## Mathematics & Science
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## Applications of Differential Equations## Electric Circuits
## A Preliminary ExampleFirst let's consider a circuit with the following values: - inductance of 0.25 henrys
- resistance of 10 ohms
- capacitance of 0.001 farads
We'll remove the voltage source from our circuit, i.e. we'll let Of course, since we're working with a -
*I*(0) = 0 amperes *q*(0) = 1 coulomb
(Note: these initial conditions imply that the initial current is zero, and
that the capacitor is holding an initial charge of 1 coulomb. Since
Shortly we'll have I and q.Since we're looking for
so to solve it uniquely, we really need to know both
Well let's look at equation (1) from the introduction again:
It relates
Now letting
*L*= 0.25, since inductance is 0.25 henrys*R*= 10, since resistance is 10 ohms*C*= 0.001, since capacitance of 0.001 farads*I*(0) = 0, given as an initial condition*q*(0) = 1, given as an initial condition*E*(*t*) = 0, mentioned in the problem's description
In the next problem you'll be on your own in gleaning this information from the problem's description. Anyway, time for you to do some work. Substitute the above list of known
values into the formula for Moving on... |
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ODE Laboratories: A Sabbatical Project by Christopher A. Barker©2017 San Joaquin Delta College, 5151 Pacific Ave., Stockton, CA 95207, USA e-mail: cbarker@deltacollege.edu |