source is Vsin(ωt). Using
these equations and substituting them in to the above equation we derive the following
equation.
𝐿(
𝑑𝑖
𝑄
) + 𝑖 ∗ 𝑅 + = 𝑉𝑠𝑖𝑛(𝜔𝑡)
𝑑𝑡
𝐶
It is also known that the derivative of charge, Q, is equal to current i. Taking the
derivative of the above equation and substituting i in for dQ/dt we derive the following.
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𝑑2𝑖
𝑑𝑖
1
𝐿 ( 2 ) + 𝑅 ( ) + ∗ 𝑖 = 𝜔𝑉𝑐𝑜𝑠(𝜔𝑡)
𝑑𝑡
𝑑𝑡
𝐶
It is important to note that this equation has the same mathematical form as the
equation for a damped oscillator. Thus we can solve the equation by first solving for the
general solution and then the particular solution. First we will solve for the general
solution.
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𝐿𝑟 2 + 𝑅𝑟 + ( ) = 0
𝐶
Using the quadratic equation, we can solve for r and see that we will get repeated roots.
−𝑅 ± √𝑅 2 − 4𝐿/𝐶 −𝑅 ± √𝑅 2 − 4𝐿/𝐶
𝑟=
,
2𝐿
2𝐿
From these repeated values, we can deduce that the general solution will be equal to
the following.
𝒚𝒄 =
−𝑹±√𝑹𝟐 −𝟒𝑳/𝑪
𝟐𝑳
𝑪𝟏 𝒆
+
−𝑹±√𝑹𝟐 −𝟒𝑳/𝑪
𝟐𝑳
𝑪𝟐 𝒕𝒆
If initial values are given, it is important to note that we cannot plug them in here. They
can only be plugged into a combined equation of yc and yp. To find a particular solution,
we will use the form shown below.
𝒚𝒑 = 𝑨𝒔𝒊𝒏(𝝎𝒕) + 𝑩𝒄𝒐𝒔(𝝎𝒕)
Taking the first and second derivatives we derive the following.
𝑦𝑝′ = 𝜔𝐴𝑐𝑜𝑠(𝜔𝑡) − 𝜔𝐵𝑠𝑖𝑛(𝜔𝑡)
𝑦𝑝′′ = −𝜔2 𝐴𝑠𝑖𝑛(𝜔𝑡) − 𝜔2 𝐵𝑐𝑜𝑠(𝜔𝑡)
Plugging these values into the left side of the original equation, we can solve for A and
B. We find the following.
𝝎𝟐 𝑹𝑽
𝑨=
𝟏
(𝝎𝟐 𝑳 − 𝑪)𝟐 + (𝝎𝑹)𝟐
𝑩=
𝟏
−𝝎𝑽(𝝎𝟐 𝑳 − 𝑪)
𝟏
(𝝎𝟐 𝑳 − 𝑪)𝟐 + (𝝎𝑹)𝟐
2
At this time, if initial values are given, we can combine the general and particular
solution and solve for C1 and C2.
From these calculations we can see that the method of undetermined coefficients can
derive an accurate mathematical formula for the current in an LRC circuit. Studying LRC
circuits using undetermined coefficients lets us not only have proof that the calculations
often assumed are correct, but also helps us understand how the mathematical
expressions are derived. However, it is not practical to use undetermined coefficients to
solve everyday circuits. While there are numerous ways to derive to current in a circuit,
one way in particular is simple and widely used; Laplace Transforms.
2. Laplace Transforms are often implemented on each component in LRC circuits to quickly
derive a desired function from the circuit. A basic LRC circuit is shown below along with
values assigned to each component of the circuit.
In order to derive an expression for current for the circuit shown above, the circuit must
be converted to what is known as the “s domain”. To do this, the Laplace Transform
must be taken for each component and the resulting circuit is shown.
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Using Kirchhoff’s Voltage Loop rule, which states that the voltages in any independent
loop will sum to equal zero, a value for the current in the circuit in the “s domain” can
be determined.
2
12
3 ∗ 𝐼(𝑠) + 𝑠 ∗ 𝐼(𝑠) + ∗ 𝐼(𝑠) −
=0
𝑠
𝑠
2
12
𝐼(𝑠) ∗ (𝑠 + 3 + ) =
𝑠
𝑠
𝑠 2 + 3𝑠 + 2
12
𝐼(𝑠) ∗ (
)=
𝑠
𝑠
𝐼(𝑠) =
12
𝑠
∗( 2
)
𝑠
𝑠 + 3𝑠 + 2
𝐼(𝑠) =
𝐼(𝑠) =
𝑠2
12
+ 3𝑠 + 2
12
(𝑠 + 1)(𝑠 + 2)
Using separation of variables, values of A and B are able to be calculated.
12
𝐴
𝐵
𝐼(𝑠) = (𝑠+1)(𝑠+2) = 𝑠+1 + 𝑠+2
12 = 𝐴(𝑠 + 2) + 𝐵(𝑆 + 1)
12 = 𝐴𝑠 + 𝐵𝑠 + 2𝐴 + 𝐵
s: 0=A+B → B=-A
constant: 12=2A+B
12=2A-A → A=12 → B=-12
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12
12
𝐼(𝑠) = (
−
)
𝑠+1 𝑠+2
Now that an equation for the current in the s domain has been found, the current in the
regular circuit can be calculated. Since the s domain was found by taking the Laplace
Transform of each component, going from the s domain back to the circuit can be done
by taking the inverse Laplace Transform. The calculations to derive the current in the
circuit are shown below.
𝑖(𝑡) = 𝐿−1 𝐼(𝑠) = 𝐿−1 (
12
12
−
) = 12𝑒 −𝑡 − 12𝑒 −2𝑡
𝑠+1 𝑠+2
By taking the Laplace Transform in the original circuit, an expression for the current in
the so domain was able to be derived. From there, the inverse Laplace Transform was
taken in order to derive an expression for the current in the initial circuit. This method
is both accurate and faster when compared to using undetermined coefficients. This
explains why Laplace Transformations are a more accepted way of analyzing circuits in
general.
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