C6-activities - ELECTRIC CIRCUITS ECSE-2010 Class 6...

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1 ELECTRIC CIRCUITS ECSE-2010 Class 6 ACTIVITY 6-1a 40 V 10 V 20 Ω 1 i 1 Find i using Superposition 20 Ω 10 Ω 10 A ACTIVITY 6-1a 40 V Short Ciruit 20 Ω 1-1 i 1-1 i due to 40 V 20 Ω 10 Ω Open Circuit 200 20 20 10 30 3 = = Ω eq 20 80 R 20 3 3 = + = Ω 40 i 1.5 A 80 3 = = i ACTIVITY 6-1a 40 V Short Ciruit 20 Ω 1-1 i 1-1 i due to 40 V 20 Ω 10 Ω Open Circuit 1-1 20 i i 1 A 20 10 = = + i Current Divider Rule i 1.5 A = ACTIVITY 6-1a 10 V Short Ciruit 20 Ω 1-2 i 1-2 i due to 10 V 20 Ω 10 Ω Open Circuit 20 20 10 = Ω 1-2 10 i .5 A 10 10 = = + ACTIVITY 6-1a 10 A Short Ciruit 20 Ω 1-3 i 1-3 i due to 10 A 20 Ω 10 Ω 20 20 10 = Ω 1-3 i 5 A = Short Ciruit
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2 ACTIVITY 6-1a 40 V 10 V 20 Ω 1 i 1 i 1 .5 5 6.5 A = + + = 20 Ω 10 Ω 10 A ACTIVITY 6-1b 40 V 10 V 20 Ω 1 i x Find v using Thevenin's Theorem 20 Ω 10 Ω 10 A x v + ACTIVITY 6-1b 40 V 10 V 20 Ω Norton Circuit 20 Ω 10 Ω 10 A x v + Replace with Thevenin Circuit ACTIVITY 6-1b 40 V 10 V 20 Ω 20 Ω 10 Ω 100 V x v + Two Voltage Sources in Series 10 100 90 V + = ACTIVITY 6-1b 40 V 20 Ω 20 Ω 10 Ω 90 V x v + Node Equation x x x v 40 v v 90 0 20 20 10 + + = x v x v 55 V = ACTIVITY 6-1b 40 V 10 V 20 Ω 1 i Check 20 Ω 10 Ω 10 A x v + 1 Found i 6.5 A using Superposition = x v 10 (6.5)(10) 55 V = − + = x v
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3 ACTIVITY 6-2
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