Ratio of effective and average voltage

AI Thread Summary
The discussion focuses on finding the ratio of effective voltage to average voltage, specifically $$\frac{U_{ef}}{U_{avg}}$$. The equations provided indicate that the effective voltage is calculated as $$U_{ef}=\frac{U_m}{\sqrt3}$$ and the average voltage as $$U_{avg} = \frac{U_m}{2}$$. A participant initially questioned the correctness of their method, which involved calculating the effective and average voltages separately for two different line equations. Ultimately, they realized their error was in not applying the correct multiplication factors for the averages and effective voltages. The correct ratio is confirmed to be approximately 1.154, as derived from the official solution.
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Homework Statement


See the diagram. Find $$\frac{U_{ef}}{U_{avg}}$$ .

Homework Equations

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$$ U_{ef}=\frac{U_m}{\sqrt3} , U_{avg} = \frac{U_m}{2} $$

The Attempt at a Solution


This one should be easy.I know that official solution probably divided $$\frac{2}{\sqrt3} \approx 1.154$$ (which is correct), but I am not sure why can we do that in this case, when we have 2 different kinds of lines, one with
u=5/3 t and second one u=-5/4t+35/4 . What did was found the effective of the first one and the second one, average of first one and second one , used $$ U_{avg}= U_{avg1}+U_{avg2}$$ , and $$U_{ef}^2 = U_{ef1}^2+U_{ef2}^2$$. My question is why is mine incorrect, and the solution shown above correct?
 

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Nevermind, I found my mistake, I forgot to multiply with 3/7 and 4/7 and with sqrt(3/7) and sqrt(4/7).
 
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