Root-mean squared values and a.c. emf

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The discussion focuses on calculating root-mean squared (rms) values for voltage and current in relation to average power supplied in an AC circuit. The peak voltage for UK mains electricity is calculated as approximately 325 V, and the rms current is determined to be about 0.115 A, leading to a peak current of approximately 0.163 A. The average power is correctly calculated as 26.45 W using the rms voltage and Ohm's Law. It is confirmed that using rms values is appropriate for finding average power, and the initial calculations for parts 1 and 2 are deemed correct. The conversation emphasizes the importance of understanding integration in power calculations, particularly in AC circuits.
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Homework Statement
Hello, I have a question which I have answered but I am unsure whether my solutions apply the suitable formula and techniques as required by the problem. I am relatively inexperienced in using root-mean squared values which is why I am a little uncertain.

A sinusoidally varying alternating EMF is described by the equation the equation below.
V = V0 sin ω t
It has been found to have a rms value of 230 V and a frequency of 50 Hz.
1. Find the values of V0 and of ω?
2. If the emf were applied across a 2000 Ω resistor, calculate the maximum and rms values of the current?
3. Find the value of the average power supplied?

My main uncertainty concerns part 3. I would really appreciate if anyone could offer any guidance or provide feedback to my solutions 👍 (Apologies for the layout of my post, I am learning LaTeX and can apply it in documents but I not sure how to write it on the physics forums yet, although I am still learning)
Relevant Equations
ω = 2πf
Vrms=V0/√2
1. We are given the root-mean squared value for the voltage at 230V. Therefore by rearranging the equation Vrms=V0/√2 one can find the value of the peak voltage; ie. V0= Vrms *√2
V0=230*√2
V0=325.269... ~ 325 V (which is the peak voltage of mains electricity in the UK).
To find ω use the formula ω = 2πf
Thus, ω = 2π*50
ω = 314.159... ~ 314 s^-1

2. Using Ohm's Law; I=V/R
Thus, Irms=Vrms/R
I rms = 230/2000
I rms = 0.115 A

Rearranging the rms formula for peak current;
I rms= I0/√2
I0=I rms * √2
I0= 0.115 *√2
I0= 0.16263... ~ 0.163 A

3. Using P=VI or P=V^2/R
P=(V rms) ^2/R
P=(230)^2/2000
P=26.45 W

My query relates to question 3, would it be correct to use the root-mean squared value for voltage (and current also if using P=VI) since we are trying to find the average power supplied. As root-mean squared values are type of averaging technique I thought this would be more appropriate than using the peak voltage and/or current.

Thank you to anyone who replies. 😁
 
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AN630078 said:
would it be correct to use the root-mean squared value for voltage (and current also if using P=VI) since we are trying to find the average power supplied.
Yes. This is why the rms voltage is interesting.
The power at any instant is V(t)2/R, so the average power is to be had by integrating over a cycle: ##(1/T)\int V^2/R.dt##. But ##V_{rms}^2=(1/T)\int V^2.dt##, so..
 
haruspex said:
Yes. This is why the rms voltage is interesting.
The power at any instant is V(t)2/R, so the average power is to be had by integrating over a cycle: ##(1/T)\int V^2/R.dt##. But ##V_{rms}^2=(1/T)\int V^2.dt##, so..
Thank you very much for reply. Oh I am so sorry I had not realized that one would integrate to find the average power.
So would T=1/f
T=1/50=0.02 s
##\int V^2.dt=V^3/3##
So does that mean P=1/0.02 * V^3/3= 50*V^3/3=50V^3/3
Substitute V rms = 230 V
P=50*230^3/3
P=881666.6667 ~ 882,000 W

This cannot be correct what have I done? 😳
 
AN630078 said:
##\int V^2.dt=V^3/3##
No, that's not how integration works.
##\int V^2.dV=V^3/3##, but to find ##\int V^2.dt## you need to substitute a function of t for V. In the present case it is V0 sin ω t.
But note that post #2 shows that with a purely resistive load the average power is Vrms2/R no matter what function V is of t.
 
haruspex said:
No, that's not how integration works.
##\int V^2.dV=V^3/3##, but to find ##\int V^2.dt## you need to substitute a function of t for V. In the present case it is V0 sin ω t.
But note that post #2 shows that with a purely resistive load the average power is Vrms2/R no matter what function V is of t.
Thank you for your reply. I am sorry I had misunderstood. So the average power supplied would be 26.45 W? Are parts 1 and 2 correct I forgot to ask 😁
 
AN630078 said:
Thank you for your reply. I am sorry I had misunderstood. So the average power supplied would be 26.45 W? Are parts 1 and 2 correct I forgot to ask 😁
Yes, all good.
 
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