How to Calculate RMS Current in a Simple Circuit

[elctronoob]

Homework Statement

Sorry for what will likely be considered a noob question but I've been stuck on it for a while and its only at the start of my assignment :(
Simple circuit
Vac = 50Sin200πt, R1 = 40Ω
Calculate the value of the RMS current

Homework Equations

Vt = Vmax sin2πft = Vmax sinωt

The Attempt at a Solution

I can't figure out how to do this when we are given no value for time at all. I think i have it worked out at say 1ms
Vt = 50sin((200π)(0.001))
= 50sin(.628318)
= 54.83V

Vrms = 54.83 * .7071 = 38.77

Irms = Vrms / R = 38.77 / 40 = 0.969 amperes

We have been told that time is intended to be an unknown variable. It's been a long time since I've look at any of these types of problems, can someone help me out on what I'm missing here please

Thanks

 

[PWiz]

I can't figure out how to do this when we are given no value for time at all.

The r.m.s. can be calculated just from the peak value if the current is sinusoidal, as is the case here. You can calculate the peak current from the peak voltage. Also remember that the average of the sin^2 function is half. Can you take it from here?

 

[elctronoob]

Thanks for the reply PWiz...any formula I can find for these calculations seem to require a time.
The only one I can find without t is Vpeak = Vrms√2...=> Vpeak/√2 = Vrms
Vrms / R = Irms

I was thinkin maybe I could half the Vac given in the question, but that still involves a t variable...so frustrating :(

 

[PWiz]

The only one I can find without t is Vpeak = Vrms√2...=> Vpeak/√2 = Vrms
Vrms / R = Irms

That's pretty much what I was talking about.

 

[elctronoob]

That's pretty much what I was talking about.

Still a bit confused, so can I say
Vpeak = 50Sin200πt / 2 = 25Sin200πt
Vrms = 25Sin200πt / √2
This still requires a value for t, no?

 

[elctronoob]

did i just miss that Sin200πt is representative of 1 volt?

Going by this conversion site, http://www.learningaboutelectronics.com/Articles/Peak-voltage-calculator.php#rms

Vrms = 25 / √2 = 17.7V
Irms = 17.7 / 40 = 0.4425 amperes

Does this look correct?

Thanks for your time and help PWiz...I've been looking at this for far too long

 

[elctronoob]

did i just miss that Sin200πt is representative of 1 volt?

Going by this conversion site, http://www.learningaboutelectronics.com/Articles/Peak-voltage-calculator.php#rms

Vrms = 25 / √2 = 17.7V
Irms = 17.7 / 40 = 0.4425 amperes

Does this look correct?

Thanks for your time and help PWiz...I've been looking at this for far too long

Could someone confirm that my thinking on this question is correct for me please? I'm basically just disregarding the Sin200πt and calculating the Vrms and Irms derived from from the 50 value at the start of the expression, i.e 50V. This seems incorrect to me, could someone just check my logic here please

 

[gneill]

electronoob,

For a sinusoid (sine or cosine) the rms value is $1 / \sqrt{2}$ its peak value. This is entirely independent of the frequency of the sinusoid. In this problem you're given a sinusoidal voltage signal with a peak value of $50 ~V$, so the rms voltage is $50 / \sqrt{2}~~V$.

It really is that simple for sinusoids. There are other relationships between peak and rms values for other signal shapes.

 

[elctronoob]

electronoob,

For a sinusoid (sine or cosine) the rms value is $1 / \sqrt{2}$ its peak value. This is entirely independent of the frequency of the sinusoid. In this problem you're given a sinusoidal voltage signal with a peak value of $50 ~V$, so the rms voltage is $50 / \sqrt{2}~~V$.

It really is that simple for sinusoids. There are other relationships between peak and rms values for other signal shapes.

Thanks for that gneill, much appreciated