# Application to AC circuits

1. Oct 18, 2015

### Calpalned

1. The problem statement, all variables and given/known data
Prove that current $I = I_0\sin{\omega t}$ can be rewritten as $I = I_0 e^{i\omega t}$

2. Relevant equations
Euler's formula $re^{i\theta} = r(\cos{\theta}+i\sin{\theta})$

3. The attempt at a solution
If $r = I_0$ and $\theta = \omega t$ then $re^{i\theta} = r(\cos{\theta}+i\sin{\theta})$ will be $I = I_0e^{i\omega t} = I_0(\cos{\omega t}+i\sin{\omega t})$. The left hand side $I = I_0 e^{i\omega t}$ is correct, but the right hand side is not equal to $I = I_0\sin{\omega t}$
Thank you

2. Oct 19, 2015

### Staff: Mentor

I think what you're looking at is the introduction of phasors to your toolkit. As you have spotted, eiωt is not formally congruent to sinωt. However, if one considers a phasor to be a rotating vector (a vector that rotates in direction around the origin with a constant angular frequency ω), then a projection of that vector on the imaginary axis does indeed follow sinωt. Similarly, its projection on the real axis follows cosωt. Now, eiωt can be considered to be a rotating vector, as t is a time variable. So, for example,$I_0 e^{i\omega t}$ would be a phasor of magnitude $I_0$ rotating at a rate of $\omega$ radians per second.

Typically (but not always) all the phasors in a given circuit have the same angular frequency, and the convention is to drop the common $e^{i \omega t}$ from the notation. Declare that $I$ is a phasor and the $e^{i \omega t}$ is implied.

Does that help?