Superposition with AC voltage source and DC current source

[eehelp150]

Homework Statement

Using the superposition principle, determine the current i(t.

2. Homework Equations
Zc = 1/jwC
Zl = jwL
V = I*R
I = V/Z

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?

 

[gneill]

Check your capacitor impedance.

 

[eehelp150]

Check your capacitor impedance.

Is it -10j instead of -0.1j?

 

[gneill]

Is it -10j instead of -0.1j?

It is indeed!

 

[eehelp150]

It is indeed!

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

 

[gneill]

That looks good!

 

[gneill]

That looks good!

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.

 

[gneill]

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

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

 

[eehelp150]

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

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

 

[gneill]

Yes, that looks good.

 

[eehelp150]

Yes, that looks good.

Thank you so much for your help!

 

[eehelp150]

Yes, that looks good.

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

 

[gneill]

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.

 

[eehelp150]

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.

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

 

[gneill]

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

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).

 

[eehelp150]

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).

So:

I(t) = sqrt(2)/2 * cos(10t - 45°) - 2

 

[gneill]

Yes. That looks better!