problemset2 - W ^/sE t/E/t t tKZ

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J±²($³'´µ3´ B¶?37´C²'³#²±=´µE´F²$·¸¹ F&±¹º)´E??G´ »¼½¾¿À´ÁÂÃĸ³Å³±Á´Å³Æ¸²±Ç´ ± ÈÉÇÊÇËÌ´Í´ ² É´ ιºÈ ± ÌÏ´ ² Ê´ ±² ȸó´ ² Ðδı³´·º¹¸´Å³Æ¸²±Î̶ ¼¿!À´ÈÄÌ´¼"´ ± #³±³´Äº´³$³Æ¸±¹Æ´%¹Î&$ÄƳ'³º¸´Å³Æ¸²±Ç´ ³ Ç´#Ãĸ´#²·$%´(³´¸Ã³´Æ²±±³Î&²º%¹º)´ÆÃı)³´%³ºÎ¹¸ÊÇ *+ Â,-.À »Ä·ÎÎд$Ä#´"²±´³$³Æ¸±¹Æ´"¹³$%ÎÀ ± ³ Í´*´Í´2 É ³ ´² Ï´2 Ê ³ ±² Ï´2 7 Ë ³ µ² Í´2 Ê ³ ±² Í´3´456' 8´ ¼¿!À´È(Ì´¼"´ ± #³±³´Ä´'Ä)º³¸¹Æ´"¹³$%´Å³Æ¸²±Ç´ ´ Ç´#Ãĸ´#²·$%´(³´¸Ã³´Æ²±±³Î& ²º%¹º)´Æ·±±³º¸´%³ºÎ¹¸ÊÇ´ µ Â,-.À´ /'&0±³Ðδ$Ä#´¹º´¸Ã³´ ¶·¸¹º±²¶·¹·¸ À´ ¼¿!À´ÈÆÌ´!²³Î´¸Ã³´'Ä)º³¸¹Æ´"¹³$%´%³¸³±'¹º³%´¹º´È(Ì´ÎÄ ¸¹Î"Ê´Ä$$´²"´9ÄÉ#³$$Ðδ³:·Ä¸¹²ºÎ+´ ¼"´º²¸Ç´#ùÆô²º³´ ¹Î´Å¹²$ĸ³%+´ Â,-.À´;ó´Æ²ºÎ¸¹¸·¸¹Å³´±³$ĸ¹²º´"²±´¸Ã³´'Ä)º³¸¹Æ´"¹³$%´:·Äº¸¹¸¹³Î´¹Î´ÄÎη'³%´ $¹º³Ä±Ç´ Í< ´ Ç´#ó±³´<´¹Î´ Ä$β´ÄÎη'³%´¸²´(³´Ä´Æ²ºÎ¸Äº¸¶´ ¼º´¸Ã¹Î´ÆÄγǴ#³´¹''³%¹Ä¸³$Ê´=º²#´¸Ãĸ´ ± 1 Í ± 1È< ´ ÌÍ< ± 1 ± >´?´"±²'´ &ı¸´ÈÄÌÇ´¹º´#ùÆôÆÄγ´»Ä·ÎÎд$Ä#´¹Î´Å¹²$ĸ³%¶´ ²ºÎ´ ¼º´Äº´,Ã'¹Æ´'³%¹·'´#¹¸Ã´¸Ã³´Æ²ºÎ¸¹¸·¸¹Å³´±³$ĸ¹²ºÇ´ µ Í@ · Ç´#¹¸Ã´¸Ã³´Æ²º%·Æ¸¹Å¹¸ÊÇ´@ǴĴƲºÎ¸Äº¸Ç´ ¸Ã³º´ · Í´ µ [email protected]Ǵĺ%´ ± É µ ÍA ² É Î¹ºÈ ± Ì´>´?Ç´¹º´#ùÆôÆÄγ´ ıÄ%ÄÊÐδ$Ä#´²"´¹º%·Æ¸¹²º´È ± É · Í´A2 62¸Ì´#²·$%´(³´ Ź²$ĸ³%´"²±´¸Ã³´Î¸³Ä%ʴθĸ³¶´ HW2 Page1
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˜ t ¯ ˜ Ee = E ¯ ¯¯ e j ( ωt + φ ) = ¯ ˜ E ¯ ( cos ( + φ )+ j sin( + φ )) , (2.3) the real part of ¯ whic ¯ his ˜ E ( t )= < ( jωt ˜ ¯ E r ¯ cos( + φ ) . (2.4) Now, turn the crank and apply trigono ¯ me ¯ tric identities to get more mean- ingful ˜˜ answers. In (a), E r =1and E i = 1, so E ( t ) = cos( ) + sin( 2 sin( +45 ) Also, the phase angle , φ , is arctan( 1) = 45 , while the magnitude is 2, so E ( t 2cos( 45 2sin 3 no ³´ ³ ´ 2.2 Problem 2 GIVEN If the electric Feld is E ( t < ˜ ,where E ˜ is a phasor, then what is E ( t )i f FIND (a) E ˜ =1 j WORK e =cos( j sin( ) by Euler’s theory. If E ˜ = E ˜ r +
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problemset2 - W ^/sE t/E/t t tKZ

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