Root-mean squared values and a.c. emf

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In summary, the peak voltage for mains electricity in the UK is 325 V and the peak current is 0.163 A. To find the average power supplied, one can use the formula P = Vrms^2/R or P = Irms^2*R. Integration is used to find the average power over a cycle, which results in the same formula. Therefore, using the root-mean squared values for voltage and current is appropriate when finding the average power supplied.
  • #1
AN630078
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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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  • #2
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..
 
  • #3
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? 😳
 
  • #4
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.
 
  • #5
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 😁
 
  • #6
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.
 

Related to Root-mean squared values and a.c. emf

1. What is the root-mean squared (RMS) value?

The root-mean squared value is a mathematical concept that represents the effective or average value of an alternating current (a.c.) signal. It is calculated by taking the square root of the mean of the squared values of the signal over a period of time.

2. How is the RMS value different from the peak value?

The peak value of an a.c. signal is the maximum value reached by the signal at any point in time, while the RMS value is the equivalent value of a steady direct current that would produce the same amount of power as the a.c. signal. In other words, the RMS value takes into account the fluctuating nature of the a.c. signal.

3. What is the significance of the RMS value in electrical engineering?

The RMS value is important in electrical engineering because it is used to calculate the power of an a.c. signal. It is also used in various calculations related to electrical circuits, such as determining the resistance of a circuit or the voltage drop across a component.

4. How is the RMS value related to a.c. emf (electromotive force)?

The RMS value of an a.c. emf is the maximum value of the emf divided by the square root of 2. This is because the RMS value represents the equivalent steady direct current that would produce the same amount of power as the a.c. emf.

5. Can the RMS value be negative?

No, the RMS value cannot be negative. It is always a positive value that represents the magnitude of the a.c. signal. However, the a.c. signal itself can be negative, which is reflected in the negative values of the squared values used to calculate the RMS value.

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