Differential equation - not sure where to start

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SUMMARY

The discussion focuses on solving a differential equation for the voltage across a capacitor in a resistor-capacitor (RC) charging circuit, where resistance varies inversely with voltage. The specific equation presented is R(t) = 1/(a + bVc(t)). The user expresses difficulty in manipulating the integral term in their equations and seeks guidance. A solution is proposed to switch from current i(t) to charge q(t), which simplifies the problem into a recognizable Riccati differential equation. Alternatively, the use of the Laplace transform is suggested, although it presents complications due to the complexity of R(t).

PREREQUISITES
  • Understanding of first-order differential equations
  • Familiarity with resistor-capacitor (RC) circuits
  • Knowledge of Riccati differential equations
  • Basic principles of Laplace transforms
NEXT STEPS
  • Study the properties and solutions of Riccati differential equations
  • Learn how to apply Laplace transforms to differential equations
  • Explore the relationship between current and charge in RC circuits
  • Review techniques for solving integral equations in electrical engineering
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Electrical engineering students, circuit designers, and anyone involved in analyzing dynamic systems involving capacitors and resistors.

marka
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Homework Statement



I need to solve for the voltage across a capacitor in a resistor-capacitor charging equation (eg. http://sub.allaboutcircuits.com/images/05175.png" ), except that the resistance varies inversely with the voltage across the capacitor. With a constant R, I'm happy solving the first order differential equation, but when R(t)=1/(a+b(Vc(t))), I don't know where to start.

Homework Equations



[tex]V_{in} = V_{r} + V_{c}[/tex]

[tex]V_{in} = i(t).R(t) + \frac{1}{C}\int_{0}^{t}i(\tau)d\tau[/tex]

[tex]R(t) = \frac{1}{a+bV_{c}(t)}[/tex]

The Attempt at a Solution



[tex]V_{in} = \frac{i(t)}{a+bV_{c}(t)} + \frac{1}{C}\int_{0}^{t}i(\tau)d\tau[/tex]

[tex]V_{in} = \frac{i(t)}{a+\frac{b}{C}\int_{0}^{t}i(\tau)d\tau} + \frac{1}{C}\int_{0}^{t}i(\tau)d\tau[/tex]

[tex]V_{in}.a + \frac{b.V_{in}}{C}\int_{0}^{t}i(\tau)d\tau = i(t) + (\frac{1}{C}\int_{0}^{t}i(\tau)d\tau)^{2}[/tex]

(differentiate wrt t, use chain rule for the last term)

[tex]0 + \frac{b.V_{in}}{C}i(t) = \frac{di(t)}{dt} + \frac{2.i(t)}{C}\int_{0}^{t}i(\tau)d\tau)[/tex]I'm a bit stuck with where to go here - I'm not sure how to get rid of the integral in the last term, or even whether I've done the right thing so far. Any hints would be greatly appreciated,

Mark
 
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It looks easier if you are to work in terms of q(t) instead of i(t) since you won't have to deal with the integral of i(t). If you write everything out in terms of q and throw the dq/dt to one side and everything else on the left, the DE is recognisable as a Riccati differential equation.

On the other hand you could make use of the Laplace transform to work with i(t) but the term in R(t) looks pretty complicated to transform.
 

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