8.4 The Natural Response of a Series/Parallel RLC Circuit

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8.3 Finding Initial Values• Example 1 (cont.)+KVL: 2A× 4 + v (0 ) + 4V = 12V+∴ v (0 ) = 0d i+∴ (0 ) = 0dt+KCL: i (0 ) = 2ACLLd v+∴ (0 ) = 0dt+iC(0 ) 2∴ = = 20 V/SC 0.1C.T. Pan 158.3 Finding Initial Values• Example 1 (cont.)4Ω 0.25Hi2Ω12V0.1F + v-t=0t →∞L is short circuitdC is open∴i( ∞ ) = 0v( ∞ ) = 12VC.T. Pan 16
• Example 28.3 Finding Initial Values4Ω3u(t)A 2Ω__ 12 F ++ vcv R --0.6H20Vi L+ d +Find : ( a) iL(0 ) , iL(0 ) , iL( ∞)dt+ d +( b) vC(0 ) , vC(0 ) , vC( ∞)dt+ d +() c vR(0 ) , vR(0 ) , vR( ∞)dtC.T. Pan 178.3 Finding Initial Values• Example 2 (cont.)4Ω3u(t)A 2Ω1__2 F ++ vcv R --0.6H20Vi Lt < 02Ω+v R-4Ω20V+v c-i L−∴ i (0 ) = 0L−v (0 ) =−20VvcR−(0 ) = 0C.T. Pan 18
Page 2 and 3: 8.1 Linear Second Order Circuits•
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Page 20 and 21: 8.5 The Step Response of a Series/P
Page 28 and 29: 8.6 State EquationStep1. Pick a tre
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Page 32 and 33: • Example 3 : Find v R48.6 State
Page 34 and 35: 8.6 State EquationR∴v = e - ( i +
Page 36: Summary• Objective 1 : Be able to

8.4 The Natural Response of a Series/Parallel RLC Circuit

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