So I had a quick conceptional question about phasors that I had trouble to understand on my own.

My book establishes this relationship:

And also we know that these forms are equal

Where the right hand side of (9.25) is in the "polar form" of (9.15). So then according to 9.15, its equivalent to the Exponential form:

$$V_m\cos{(\omega t + \phi)} = V_{m} \angle\phi = V_m e^{j\phi}$$ and since

$$e^{j\phi} = \cos{\phi} + j\sin{\phi}$$

doesn't this mean

$$V_m\cos{(\omega t + \phi)} = V_m(\cos{\phi} + j\sin{\phi})$$

So my question is, where is the sine term in the above equation? Since on the RHS we have cosine and sine, but on the LHS we have cosine only. Like I'm just confused on whether the equation is true or not or if I made a mistake. Seems to me the sine is "missing" and when we represent a cosine $$V_m\cos{(\omega t + \phi)}$$ as $$V_m e^{j\omega}$$

it is wrong since the $$e^{j\omega}$$ has both a cosine and sine component to it.

• I'd change your book because the 1st picture is wrong. In the phasor domain, the representation of a voltage is with $V_{RMS}$ and not $V_M$. Jul 14 at 8:48
• It's $V = V_m\cdot\exp\left(j\left(\omega t+\phi_v\right)\right)$ and, by convention, only the real part is used and the $\exp\left(j\omega t\right)$ part is dropped, for phasors. (When computing power, to secure $\phi_v-\phi_i$, the conjugate of the current is used when multiplying it by voltage.) Jul 14 at 9:34

$$v(t)=V_m\cos\left(\omega t+\phi\right)\Leftrightarrow {\bf V}=V_m\angle\phi$$

You have interpreted this as meaning $$V_m\cos\left(\omega t+\phi\right)=V_m\angle\phi$$ but this is not correct.

The $$\\Leftrightarrow\$$ symbol doesn't mean that all the terms on its left side are numerically equal to the terms on its right side. It only means that the equation on its left implies the equation on its right and vice versa. This relationship is called an if and only if or necessary and sufficient logical relationship.

That is, the equations on the two sides give equivalent information.

It doesn't mean we can equate the rhs of the equation on the left with the rhs of the equation on the right.

The complex representation of voltage is:

$$\textbf{V}=V_m\cdot\exp\left(j\left[\omega\,t+\phi_v\right]\right)=V_m\cdot\exp\left(j\,\phi_v\right)\cdot\exp\left(j\,\omega\,t\right)\tag{1}\label{eq1}$$

The vector length is always just $$\V_m\$$. But the instantaneous value is taken to be the real part, or $$\V_t=\mathscr{Re}\left(\textbf{V}\right)=V_m\cdot\cos\left(\omega t+\phi_v\right)\$$. The imaginary part simply retains the magnitude throughout the rotation.

For phasors, it's assumed that only one specific angular frequency is under discussion ($$\\omega\$$.) (Everything is rotating at the same rate. They only may have different phase offsets.) So the last factor in eq. $$\\ref{eq1}\$$ is dropped, by convention, leaving only $$\V_m\,\angle\, \phi_v\$$ (as the dropped factor is assumed.) Recovering the fuller Euler's expression is just a matter of adding back the dropped factor.

I used your term, $$\V_m\$$. Sometimes, this constant is specified as the RMS value and not the peak. (Context is the only way I know about how one may discern which is the one under discussion at the time.)