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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?

pico
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2 Answers2

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Start from the KCL equation for the common top node $$i_{R} + i_{L} + i_{C} = 0.$$ Substitute the equations for the capacitor current and resistor current in terms of the inductor voltage $$v_{L}/R + i_{L} + C\cdot \dot{v}_{L} = 0.$$ The last step relies on substitung the expression for the inductor voltage (i will drop the subscript \$L\$) $$\frac{L\cdot \dot{i}}{R} + i + C\cdot L\cdot \ddot{i} = 0.$$

Yggdrasil
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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.

pico
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  • What are the dimensional units for the first term of your "corrected" version of your first equation? – jonk Jan 13 '20 at 01:42
  • They didn’t really specify... so let’s call it volts... – pico Jan 13 '20 at 01:44
  • So you've never been trained in dimensional analysis? – jonk Jan 13 '20 at 01:45
  • yes... but i'm being lazy... what's your take on the units? i was just saying voilts because it matches my KCL equation... – pico Jan 13 '20 at 01:46
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    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. – jonk Jan 13 '20 at 01:48
  • You will find that your uncorrected version *can* be summed without a problem. You will find that your "corrected" version fails this simple test. – jonk Jan 13 '20 at 01:50
  • what test is that? – pico Jan 13 '20 at 01:51
  • The same units being added. You cannot add volts to amps, for example. But you are doing worse than that. Do you not see it? – jonk Jan 13 '20 at 01:55
  • do you agree with: $$I_R + I_C + I_L = 0$$

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

     – pico Jan 13 '20 at 01:55
  • I'm talking about your first two equations at the top of this answer. You are directing me elsewhere. You cannot do this: $L~C~ v'' + \frac{L}{R} v' + v = 0$. The rest I don't care about, right now. – jonk Jan 13 '20 at 01:56
  • now if i differentiate that equation i get:

    $$\frac{V'}{R} + C V'' + \frac{1}{L}V = 0$$

     – pico Jan 13 '20 at 01:59
  • then multiply both sides by L... its the same thing... if you agree with KCL – pico Jan 13 '20 at 01:59
  • I don't see an error in my math. – pico Jan 13 '20 at 02:02
  • Sorry, pico. My fault, entirely. I was merely looking at the change in units from current to voltage (dimensional unit change) and not looking at the actual correction, itself. The way I'd write it is instead in standard form, which is: $\frac{\text{d}^2}{\text{d}t^2}V+\frac{1}{R:C}\frac{\text{d}}{\text{d}t}V+\frac1{L:C}V=0:\text{A}$. But we agree. I was momentary worried about your unit change but also very busy and distracted by and focusing on something else at the time. I should have held my tongue. Please forgive me for my error. – jonk Jan 13 '20 at 04:51
  • glad somebody was checking... because i was just having faith in the SI system... – pico Jan 13 '20 at 13:27