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cmathis
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Homework Statement
http://img202.imageshack.us/img202/8355/screenshot20130330at234.png
a) The switch closes at time t=0. What is the current flowing down through the inductor at time t=2ms?
b) What time will it take for the inductor to reach maximum current?
c) How long will it take for the inductor to reach current 2mA?
Homework Equations
i(t) = Ifinal + (Iinitial - Ifinal)[itex]e^{-t/\tau}[/itex]
The Attempt at a Solution
Current before the switch closes at t=0 is obviously 0A, so that means Iinitial = 0A.
After the switch closes and the circuit has come to steady state, you can represent the inductor by a short-circuit. The short circuit being in parallel with the two 3k[itex]\Omega[/itex] (equivalent to a 1.5k[itex]\Omega[/itex] resistor) effectively cancels out the two parallel 3k[itex]\Omega[/itex] resistors. So you're left with basically a single-loop circuit, so Ifinal would be equal to the current through the 3k[itex]\Omega[/itex] resistor in series with the voltage source, so Ifinal=0.001A.
[itex]\tau[/itex] = [itex]\frac{L}{R}[/itex] = [itex]\frac{1 H}{3000 \Omega}[/itex] = 3.33 x 10-4
That gives you equation:
i(t) = 0.001 - 0.001[itex]e^{-t/3.33 x 10^{-4}}[/itex]
Solving the equation at t=2ms give you i(2ms) = 0.998mA.
Am I correct so far?
For part b, wouldn't it take t=∞ amount of time since the current never really reaches 0.001A? The professor seems to expect a definite answer though...
For part c, is this a trick question? If I've done everything right, the inductor current never reaches 2mA.
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