PPE25_ResistorCircuits

# PPE25_ResistorCircuits - Now expand R A into R 1 and R 2 ....

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Current, Resistance, Current, Resistance, and Circuits and Circuits Resistor Circuits © RHJansen

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Example 1 © RHJansen V I R ε 6 R 1 2 R 3 2 R 3 2 R s = ΣΡ ι = 2 + 2 = 4Ω R 3 R 2 R 1 Solve between junctions first. Reduce each path to one resistor per path between junctions. R A 4 R A S
Example 1 © RHJansen V I R ε 6 R 1 2 R 3 2 R 3 2 R A 4 1 R P = Σ 1 Ρ ι = 1 4 + 1 2 = 3 4 R 3 R 2 R 1 Now add the two parallel paths. R A S P R P = 4 3 = 1.33Ω 1.33

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Use Ohm’s Law Example 1 © RHJansen V I R ε 6 1.33 R 1 2 R 3 2 R 3 2 R A 4 V = ΙΡ R 3 R 2 R 1 R A S P I = ς Ρ = 6 1.33 = 4.5Α 4.5 Once the top row is done follow the arrows backwards to fill in the rest.
In parallel potential is the same. Use Ohm’s Law to find current.

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Unformatted text preview: Now expand R A into R 1 and R 2 . Example 1 © RHJansen V I R ε 6 4.5 1.33 R 1 2 R 3 2 R 3 2 R A 4 R 3 R 2 R 1 S P 6 6 1.5 3 R A In parallel potential is the same. Use Ohm’s Law to find current. Now expand R A into R 1 and R 2 . Example 1 © RHJansen V I R ε 6 4.5 1.33 R 1 2 R 3 2 R 3 2 R A 4 R 3 R 2 R 1 S P 6 6 1.5 3 1.5 1.5 3 3 In series current is the same. Use Ohm’s Law to find potential....
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## This note was uploaded on 07/21/2009 for the course PHYSICS 7B taught by Professor Packard during the Spring '08 term at University of California, Berkeley.

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PPE25_ResistorCircuits - Now expand R A into R 1 and R 2 ....

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