# Confusion on Alternating Current [duplicate]

I was studying about alternating current, and I stumbled upon RMS. I clearly cannot understand what it means, the only thing my textbook says is: RMS current $I$ is related to the peak current $i_m$ by $I=0.707 i_m$

Why are we multiplying AC current by $0.707$?

Are there any sort of things I've missed learning?

• RMS in general means the effective value that the current/voltage takes. It has a definition. For any function $f(t)$ with period $T$, $f_{rms} = \sqrt{ \frac1T \int_0^T f^2(t) dt}$. In general the current is a $\sin$ or a $\cos$ function and hence the $\frac 1 {\sqrt2}$ factor. – sbp Feb 18 '18 at 9:43
• The answer is here: physics.stackexchange.com/a/41785/4962 – Steeven Feb 18 '18 at 9:49

"RMS" refers to "root-mean-square." It's one particular way of taking an average magnitude of something. In particular, the RMS of x is given by:

$$x_{rms}=\sqrt{\langle x^2\rangle}$$

Where $\langle a\rangle$ denotes the average of $a$. Or, to put that in words, it's the square root of the average of $x$ squared.

For a sinusoidal wave with current $I=i_m\sin\omega t$, the RMS current is given by:

$$I_{rms}=\sqrt{\langle i_m^2\sin^2\omega t\rangle}$$

which simplifies to:

$$I_{rms}=i_m\sqrt{\langle\sin^2\omega t\rangle}$$

$\langle\sin^2\omega t\rangle={1\over2}$. You can see this in a number of ways, for instance by considering $\sin^2\omega t+\cos^2\omega t=1$ and that $\sin$ and $\cos$ are just shifted versions of one another, so they have the same average, or just brute force the math. Putting this together, you get:

$$I_{rms}={i_m\over\sqrt2}\approx.707i_m$$

Note that this means the expression given in your book is only correct for a sinusoidal current.