Math 302, Differential Equations JHU

An example of a circuit

This circuit has a diode in it which acts as a resistor with resistance R1=10 ohms if the current through it is positive and R2=10000 ohms otherwise. The voltage is E(t)=110*cos(w*t) where w=2*PI*60. From Kirchoff's laws we obtain:

I1 + I2 + I3 = 0
L*dI1/dt - RD*I2 = E(t)
Q/C + RD*I2 + R * I3 = 0
dQ/dt = I3
RD = R1 if I2>0 and R2 if I2 < 0
Thus
I2 = -I1 - I3 = -I1 - dQ/dt
and we can eliminate I2 from the second and third equation:
L*dI1/dt - RD*(-I1 - dQ/dt) = E(t) 
Q/C + RD * (-I1 - dQ/dt) + R * (dQ/dt) = 0

View Instructions on using the Applet

Marek Rychlik (rychlik@u.arizona.edu)

Author's Home Page: http://alamos.math.arizona.edu

 


This page last modified Sun Sep 12 12:32:06 2004
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