How do I find voltage at time t given a voltage phasor?

[BryanDorais]

This is a problem from my midterm that I'm trying to understand before my final. The question is;

A voltage phasor is given by V = 12∠70° and the frequency is 60 Hz. What is the time-domain voltage at t = 1 ms?
a. 1.47 V
b. 5.11 V
c. -0.335 V
d. -4.73 V

As far as i know, the only equation relevant to this is V(t) = V_max*sin(ωt + φ). When i plug in the values, V(t = 0.001) = 12 sin(120π(0.001) + 7π/18), i get V = 12(0.9996), which does not match any of the possible results.

What am i getting wrong?

 

[magoo]

Your math looks fine to me.

 

[berkeman]

Agreed. As a sanity check, (1ms/16.7ms) * 360 degrees is 21.6 degrees, added to 70 degrees is very close to 90 degrees, so the sin() is very close to 1. So it seems like there should be an answer close to the max amplitude of 12 in the choices...

 

[BryanDorais]

Agreed. As a sanity check, (1ms/16.7ms) * 360 degrees is 21.6 degrees, added to 70 degrees is very close to 90 degrees, so the sin() is very close to 1. So it seems like there should be an answer close to the max amplitude of 12 in the choices...

Among the choices was "None of the above." I did not include that option since it was what i chose during the exam and it was marked as incorrect

 

[berkeman]

V(t) = Vmax*sin(ωt + φ).

What do you get if you use cos() instead...?

 

[berkeman]

https://en.wikipedia.org/wiki/Phasor

Definition
Euler's formula indicates that sinusoids can be represented mathematically as the sum of two complex-valued functions:

A ⋅ cos ⁡ ( ω t + θ ) = A ⋅ e i ( ω t + θ ) + e − i ( ω t + θ ) 2 , {\displaystyle A\cdot \cos(\omega t+\theta )=A\cdot {\frac {e^{i(\omega t+\theta )}+e^{-i(\omega t+\theta )}}{2}},}

[a]
or as the real part of one of the functions:

A ⋅ cos ⁡ ( ω t + θ ) = Re ⁡ { A ⋅ e i ( ω t + θ ) } = Re ⁡ { A e i θ ⋅ e i ω t } . {\displaystyle {\begin{aligned}A\cdot \cos(\omega t+\theta )=\operatorname {Re} \{A\cdot e^{i(\omega t+\theta )}\}=\operatorname {Re} \{Ae^{i\theta }\cdot e^{i\omega t}\}.\end{aligned}}}

The function A ⋅ e i ( ω t + θ ) {\displaystyle A\cdot e^{i(\omega t+\theta )}}

is the analytic representation of A ⋅ cos ⁡ ( ω t + θ ) . {\displaystyle A\cdot \cos(\omega t+\theta ).}

Figure 2 depicts it as a rotating vector in a complex plane. It is sometimes convenient to refer to the entire function as a phasor,[11] as we do in the next section. But the term phasor usually implies just the static vector , A e i θ . {\displaystyle ,Ae^{i\theta }.}

An even more compact representation of a phasor is the angle notation: A ∠ θ . {\displaystyle A\angle \theta .\,}

See also vector notation.

 

[BryanDorais]

What do you get if you use cos() instead...?

If i use Cosine the answer is -0.355(C). How did you know to use Cos? The only formula i have uses Sin only

 

[gneill]

A couple of points to ponder:

Phasors are usually assumed to be rms values. Time domain versions are generally peak values.

You have a choice of basis functions for the time domain, either cosine or sine. Cosine may be more common

 

[berkeman]

If i use Cosine the answer is -0.355(C). How did you know to use Cos? The only formula i have uses Sin only

It's been so long since I used them, I don't remember if there is a reason that you would choose one over the other. There is the wikipedia definition above, and it kind of makes sense to use the cos() form because you'd like the value to be on the Re axis for an argument of zero, I would think.

 

[BryanDorais]

Thank you very much for the assistance.

 

[Drakkith]

If i use Cosine the answer is -0.355(C). How did you know to use Cos? The only formula i have uses Sin only

Note that you can convert between sin and cos if you're given one and need the other (my instructor likes to give us things in terms of sines, forcing us to convert to cosine):
$sin(ωt+θ)=cos(ωt+θ-\frac{π}{2})$
$cos(ωt+θ)=sin(ωt+θ+\frac{π}{2})$
(θ may be zero or negative)