# am I wrong or my teacher is wrong?

Discussion in 'Homework Help' started by ][ Shocked ][, Dec 22, 2009.

1. ### ][ Shocked ][ Thread Starter Member

Apr 13, 2009
22
0
hello everybody

I had a problem with a problem > how come

here it's

I think his answer is wrong !

because s = -10 + j2

and when you calculate I = V/Z

you will never get a degree of - 30 !

I hope someone corrects me if im wrong

2. ### mik3 Senior Member

Feb 4, 2008
4,846
63
The current is:

I=V/ZL

V=240<60

ZL=4<90

Thus I=A<ω=60<-30

A=240/4

ω=60-90

3. ### MasterSnow Member

Jan 18, 2009
22
0
Am I the only one that finds this problem a little strange? The input is certainly an exponetially decaying sinusoid...but the output response implies phasor analysis and therefore ignores the decaying exponential. If this is the case, then I agree with mik3's answer.

4. ### steveb Senior Member

Jul 3, 2008
2,433
469
Nope. I find it a little strange too.

5. ### steveb Senior Member

Jul 3, 2008
2,433
469
I had a couple of spare minutes, so I thought I would expand on this. Please see attached PDF for a mathematical verification that you are correct.

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6. ### ][ Shocked ][ Thread Starter Member

Apr 13, 2009
22
0
Z of L

is it = Ls or to L(wj) only?

there is no effect of s in the value of I?

because we had it depending on s

7. ### steveb Senior Member

Jul 3, 2008
2,433
469
It depends on what you are trying to do. Using s is more general and more appropriate in this case.

It's not clear what you mean that there is no effect of s in the value of I. The complex frequency is s, and the idea is that a linear system will have a response at the same complex frequency as the input. In this case the voltage is input with complex frequency s=-10+j2, so the current response is at the same complex frequency. An inductor is a linear device that has impedance which is a function of the complex frequency, i.e. Z=sL. If you know Z and V in the complex frequency domain, you can easily calculate the AC current response, i.e. I=V/Z.

Note that I chose to do an analysis in the time domain because I wanted to prove that the given problem was flawed, misleading or confusing. Once you see the time response, there is no way to relate it to any of the given answers, and it becomes clear that the question itself is ill-conceived. Phasors are all about response to pure sinusoids. In other words, they relate to Fourier Analysis. However, the problem talks about exciting the circuit with a decaying sinusoid, which puts the problem in the realm of Laplace Analysis.

8. ### ][ Shocked ][ Thread Starter Member

Apr 13, 2009
22
0
ok I got

thanks alot :")