Current Through an Inductor

1. Oct 28, 2017

RoyalFlush100

1. The problem statement, all variables and given/known data
"The switch in the circuit in the figure has been closed for a long time and is opened at t = 0. Find i(50 ms)."
I attached an image of the circuit below.

2. Relevant equations
KVL:
Ri + q(t)/C - q(0)/C + vc(t) + Li' = vs(t)
Li'' + Ri' + i/C = vs'(t)
Where vc is voltage across the capacitor and vs is the source voltage. Q is the charge flowing through the current.

3. The attempt at a solution
First I found i(t) through the inductor and the vc prior to the switch opening:
30 = 10i --> i = 3 A
20-vc = 5i --> vc = 5 V

Now, I applied the second equation and this is where I become confused. It has to do with the application of Kirchhoff's Voltage Law. i goes through the bottom of the capacitor, so I don't know how to set up the equation.
Li'' + Ri' + i/C = 0
i'' + 10i' + 25i = 0 This?
i'' + 10i' - 25i = 0 Or this?
Likewise, I'll need to find i'(0) after. Would that be:
Ri + q(t)/C - q(0)/C + vc(t) + Li' = vs(t)
10 + 5 + i' = 10 This?
10 - 5 + i' = 10 Or this?

I think it should be minus in both cases, but the numbers behave weirdly when I try that.

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• Circuit.gif
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2. Oct 29, 2017

scottdave

It looks like you just took the derivative of both sides of your first equation to get the second equation:
Li'' + Ri' + i/C = vs'(t)
Why did you do this? You will need to know an extra initial condition to solve the second order equation.
How about just solve for an expression for q(t), then take the derivative of that to get I(t) ?
What methods are you familiar with in solving this type of differential equation?

3. Oct 29, 2017

RoyalFlush100

This is the way I typically do these types:
1) Take the derivative, giving a second order differential equation. Find i (the roots of which will be real and unequal, real and equal, or complex).
2) Solve for i(0), which will give an equation involving the two constants.
3) Take the first equation and solve for i'(0). Then, take the derivative of the equation for i and solve it for i'(0). With both these equations, the two constants in the equation for i can be solved.

The main thing I'm unsure of though is current travelling through the bottom of the capacitor. I'm assuming that means the voltage drop will be negative (assuming vc is positive), but the numbers of working out oddly when I try that.

Also, the reason I didn't solve directly for q is because that would just be a more difficult second order differential equation, since i' would just be q''. It would also require me to know q(0) and there'd be more constants in the equations.

4. Oct 29, 2017

scottdave

Ok. I just wasn't sure what you were asking. Yes it is still 2nd order as you have q' is I and i' is q''. If current is flowing into the positive end of a device, then that device is using energy. If current is flowing out of the positive terminal, then the device is providing energy, if that helps you.

5. Oct 29, 2017

RoyalFlush100

Okay, so in that case it'd be minus then. So:
i'' + 10i' - 25i = 0
s^2 + 10s - 25 = 0
s = -10/2 +/- sqrt(100-4(-25))/2
s = 5 +/- sqrt(200)/2
s = 12.071 or -2.0711

i(t) = Ae^(12.071t) + Be^(-2.0711t)
i(0) = 3 = A + B
B = 3 - A

Now:
10 - 5 + i' = 10
i' = 5

i'(t) = 12.071Ae^(12.071t) - 2.0711Be^(-2.0711t)
i'(0) = 5 =12.071A - 2.0711B
5 = 12.071A - 2.0711(3-A)
5 = 12.071A - 6.2133 + 2.0711A
11.2133 = 14.1421A
A = 0.7929
B = 3 - 0.7929 = 2.2071

i(t) = 0.7929e^(12.071t) + 2.2071e^(-2.0711t)

The problem is, this makes no sense. As t approaches infinity, the problem will approach infinity, but just by looking at the circuit I can tell it should approach 0.

6. Oct 29, 2017

scottdave

You know it needs to be a damped oscillation, but you are not getting that (in fact the exponential has positive exponent, which is an unstable system). So you know you have a problem somewhere.

When you take the Laplace Transform of the second derivative, you get
L(i’’(t)) = s²*I(s) – s*i(0) – i’(0)
Laplace of the first derivative is L(i’(t)) = s*I(s) – i(0)
See this:
http://mathfaculty.fullerton.edu/mathews/c2003/laplacetransformmod.html
You already determined that i(0) = 3 A, but that is not reflected in your formulas.

7. Oct 30, 2017

At first I think the voltage across capacitor VC=q/C. Since q=qo-integral(idt)|t=0 to t the voltage drop in the last loop has to be :
Ldi/dt+Ri+[(qo-integral(idt)]/C+10=0 then take the derivate:
Li”+Ri’-i/C=0
The solution-see the attached link-will be i=Ae^(x1.t)+Be^(x2.t) and if i=0 for t=infinite
Then A=0
Solve the above equation by using y=Be^(x2.t) for t=0 y=B and find the B
https://www.wikihow.com/Solve-Differential-Equations

8. Oct 31, 2017

Correction:
Since eventually the voltage across the capacitor has to be 10 V the correct capacitor charge will be:

Then the voltage drop equation has to be:

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File size:
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Views:
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9. Oct 31, 2017

If we take 10 V [of the voltage source] as positive the initial Vc=-5 V.[qo=-5*.04] .The integral i.dt has to be in the limits 0 and t.

10. Nov 1, 2017

I think you-almost-solved the problem. You have only to revise some points.
The voltage on capacitor terminal before switch opening is +5 volt-indeed as you said. At t=infinite the voltage will be -10 volt. At first qo=5*.04=0.2 coulomb will decrease up to 0 and the current will continue to flow in the same direction.
If we'll consider the last -10 V as positive [+10 V] then qo=-0.2 coulomb. So the KVL in the loop will be:
L.i'+10*i+[-0.2+(∫i/dt)]|t=0 to t]/C=10
L.i''+10*i'+i/C=0
s1,2=[-10+/-√(100-4/.04)]/2
s1=s2=-5
i=Io.e^(-5.t) i'=-5.Io.e^(-5.t) (∫i/dt)|t=0 to t|=(-1/5.e^(-5.t)+1/5).Io
if t=0 -5.Io+10.Io-0.2/0.04=10 5.Io=15 Io=3
i=3.e^(-5.t)
Check:
t=infinite q(infinite)= -5*0.04+(-0+3/5)=0.4 Vc=0.4/.04=10 V