Root-mean squared values and a.c. emf

[AN630078]

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.

 

[haruspex]

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..

 

[AN630078]

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?

 

[haruspex]

$\int V^2.dt=V^3/3$

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 V_rms²/R no matter what function V is of t.

 

[AN630078]

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 V_rms²/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

 

[haruspex]

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.