EE411_HW11_Soln - 1 = + = + + Ans:Assume the current...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
EE411 Circuit Theory, HW 11 Prof. R. Gharpurey Due Date: Apr. 20, 2010 Ans:For load 1, we have cos(θ 1 )=0.8 b θ 1 =36.87 o . For load 2, we have cos(-θ 2 )=0.95 b θ 2 =- 18.19 o . So we have ) ( 9982 . 0 ) 45 . 3 cos( 19 . 39 7 . 936 8 . 13 3 3 45 . 3 7 . 936 450 19 . 18 300 87 . 36 250 3 2 1 lagging factor Power A I I I V S S S o L L L L o o o = = = = × = = + + = + + Ans:We have mH L L k M 485 . 8 10 5 10 40 6 . 0 3 3 2 1 = × × × × = =
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
) cos( 2 10 5 ) ( 10 485 . 8 ) ( ) ( ) ( ) ( ) cos( 2 10 485 . 8 ) ( 10 40 ) cos( 10 ) ( ) ( ) ( 3 1 3 2 2 2 1 2 3 1 3 2 1 1 1 t dt t di t v dt t di L dt t di M t v t dt t di t dt t di M dt t di L t v ω × × + × × = + = × × × × = = So it can be solved to obtain ) sin( 5493 . 0 ) ( ) 2000 cos( 10 0985 . 1 ) cos( 10 0985 . 1 10 40 ) cos( 2 10 485 . 8 ) cos( 10 ) ( ) cos( 679 . 10 ) ( 1 3 3 3 3 1 2 t t i t t t t dt t di t t v = × = × = × × × + = = Ans:(a) 3536 . 0 2 4 1 = × = k k (b)We have 4 = , the impedance of the 1/4F is 1/j. So ) 59 . 57 4 cos( 3216 . 0 59 . 57 3216 . 0 41 . 77 4548 . 0 4439 . 0 0991 . 0 23 . 81 8547 . 0 8477 . 0 1304 . 0 ) 1 / 1 1 / 1 ( 2 4 1 4 1 4 ) 4 4 2 ( 12 2 1 2 2 1 2 1 o o o o o o t v j i j i i j j i j i j v i j i j + = = = + = = = + × = × × × × × × = × × × + × × + =
Background image of page 2
(c) J t i t Mi t i L t i L 168 . 1 ) 180 41 . 77 2 4 cos( ) 180 23 . 81 2 4 cos( 04548 8547 . 0 1 ) 180 41 . 77 2 4 ( cos 04548 2 2 1 ) 180 23 . 81 2 4 ( cos 8547 . 0 4 2 1 ) ( ) ( ) ( 2 1 ) ( 2 1 2 2 2 2 2 1 2 2 2 2 1
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 = + = + + Ans:Assume the current flowing through the first coil is i 1 and the current flowing through the first coil is i 2 . Then we have ( ) 0769 . 9 6154 . 1 0952 . 0063 . ) 12 8 ( 4 4 ) 10 1 ( 4 ) ( 2 6 ) 10 8 ( ) ( 2 6 ) 12 1 ( 1 1 1 1 2 1 1 2 1 2 2 1 1 2 1 2 1 2 1 j i v Z v j i So i j i j v i j i j i j i i j j i j i v i i j j i j i + = = = = + + = + + = + + + = + + +...
View Full Document

This note was uploaded on 02/24/2011 for the course EE 411 taught by Professor Lee during the Spring '07 term at University of Texas at Austin.

Page1 / 3

EE411_HW11_Soln - 1 = + = + + Ans:Assume the current...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online