Euler's identity for phasor (1 Viewer)

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for phasor, v = 20V e^(-j60) and i = 0.5A e^(-j30)
how can i write them in v(t) and I(t) ??
pls help


thanx
 

chroot

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Use Euler's identity and take only the real part.

[itex]e^{j \theta} = \cos \theta + j \sin \theta[/itex]

- Warren
 
416
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i got v(t) = 20V cos (wt - 60 ) and i(t) = 0.5A cos(wt - 30)
from here,how to find p(t)??
a hint is given but i don't understand : coa A cos B = 1/2[cos(A+B) + cos(A-B)]
 

chroot

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Power is voltage * current, yes?

Multiply your v(t) and i(t) to get p(t).

The cosine identity was given to you to help you with the simplification.

- Warren
 
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so what is A and B?? is it -60 and -30??
 

chroot

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A is the argument of the one cosine function; B is the argument of the other.

When you multiply two cosine functions, with arguments A and B, you can use the identity you provided to simplify.

In this case, A = wt - 60, and B = wt - 30.

- Warren
 

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