Begin Usage 5: Engineering Applications

DifferentialEquationsEngineeringApplicationsEx1.html

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Engineering Application Ex1

Properties:

  • * .... remember how to originally solved equations.
  • * .... have method that we know to solve these equations directly.
    •

    The following example is of an Electrical Engineering differential equation method:

    Implicit Differentiation is used!

    What's the problem? As seen below in the GENERAL DIAGRAM a circuit contains a DC source of 14V, a switch, two resistors (both 4k ohms), and a capacitor having the value of 90 micro farads (or 0.000090 farads). In the circuit depiction directly beneath it (on the same page), we see that before the switch is thrown the capacitor has become full and the current therefore stops in that branch of the circuit. The voltage Vc of that branch can therefore be found via voltage division. The value is found to be seven volts (7V).

    In the diagram directly beneath that one (on the bottom of the page) the switch is thrown, t>0,(separating the circuit from the source); but, allowing the capacitor to discharge to what's left of the circuit.

    As can be seen in the current vs. time diagram below the current is being discharged at an exponential rate. Now using KCL we can find the differential equation for the capacitor. Plugging in the values of the components gives us the equation dv/dt + 5.56v(t) = 0.

    Assuming the solution has the form K2 = e^(-t/T) and plugging that into the differential equation allows us to solve for the time constant T = 0.2 seconds.(seen below) Finally at the bottom of the page we have the final solution using ohms law to solve for the current!

    Use the link below to get back to Differential Equations Part 1 for the next problem on the table.

    Back to the start ... D.E. Ex1.

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