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EE411_HW4_Soln

EE411_HW4_Soln - 5 7 5 3 12 4 4 5 3 − − − = − = ⇒...

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Ans:When t<0, the switch is closed and the capacitor can be treated as open circuit. From KCL, we know that the current flowing through the 40 resistor is 0.5i. We have A i i i 8 . 0 ) 50 30 ( 40 5 . 0 80 = + × + × = When t>0, the switch is opened. And we know the capacitor voltage v c (0) at t=0 is i(0)*(30+50)=64V. So we have 0 ) 50 30 ( ) ( 5 . 0 ) ( 5 . 0 ) ( 5 . 0 ) ( = + + = + = + t v dt t dv C i dt t dv C i i dt t dv C c c c c So 480 ) ( 0 480 ) ( ) ( t c c c Ae t v t v dt t dv = = + Using the initial condition that v c (0)=64V, we can obtain A=64. So A e t v t i e t v t c t c 480 480 8 . 0 ) 50 30 ( ) ( ) ( 64 ) ( = + = =
Ans: (a)Before t=0, I is obtained by current division A t i 5 . 1 3 4 4 4 ) ( = + = After t=0 A e e t i i i R L e i i i t i t t eq t 2 5 . 0 / / 21 . 0 29 . 1 ] 7 9 5 . 1 [ 7 9 ) ( 7 9 ) 3 ( 12 // 4 4 12 // 4 ) ( , 1 ) 0 (

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Unformatted text preview: 5 . 7 5 . 3 12 // 4 4 5 . 3 )] ( ) ( [ ) ( ) ( − − − + = − + = ⇒ = + = ∞ = = = + = = ∞ − + ∞ = τ (b) Before t=0, I is obtained by A t i 4 3 2 20 ) ( = + = After t=0 A e e t i i i R L e i i i t i t t eq t 25 . 2 25 . 2 / 2 6 ] 6 4 [ 6 ) ( 6 6 3 6 6 // 3 2 20 3 2 2 2 // 3 6 48 ) ( , 4 ) ( 5 . 4 2 2 // 6 3 2 )] ( ) ( [ ) ( ) ( − − − − = − + = ⇒ = + × + + + × + = ∞ = = + = = ∞ − + ∞ = τ...
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EE411_HW4_Soln - 5 7 5 3 12 4 4 5 3 − − − = − = ⇒...

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