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

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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/ CC S S C 0 C S S v( t ) v (v v ) e τ =+− t/ SS 0 SS i(t) i (i i ) e C0 0 CSS SS Need to find: v , i , v , i , Follow a 4 Step Approach 0 t0 = 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 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 14 == + C v (0 ) 10 x 4 40 V =− C + DC Steady State i(0 ) Step 1 = C + = ACTIVITY 14-1 Step 2 Draw Circuit at t 0 + = 0C 0 Find i , v 40 V C0 C vv ( 0 ) 4 0 V 40 50 V 3 k Ω 1 k Ω 4 k Ω 5 F C0 v + 0 i 0 40 Open 0 40 ( 40) i2 0 m A 31 + ACTIVITY 14-1 ± v c Cannot Change Instantaneously: ² => v c (0 + ) = v co = v c (0 - ) = - 40 V ± i Can Change Instantaneously: ² Changed from -10 mA at t = 0 - to + 20 mA at t = 0 + ² 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 SS i CSS v + SS 40 i5
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C14-activities - ACTIVITY 14-1 3 k ACTIVITY 14-1 i + 5 F 1...

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