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Lecture23 - Chapter 4

Lecture23 - Chapter 4 - – Remove R L – Short circuit...

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1 EE216 Circuit Analysis I Lecture # 23– Chapter 4
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R 3 3 ohms C D A B R 2 4 ohms R 4 6 ohms R 1 6 ohms R L 2 ohms 30 V 2 Thevenin’s Equivalent Example 1 Voltage& Current through R L ? To calculateThevenin’s Voltage RemoveR L Assume C as referencenode. Voltagedivision across R 3 and R 4 yields V A = 10 V Voltagedivision across R 1 and R 2 yields V B = 18 V So, V TH = V AB = 10 – 18 = -8 V
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3 Thevenin’s Equivalent Example 1 To calculateThevenin’s Resistance
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Unformatted text preview: – Remove R L – Short circuit voltage supply • Equivalent resistance at AB is = (3 | | 6) + (6 | | 4) = 4.4 Ω R 3 3 ohms C D A B R 2 4 ohms R 4 6 ohms R 1 6 ohms R L 2 ohms 30 V 4 Thevenin’s Equivalent Example 1 • Current through R L is = 8/(4.4 + 2) = 1.25 A • Voltage across R L is = 2(1.25) = 2.5 V R TH 4.4 ohms R L 2 ohms 8 V...
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