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Question about: "Circuit Analysis Demystified", David McMahon, 2008, Chapter 6, page 131, Quiz question 7.

How to derive differential equation for current? Example:

$$L~C~ i'' + \frac{L}{R} i' + i = 0$$

where:

$$\begin{matrix}i(0) = 1 & v(0) = 0\end{matrix}$$

for this circuit:

RLC circuit, zero input

KCL with all currents leaving node:

$$I_R + I_C + I_L = 0$$

$$\frac{V}{R} + C\frac{dV}{dt} + \bigg(I_L(0) + \frac{1}{L} \int \limits_{0}^{t} V(\tau) d\tau \bigg) = 0$$

Then I'm not entirely sure how to get rid of the integral. I suppose I could differentiate the entire equation with respect to d/dt and hope it act as the inverse of the integral..

maybe another problem, is that if I use KCL i'm getting the differential equation for voltage instead of the differential equation for current...

They didn't really specify which current I'm looking at in the Differential equation... maybe its the i_r, i_c, or i_L?

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    \$\begingroup\$ See: Source-free, under-damped, parallel RLC with 2 intial conditions. The remaining two conditions, critically damped and over-damped, are trivial by comparison. I'm sure they will be no difficulty to you. \$\endgroup\$ – jonk Jan 12 '20 at 21:47
  • \$\begingroup\$ Good link @jonk lol \$\endgroup\$ – Andy aka Jan 12 '20 at 22:59
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    \$\begingroup\$ @Andyaka I sure don't want to do that one, twice! ;) \$\endgroup\$ – jonk Jan 13 '20 at 1:16
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They made a mistake in the book.

Change this equation:

$$L~C~ i'' + \frac{L}{R} i' + i = 0$$

To This equation:

$$L~C~ v'' + \frac{L}{R} v' + v = 0$$

That's why they didn't specify which current I was looking at. For example, is i(t) referring to the register, the inductor, or the capacitor..they didn't specify because the differential equation is really suppose to be a function of v instead of i.

I think they made a mistake with the initial conditions as well...

$$v(0) = 1$$

$$v'(0) = 0$$

When I make these changes, i get an answer of:

$$v(t) = e^{-t/4}\cos\bigg(\frac{\sqrt{15}}{4}t\bigg) + \frac{1}{\sqrt{15}}e^{-t/4}\sin\bigg(\frac{\sqrt{15}}{4}t\bigg)$$

well anyways... that makes the question clean.

one further note, i can rationalize the denominator of:

$$\frac{1}{\sqrt{15}}$$

by multiplying by:

$$\frac{\sqrt{15}}{\sqrt{15}}$$

this yields:

$$\frac{1}{\sqrt{15}} = \frac{\sqrt{15}}{15}$$

so i can rewrite the expression for v(t) as:

$$v(t) = e^{-t/4}\cos\bigg(\frac{\sqrt{15}}{4}t\bigg) + \frac{\sqrt{15}}{15} e^{-t/4}\sin\bigg(\frac{\sqrt{15}}{4}t\bigg)$$

$$v(t) = \frac{1}{15}\Bigg(15e^{-t/4}\cos\bigg(\frac{\sqrt{15}}{4}t\bigg) + \sqrt{15} e^{-t/4}\sin\bigg(\frac{\sqrt{15}}{4}t\bigg)\Bigg)$$

And that answer exactly matches the answer key in the back of the book, which confirms what i was saying about the mistakes in the question.

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  • \$\begingroup\$ What are the dimensional units for the first term of your "corrected" version of your first equation? \$\endgroup\$ – jonk Jan 13 '20 at 1:42
  • \$\begingroup\$ They didn’t really specify... so let’s call it volts... \$\endgroup\$ – pico Jan 13 '20 at 1:44
  • \$\begingroup\$ So you've never been trained in dimensional analysis? \$\endgroup\$ – jonk Jan 13 '20 at 1:45
  • \$\begingroup\$ yes... but i'm being lazy... what's your take on the units? i was just saying voilts because it matches my KCL equation... \$\endgroup\$ – pico Jan 13 '20 at 1:46
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    \$\begingroup\$ Then you are really being lazy. When you add three things together, it's important that all three things have the same units, right? You can't add apples and oranges, so to speak, right? Look at your first version with \$i\$ and then look at your "corrected" version with \$v\$. Examine the units of each term. \$\endgroup\$ – jonk Jan 13 '20 at 1:48

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