Laplace transforms for the transient analysis of networks

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  • #1
agata78
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Homework Statement



Use Laplace transforms to derive an expression for the current flowing in the circuit shown in the figure, given that i = 0 when t=0

Homework Equations



Expression for the current in an LR series circuit
I(t) = V / R (1-e-Rt/L)

V - Volts (6V)
R - Ohms (10Ω)
L - Henries (2H)
t - Seconds
e - The base of the Natural Logarithm = 2.71828

Steady State Current
I = V / R
I = 6 / 10
I = 0.6 A

Time Constant
τ = L / R
τ = 2 / 10
τ = 0.2 seconds

Instantaneous Current
I(t) = Vs / R (1 - e-Rt/L)
I(t) = 6 / 10 (1 - e-10x0.2/2)
I(t) = 6 / 10 (1 - e-2/2)
I(t) = 6 / 10 (1 - 0.36787)
I(t) = 0.6 (0.63213)
I(t) = 0.379
I(t) = 0.38 A
 

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Answers and Replies

  • #2
gneill
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I don't know what you were calculating for I(t) in the Relevant Equations section above since no value for t was given in the problem statement, nor were you asked to calculate I for a specific value of t. But anyways, you haven't addressed the question as stated...
 
  • #3
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The first thing you need to do is write the differential equation for the circuit. This differential equation should involve I, dI/dt, V, R, and L. After you do that, you will be able to take the Laplace Transform of the equation. So.... what is your differential equation?

Chet
 
  • #4
agata78
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I found a very similair example, but dont know how to use my numbers. Im running out of the time now. Can you help?

The voltage, V, will be the sum of the two voltages, VR and VL. Where VR = Ri(t) and VL = L(di(t)/dt).

Thus, 6 = VR + VL = 10i(t) + 2 di(t) /dt


Applying the Laplace Transforms gives:

L(V) = L[Ri(t) + Ldi(t)/dt] = L(Ri(t) + L[L di(t)/dt] )


L(V) = RL(i(t)) + LL[di(t)/dt]


Using the table of standard Laplace transforms gives:

V / s - RI(s) + sL[I(s) - i(0)]


V / s = RI(s) + sL[I(s) - 0] = RI(s) + sLI(s) = I(s) x (R + sL)

or I(s) = V / s(R+sL)
 
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  • #5
gneill
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Just plug in your component values. You've got R = 10 and L = 2.

Reduce the (R + sL) term to the form (s + n) by a suitable manipulation.
 
  • #6
agata78
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ok,

I(s)=6/ (s(10+2s))= 3/ (s(s+5))
I(t)= 3+ 3e -5t = 3(1+e -5t)

Am i right ? And what next? I think i will have to deal with partial fractions.

Is= 3/ s(s+5)= A /s + B / (s+5)
 
  • #7
gneill
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ok,

I(s)=6 (s(10+2s))= 3/ (s(s+5))
I(t)= 3+ 3e -5t = 3(1+e -5t)

Am i right ? And what next? I think i will have to deal with partial fractions.

Is= 3/ s(s+5)= A /s + B / (s+5)

You need to do the partial fractions first, before you try to write the time domain function! Your I(t) above is not correct (largely due to it ending up with the wrong constant "3").

So do your partial fractions, then transform term by term.

If you're working with good tables you'll probably find an entry for the form ##\frac{\alpha}{s(s + \alpha)}## which you can use directly and skip the partial fractions.
 
  • #8
agata78
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I had to calculate partial fractions cause my table is rubbish.

A= -6/5
B= 3/5

What to do next?

(-6/5 )/s + (3/5) (10+2s)

i(t) = -6/5 +( 3/10)e -2t


The answer should be 0.6(1-e -5t)

Could you tell me where i made mistake?
 
Last edited:
  • #9
agata78
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ok i know where i made a mistake. I really need online calculator!

thank you!
 
  • #10
gneill
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Sorry abut that, I was off to watch a football game. As you've no doubt determined, your value for A in the partial fraction decomposition of the expression was incorrect. Both values A and B should have been equal to 3/5.

You can find decent Laplace transform tables on the web. Google "Laplace Transform".
 

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