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241-26%2b27_Lec-12

# 241-26%2b27_Lec-12 - Physics 241 Lecture 12 Y E Kim Chapter...

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Physics 241 Lecture 12 Y. E. Kim September 30, 2010 Chapter 26, Section 3 Chapter 27, Sections 1 - 2 September 30, 2010 University Physics, Chapter 26 and 27 1

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September 30, 2010 University Physics, Chapter 26 2 Chapter 26 Section 3 Multi-Loop Circuits To analyze a multi-loop circuit, identify all complete loops and all junction points in the circuit and apply Kirchhoff’s Rules to these parts of the circuit separately Analyze the single loops in a multi-loop circuit with Kirchhoff’s Loop Rule and the junctions with Kirchhoff’s Junction Rule, and obtain a system of coupled equations in several unknown variables These coupled equations can be solved in several ways Solution with matrices and determinants Direct substitution
September 30, 2010 University Physics, Chapter 26 3 Example: Kirchhoff’s Rules (1) The circuit here has three resistors, R 1 , R 2 , and R 3 and two sources of emf, V emf,1 and V emf,2 This circuit cannot be resolved into simple series or parallel structures To analyze this circuit, we need to assign currents flowing through the resistors We can choose the directions of these currents arbitrarily

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September 30, 2010 University Physics, Chapter 26 4 At junction b the incoming current must equal the outgoing current At junction a we again equate the incoming current and the outgoing current But this equation gives us the same information as the previous equation! We need more information to determine the three currents 2 more independent equations Example: Kirchhoff’s Laws (2) 2 1 3 i i i  1 3 2 i i i 
September 30, 2010 University Physics, Chapter 26 5 To get the other equations we must apply Kirchhoff’s Loop Rule. This circuit has three loops. Left R 1 , R 2 , V emf,1 Right R 2 , R 3 , V emf,2 Outer R 1 , R 3 , V emf,1 , V emf,2 Example: Kirchhoff’s Laws (3)

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September 30, 2010 University Physics, Chapter 26 6 Example: Kirchhoff’s Laws (4) Going around the left loop counterclockwise starting at point b we get Going around the right loop clockwise starting at point b we get Going around the outer loop clockwise starting at point b we get But this equation gives us no new information! 1 1 ,1 2 2 1 1 2 2 0 0 emf i R V i R i R V i R 3 3 ,2 2 2 3 3 ,2 2 2 0 0 i R V i R i R V i R 3 3 ,2 1 1 0 i R V V i R
September 30, 2010 University Physics, Chapter 26 7 We now have three equations And we have three unknowns i 1 , i 2 , and i 3 We can solve these three equations in a variety of ways Example: Kirchhoff’s Laws (5) 1 3 2 i i i  1 1 ,1 2 2 0 emf i R V i R 3 3 ,2 2 2 0 emf i R V i R 2 3 2 ,2 1 1 2 1 3 2 3 3 1 2 1 2 1 3 2 3 2 1 2 3 1 2 1 3 2 3 () emf emf emf emf emf emf R R V R V i R R R R R R R V RV i R R R R R R R V R R V i R R R R R R

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241-26%2b27_Lec-12 - Physics 241 Lecture 12 Y E Kim Chapter...

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