Superposition with AC voltage source and DC current source

1. Oct 10, 2016

eehelp150

1. The problem statement, all variables and given/known data
Using the superposition principle, determine the current i(t.

2. Relevant equations

Zc = 1/jwC
Zl = jwL
V = I*R
I = V/Z

3. The attempt at a solution
First, I converted inductor/capacitor to impedance:
L = 1.5H -> jwL = j * 10 * 1.5 = 15j
C = 10mf = 1/(jwC) = 1/(j*10*10*10^-3) = -0.1j

Then, I transformed the 3A DC current source and 10 ohm resistor into a DC voltage source:

I then shorted the 30V DC source to find i(t) with only Vs(t) = 10cos(10t)V
After shorting the 30V DC source, C2 and R3 are in parallel. Their impedance is:

(0.1j * 10)/(10-0.1j) = -j/(10-0.1j)
After rationalizing: (0.1 - j)/100.01

Now that, L2 and R4 are in series, total impedance:
15j + 5 + (0.1 - j)/100.01 = (500.15-1499.15j)/100.01 = 15.8 < -71.55°.
I = V/Z = (10<0°)/(15.88<-71.55° ) = 10/15.88<71.55°A

Now I short the AC source:

The DC source had a frequency of 0 rad/s, so the impedance of C3 is infinity (1/0) and L3 is 0 (j*0*L).
The inductor is shorted and the capacitor is opened. There is a voltage source of 30V and a combined resistance of 15ohms. V=IR -> I = 30/15 = 2A<0°
[(500.15/100.01 + 2) + (1499.15j/100.01)]
Is 16.54<64.96° A the correct answer?

Last edited: Oct 10, 2016
2. Oct 10, 2016

Staff: Mentor

3. Oct 10, 2016

eehelp150

Is it -10j instead of -0.1j?

4. Oct 10, 2016

Staff: Mentor

It is indeed!

5. Oct 10, 2016

eehelp150

My new current for part 1 is 10/14.14 <-45°
Is the final answer (2.5 + 0.5j) A?

6. Oct 10, 2016

Staff: Mentor

That looks good!

7. Oct 10, 2016

Staff: Mentor

Oops, hang on. I think your signs may be incorrect. The diagram defines the current direction as flowing out of the AC source. So the DC contribution to that must be negative.

8. Oct 10, 2016

Staff: Mentor

Also, if the AC current angle is -45°, its imaginary part should be negative...

9. Oct 10, 2016

eehelp150

It was negative on my paper, I don't know why I typed positive.
(-1.5 - 0.5j)A, correct?

10. Oct 10, 2016

Staff: Mentor

Yes, that looks good.

11. Oct 10, 2016

eehelp150

Thank you so much for your help!

12. Oct 13, 2016

eehelp150

the final answer is -1.5 - 0.5j amps
If w = 10, would the answer in time domain be:
$$\sqrt{2.5} * cos(10t + 18.435°)A$$
?

13. Oct 13, 2016

Staff: Mentor

The DC and AC components must be kept separate. There will be a constant DC current with an AC component superimposed on it, or if you like, the AC signal will be offset by a DC bias.

14. Oct 13, 2016

eehelp150

How would I write it out? I need the answer in the time domain.

15. Oct 13, 2016

Staff: Mentor

A DC current is just a fixed value in the time domain; a number. So your result should have the form:

$I(t) = I_{dc} + I_{ac}$

where $I_{ac}$ is your cosine expression (the part that varies with time).

16. Oct 13, 2016

eehelp150

So:

I(t) = sqrt(2)/2 * cos(10t - 45°) - 2

17. Oct 13, 2016

Staff: Mentor

Yes. That looks better!