- #1
Simon.T
- 15
- 0
Ok guys this one is straight out of a textbook, (Tooley, M. and Dingle, L. (2008) Higher National Engineering, 2nd ed. Oxford: Elsevier, p.353.)
[tex]v=50sin(\omega t)+10sin(3\omega t + \pi /2)[/tex]
[tex]i=3.54sin(\omega t + \pi /4) + 0.316sin(3\omega t +0.321)[/tex]
[tex]\sum_{n=1}^{n=m} \frac {V_nI_ncos\phi_n} 2[/tex]
The solution given is
[tex]P_{TOT} =\frac {50\times 3.54} 2 cos(-\pi /4) + \frac {10\times 0.316} 2 cos(\pi /2 - 0.321)[/tex]
I don't understand how the solution in part 3 relates to the equation in part 2. The solution implies [tex]cos(a-b) = sin(wt + \phi_1)\times sin(wt+\phi_2) [/tex] but I cannot find this written anywhere apart from here
[tex]cos(a-b)=cos(a)cos(b)+sin(a)sin(b)[/tex] So we are just ignoring the cos terms?
I'm really quite tired so I am sorry if this seems like a stupid question, I swear this textbook is useless.
Homework Statement
[tex]v=50sin(\omega t)+10sin(3\omega t + \pi /2)[/tex]
[tex]i=3.54sin(\omega t + \pi /4) + 0.316sin(3\omega t +0.321)[/tex]
Homework Equations
[tex]\sum_{n=1}^{n=m} \frac {V_nI_ncos\phi_n} 2[/tex]
The Attempt at a Solution
The solution given is
[tex]P_{TOT} =\frac {50\times 3.54} 2 cos(-\pi /4) + \frac {10\times 0.316} 2 cos(\pi /2 - 0.321)[/tex]
I don't understand how the solution in part 3 relates to the equation in part 2. The solution implies [tex]cos(a-b) = sin(wt + \phi_1)\times sin(wt+\phi_2) [/tex] but I cannot find this written anywhere apart from here
[tex]cos(a-b)=cos(a)cos(b)+sin(a)sin(b)[/tex] So we are just ignoring the cos terms?
I'm really quite tired so I am sorry if this seems like a stupid question, I swear this textbook is useless.
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