Adding Voltages: Calculate Peak Value of U3

[Lindsayyyy]

Homework Statement

Hi,

Two given Voltages:
U1(t)= 40V * cos(wt)
U2(t)= 120V * sin(wt)

wheres w means omega.

Now I have to sum them up to U3= U1+U2

and calculate the peak value of U3.




2. The attempt at a solution


My problem is, I'm not sure if the following is allowed:

calculate the root mean square(RMS) of U1 and U2, sum them up.

And solve the Equation:

U(RMS) = U0/sqrt(2)


Thank you very much for your help in advance.

bb

 

[tiny-tim]

Hi Lindsayyyy!

You must learn all your trigonometric identities …

what is Acosx + Bsinx ? ​

 

[Lindsayyyy]

Thanks for your fast reply. I'll found the one I need. I didn't find it when I was looking for it. If I have any further troubles, I'll ask you guys.

But do you know if it's allowed the way I tried it? I'm note sure because the voltages have a different phase.

 

[MisterX]

Your solution technique from the first post in this thread is incorrect.

If the signals were in phase, you could just add the peak values.

Converting to from peak amplitude to RMS, adding the RMS values, and then converting back to peak in this situation is pointless. The RMS is proportional to the peak magnitude for these waveforms. So what you were describing would be like calculating

$$\frac{ax + ay}{a}$$

which is unecessary as the expression is equal to x + y.

 

[tiny-tim]

If the signals were in phase, you could just add the peak values.

But they're 90° out of phase.

 

[Lindsayyyy]

Ok, I get the point. Solved it via the identity given bei tiny-tim. I didn't find it in the first place, looks like this identity is quite rare?

my solution was about 126,49 V if I remember correctly. I also tried to derive U3 but I don't know how to find the nulls of the function.

Thanks for your help :)

 

[tiny-tim]

Ok, I get the point. Solved it via the identity given bei tiny-tim. I didn't find it in the first place, looks like this identity is quite rare?

Yes, I think it's rarely listed because it's derivative, ie it can be derived from the basic trigonometric identities …

but it's not obvious, and it's very useful, so I think it should be better publicised! ​