Engineering Simple R/L Circuit Current Problem

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The discussion revolves around a simple R/L circuit problem involving an inductor and a switch. The initial current is zero, and after closing the switch, the final current is calculated to be 1mA. The time constant (τ) is determined to be 3.33 x 10^-4 seconds, leading to the equation for current over time. It is noted that the inductor never truly reaches the maximum current of 1mA, suggesting an infinite time to reach that point. The conversation also touches on the time required for the inductor to reach a current of 2mA, indicating that it is not achievable under the given circuit conditions.
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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)e^{-t/\tau}



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\Omega (equivalent to a 1.5k\Omega resistor) effectively cancels out the two parallel 3k\Omega resistors. So you're left with basically a single-loop circuit, so Ifinal would be equal to the current through the 3k\Omega resistor in series with the voltage source, so Ifinal=0.001A.

\tau = \frac{L}{R} = \frac{1 H}{3000 \Omega} = 3.33 x 10-4

That gives you equation:
i(t) = 0.001 - 0.001e^{-t/3.33 x 10^{-4}}

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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R2 and R3 will influence the value of the time constant. Try turning the voltage source and resistor network into its Thevenin equivalent, with the inductor serving as the 'load'.

Your query about part c has merit.
 
(oops... ignore this post. I'll fix it soon.)
 
I'd get:

http://img145.imageshack.us/img145/5180/screenshot20130330at324.png

right?

So the above equation is right, but now \tau = 0.001.

So i(2ms) = 0.865mA.
 
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Yup. That looks good.
 
Ok great. But what about part b? The limiting current in the inductor is 1mA, but it never really reaches that current, so t=infinity.
 
cmathis said:
Ok great. But what about part b? The limiting current in the inductor is 1mA, but it never really reaches that current, so t=infinity.

True. (Although as an engineering rule of thumb, it's common practice to assume that all the excitement is essentially over after five time constants have elapsed)
 

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