# RLC circuit

Discussion in 'Homework Help' started by yoamocuy, Jan 23, 2010.

1. ### yoamocuy Thread Starter Active Member

Oct 7, 2009
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0
Find ic(t) for t>0.

ok, so this is what I've done so far.

1) I evaluated at t=0- and found current across the inductor to be 2A because the capacitor is open and the inductor is short so all the current will go across it rather than go across the resistor. That also means that the current across the capacitor is 0.

2) I evaluated at t=0+. Now the original current source is off but the inductor is acting as a 2 A current source. The current flows through the inductor rather htan the resistor so Il is still 2A but Ic is now -2 A. Vc(0+) is still 0 V.

3) use the forumala α=1/(2*R*C) and ωo=1/sqrt(L*C). α=4000 and ωo=4472.14. Since α<ωo we have to use the formula:
ic(t)=e^(-α*t)*(B1cos(ωd*t)+B2*sin(ωd*t)) where ωd=sqrt(ωo^2-α^2)

4) I use this formula at t=0+ and everything cancels out down to ic(0+)=B1 and ic(0+) was found to be -2 A so B1=-2.

5) I take the derivative of this equation to get:
di/dt=α*e^(-α*t)*(cos(ωd*t)*(B2*ωd-α*B1)-sin(ωd*t)*(B1*ωd+α*b2))
di/dt is also = Vc/L

6) By using this formula at t=0+ I get the formula α*(2000*B2-4000*B1)=Vc/L where Vc is =0. After plugging in values for α and B1 I get B2=-4.

7) By plugging all these values back into the base equation for ic(t) I get:
ic(t)= e^(-4000t)(-2cos(2000t)-4sin(2000t))

The answer at the back of the book is ic(t)= e^(-4000t)(-2cos(2000t)+4sin(2000t))

I can't figure out why B2=4 rather than -4... help please :/

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
If the input source is a 2Amp step.....

The inductor current cannot be 2A at t=0+ since it was zero at t=0-. All the 2A current initially (at t=0+) flows in the capacitor. It is not until t=∞ that the inductor current is a constant 2A.

If the source is rather representing an initial condition in L of 2A then your analysis approach looks OK with the exception that the current polarity would be such that it is flowing out of the inductor top terminal (and into C top terminal) and hence the first unknown constant (B1?) would be +2 rather than -2. The "equivalent" current source direction tends to confuse matters when translating between the initial condition view and the equivalent view.

I'll have a closer look and try to solve the problem completely.

Last edited: Jan 24, 2010
3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Based on the circuit configuration as shown in your diagram the capacitor current will be

ic(t)=2exp(-4000t)(cos(2000t)-2sin(2000t))

which is essentially the same as your book solution but differing in the overall sign.

If the book circuit diagram is identical to that shown in your diagram then I would dispute the book solution.