# 1.21 - 2 4 t(W p = 5 W t = p 25 s = 5cos 2 4 × 25 = 5 W t...

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Problem 1.21 The voltage across and current through a certain device are given by v ( t ) = 5cos ( 4 π t ) V , i ( t ) = 0 . 1cos ( 4 π t ) A . Determine: (a) The instantaneous power p ( t ) at t = 0 and t = 0 . 25 s. (b) The average power p av , defined as the average value of p ( t ) over a full time period of the cosine function (0 to 0.5 s). Solution: (a) p ( t ) = vi = ( 5cos4 π t )( 0 . 1cos 4 π t ) = 0 . 5cos 2 4 π t (W)
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Unformatted text preview: 2 4 t (W) . p ( ) = . 5 W @ t = p ( . 25 s ) = . 5cos 2 ( 4 × . 25 ) = . 5 W @ t = . 25 s . (b) p av = 1 T i T p ( t ) dt = 1 . 5 i . 5 . 5cos 2 4 t dt = 1 8 [ sin4 t cos4 t + 4 t ] | . 5 = 1 4 (W) ....
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