p33_042

# p33_042 - 42(a We note that we obtain the maximum value in...

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42. (a) We note that we obtain the maximum value in Eq. 33-28 when we set t = π 2 ω d = 1 4 f = 1 4(60) =0 . 00417 s or 4 . 17 ms. The result is E m sin( π/ 2) = E m sin(90 )=3 6 . 0 V. We note, for reference in the subsequent parts, that at t =4 . 17 ms, the current is i = I sin( ω d t φ )= I sin(90 ( 29 . 4 )) = (0 . 196 A) cos(29 . 4 )=0 . 171 A using Eq. 33-29 and the results of the Sample Problem. (b) At t =4 . 17 ms, Ohm’s law directly gives v R = iR =( I cos(29 . 4 )) R (0 . 171 A)(160 Ω) = 27 . 3V . (c) The capacitor voltage phasor is 90 less than that of the current. Thus, at t =4 . 17 ms, we obtain v C = I
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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