- #1

AN630078

- 242

- 25

- 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

Thank you to anyone who replies.

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.