# Homework Help: Phasor Analysis of a Circuit

1. Sep 10, 2014

### PenDraconis

1. The problem statement, all variables and given/known data

I have $V = 25cos(1000t)$ and $I = Acos(1000t)$ for a circuit, I'd like to make those into phasors.

2. Relevant equations
$V = Acos(\omega t+\theta)$ $\leftrightarrow$ $V = Ae^{j\theta}$

3. The attempt at a solution
It seems like this is relatively simple, but that seems to be my problem, these are my two confusions:

$V = 25cos(1000t) = 25e^{j(1000t)}$ ?​

Or is it:

$V = 25cos(1000t) = 25e^{0}$ ?​

I'm not quite sure which route to go towards, could anyone possible point me towards the right direction or explain why it's one over the other? My main confusion is that if I change $V = 25e^{j(1000t)}$ back into it's cosine and sine components (using Euler's Identity) I end up with:
$V = 25cos(1000t)+jsin(1000t)$​
...is the imaginary part ignored?

Last edited: Sep 10, 2014
2. Sep 10, 2014

### da_nang

$V = Acos(\omega t+\theta)$ $\leftrightarrow$ $V = Ae^{j\theta}$

3. Sep 10, 2014

### nrqed

I am not sure what is your reasoning here. These two cannot be equal as the left side is a function of time and not the right side.
You are right that there is a point that sometimes people do not explain well enough (or not at all). The key point is that the physical voltage is the real part of the corresponding phasor . So yes, in going from the phasor back to the actual voltage, one must drop the imaginary part.

Last edited: Sep 10, 2014
4. Sep 10, 2014

### PenDraconis

Thank you! I was just confused as to why this point was glossed over whenever I mentioned it in class, I wasn't able to ever get a straight answer.

The second part of the circuit I seem to understand, basically I have a RLC circuit and I must used phasor analysis to find the capacitance given R = 3, L = $j4$, C =$\frac{-j}{1000C}$, $V = 25e^{j1000t}$, $I = Ae^{j1000t}$. It's simply finding the total resistance of the circuit and plugging that in to a V = IR equation to find the value of C.

Last edited: Sep 10, 2014
5. Sep 10, 2014

### nrqed

Well, you cannot simply use Ohm's law. You also need to add the impedance of the inductor and of the capacitor. Can you calculate those? If the three elements are in series, you simply have
$V = (R+ X_C + X_L) I$ and solve. Watch out for the units, they must be consistent.

6. Sep 10, 2014

### PenDraconis

Yup! You're right.

Those are the impedances I gave. The resistor and inductor are in series and they're both in parallel with the capacitor.