# How do you differentiate between rms and peak voltage?

In the circuit above I've been asked to calculate the average power absorbed by the 10 Ω resistor. I managed to calculate a value for $$\V_0=40\sqrt2\,\angle{-25}\$$.

But I'm not sure if this is peak or rms voltage. I've spoken to some people and they say that the rms value is simply $$\40\sqrt2\$$, which doesn't really make sense to me, and I've spoken to other people who say to follow the $$\\frac{V_p}{\sqrt2}\$$ formula, which makes a lot more sense to me.

I'm not really sure who to believe, so I've come here seeking help.

• You should just ask the instructor for clarification. – Elliot Alderson May 9 at 13:50
• I love how the title of your question and your username form a pun. – Marcus Müller May 9 at 20:51
• That was a mean thing to say @MarcusMüller lol – Andy aka May 10 at 9:00

Noting this: -

When your circuit says this (my words in red): -

In the absence of any other information, AC voltages are always presumed to be RMS.

See this Wiki reference: -

the magnitudes of the voltage and current phasors V and I are the RMS values of the voltage and current, respectively).

And...

I managed to calculate a value

Well, unless you have decided to convert to sinusoidal peak values, you will have calculated a value based on an input of 8 volts RMS.

RMS of a sinusoid is the square root of 2 smaller than the peak amplitude ($$\\frac{V_m}{\sqrt2}\$$ ).This link should help.

In power system analysis texts we almost always use rms magnitudes. e.g. A current of 5@45 degrees amps is 5A rms or 7.07A peak.