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August 28th, 2017, 05:30 AM  #1 
Newbie Joined: Aug 2017 From: South Africa Posts: 3 Thanks: 0  How to obtain an RLC circuit from this differential?
(D^2 + 4D + 13)x(t) = (4D + 2)y(t). I know the term that is squared related to the inductance, the 4D term is related to the capacitance and the term with no D is the resistance. But I am unsure whether I have to solve for y(t) as x(t) is the input and y(t) is the output. Thanks 
August 28th, 2017, 06:13 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,530 Thanks: 1750 
What instructions came with the equation?

August 28th, 2017, 06:45 AM  #3 
Newbie Joined: Aug 2017 From: South Africa Posts: 3 Thanks: 0 
The instructions were to model an electric circuit using the differential equation above. So as far as I understand, it would be to divide both sides by (4D +2) to get y(t) by itself and then solve from there.

August 28th, 2017, 07:08 AM  #4  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,127 Thanks: 716 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Quote:
Consider a circuit with an ohmic resistor, an inductor and a capacitor. The voltage, $\displaystyle V=V(t)$, for each of these components relates to the charge, $\displaystyle q = q(t)$, using the following equations: Resistor: $\displaystyle V_R = R \frac{dq}{dt}$ Inductor: $\displaystyle V_L = L \frac{d^2q}{dt^2}$ Capacitor: $\displaystyle V_C = \frac{q}{C}$ In a series RLC circuit, $\displaystyle V_{total} = V_R + V_L + V_C$ $\displaystyle = L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{q}{C}$ If we introduce a new operator 'D' such that $\displaystyle Dq = \frac{dq}{dt}$ and $\displaystyle D^2q = \frac{d^2q}{dt^2}$, then $\displaystyle V_{total} = (LD^2 + RD + C)q$ Therefore, by comparison: L = 1 H R = 4 $\displaystyle \Omega$ C = 13 F $\displaystyle V_{total} = (4D + 2) y(t)$ To solve, consider dividing through by L and then investigate solutions to second order differential equations  
August 28th, 2017, 07:18 AM  #5 
Newbie Joined: Aug 2017 From: South Africa Posts: 3 Thanks: 0 
Thank you for the detailed reply Benit13. Howcome you suggest to divide by L? I stuck on having to solve it as I did Laplace transforms many moons ago.


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circuit, differential, obtain, rlc 
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