C14-activities - ACTIVITY 14-1 3 k ACTIVITY 14-1 i 5 F 1 k...

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1 ACTIVITY 14-1 C v + i C v + 40 V 50 V 3 k Ω 1 k Ω 4 k Ω 5 F μ Switch closed for a long time Opens at t 0 = i C Find i(t) and v (t) for t 0 ACTIVITY 14-1 Write Down v c (t), i(t) for t > 0: t/ C CSS C0 CSS v (t) v (v v ) e τ = + t/ SS 0 SS i(t) i (i i ) e τ = + C0 0 CSS SS Need to find: v , i , v , i , τ Follow a 4 Step Approach 0 t 0 = ACTIVITY 14-1 Step 1 40 V 50 V 3 k Ω 1 k Ω 4 k Ω Draw Circuit at t 0 = C Find i(0 ), v (0 ) i(0 ) C v (0 ) + 0 50 40 V source and 3 k resistor do not affect circuit Ω Open Circuit Switch Closed ACTIVITY 14-1 50 V 1 k Ω 4 k Ω i(0 ) 50 i(0 ) 10 mA 1 4 = = − + C v (0 ) 10 x 4 40 V = − = − C v (0 ) + DC Steady State i(0 ) Step 1 t 0 = C v (0 ) + = ACTIVITY 14-1 Step 2 Draw Circuit at t 0 + = 0 C0 Find i , v 40 V C0 C v v (0 ) 40 V = = − 40 50 V 3 k Ω 1 k Ω 4 k Ω 5 F μ C0 v + 0 i 0 40 Open 0 40 ( 40) i 20 mA 3 1 − − = = + ACTIVITY 14-1 v c Cannot Change Instantaneously: => v c (0 + ) = v co = v c (0 - ) = - 40 V i Can Change Instantaneously: Changed from -10 mA at t = 0 - to + 20 mA at t = 0 + i is Not the Current through an Inductor
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2 ACTIVITY 14-1 Step 3 40 V 50 V 3 k Ω 1 k Ω 4 k Ω Open Ckt
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