Phasor Analysis of a Circuit

[PenDraconis]

Homework Statement

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

Homework Equations

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

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?

 

[da_nang]

$V = Acos(\omega t+\theta)$ $\leftrightarrow$ $V = Ae^{j\theta}$
What does Euler's identity say about this part?

 

[nrqed]

Homework Statement

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

Homework Equations

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

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 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.

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?

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.

 

[PenDraconis]

You are right that there is a point that sometimes people do not explain this point 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.

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 into a V = IR equation to find the value of C.

 

[nrqed]

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 into a V = IR equation to find the value of C.

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.

 

[PenDraconis]

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.

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.