User:Egm4313.s12.team18.hollon

=William Hollon=

Useful Links
Team 18 Course Wiki Course Website Report Table

Problem
The goal of this exercise is to derive the alternative formulations of the circuit equation seen in (3) and (4) using the circuit equation from (2). Where:


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$$  \displaystyle Q=Cv_C \Rightarrow \int i\,dt=Cv_c \Rightarrow i = C \frac{dv_C}{dt} $$     (1a,1b,1c)
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$$  \displaystyle V=LC\frac{d^2v_C}{dt^2}+RC\frac{dv_C}{dt}+v_C $$     (2)
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$$  \displaystyle LI''+RI'+\frac{1}{C}I=V' $$     (3)
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$$  \displaystyle LQ''+RQ'+\frac{1}{C}Q=V $$     (4)
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Deriving (3) from (2)
In order to derive first alternate version, (3), of the circuit equation, (2), we begin by taking the derivative of (2) with respect to time, the result of which can be seen below in (5)


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$$  \displaystyle V'=LC \frac{d^3 v_C}{dt^3} + RC \frac{d^2 v_C}{dt^2} + \frac{d v_C}{dt} $$     (5)
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Next, by rearranging the right hand equation from (1) we obtain the following relation:


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$$  \displaystyle \frac{I}{C}=\frac{d v_C}{dt} $$     (6)
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And by taking subsequent derivatives, we also get:
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$$  \displaystyle \frac{I'}{C}=\frac{d^2 v_C}{dt^2} \, \, \& \, \, \frac{I''}{C}=\frac{d^3 v_C}{dt^3} $$     (7) & (8)
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By plugging the relations from (6), (7), and (8) into equation (2) and reducing the fractions we obtain the variant seen in (3)
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$$  \displaystyle LI'' + RI' + \frac{1}{C}I=V' $$     (3)
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Deriving (4) from (2)
We can derive the second alternate equation, (4), from circuit equation (2) using a method similar to the one used to derive (3) as shown above.

By taking the derivatives of (1a) we obtain the following relations:


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$$  \displaystyle Q=Cv_C \, \, \& \, \, Q'=C \frac{dv_C}{dt} \, \, \& \, \, Q''=C \frac{d^2 v_C}{dt^2} $$     (9a, 9b, 9c)
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Once again, plugging (9a, 9b, 9c) into (2) and reducing the results we can obtain the second variation of the circuit equation: (4)
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$$  \displaystyle LQ''+RQ'+\frac{1}{C}Q=V $$     (4)
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