marka
- 1
- 0
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
V_{in} = V_{r} + V_{c}
V_{in} = i(t).R(t) + \frac{1}{C}\int_{0}^{t}i(\tau)d\tau
R(t) = \frac{1}{a+bV_{c}(t)}
The Attempt at a Solution
V_{in} = \frac{i(t)}{a+bV_{c}(t)} + \frac{1}{C}\int_{0}^{t}i(\tau)d\tau
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
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}
(differentiate wrt t, use chain rule for the last term)
0 + \frac{b.V_{in}}{C}i(t) = \frac{di(t)}{dt} + \frac{2.i(t)}{C}\int_{0}^{t}i(\tau)d\tau)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
Last edited by a moderator: