teng125 Messages 416 Reaction score 0 Thread starter Feb 28, 2006 #1 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
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 Staff Emeritus Science Advisor Gold Member Messages 10,270 Reaction score 45 Feb 28, 2006 #2 Use Euler's identity and take only the real part. [itex]e^{j \theta} = \cos \theta + j \sin \theta[/itex] - Warren
Use Euler's identity and take only the real part. [itex]e^{j \theta} = \cos \theta + j \sin \theta[/itex] - Warren
teng125 Messages 416 Reaction score 0 Feb 28, 2006 #3 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)]
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 Staff Emeritus Science Advisor Gold Member Messages 10,270 Reaction score 45 Feb 28, 2006 #4 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
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
chroot Staff Emeritus Science Advisor Gold Member Messages 10,270 Reaction score 45 Feb 28, 2006 #6 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
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