{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

EE411_HW11_Soln - 1 = − × × − × × × × − − ×...

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 barb2right θ 1 =36.87 o . For load 2, we have cos(-θ 2 )=0.95 barb2right 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 Document Right 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 ω ω ω 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 ω ω ω ω ω ω = × = × = × × × + = = Ans:(a) 3536 . 0 2 4 1 = × = k k (b)We have 4 = ω , 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
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 = − × × − × × × × − − × × × × + − × × × × = + + π 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

{[ snackBarMessage ]}