241-25%2b26_Lec-11 (1) - Physics 241 Lecture 11 Y. E. Kim...

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Physics 241 Lecture 11 Y. E. Kim September 28, 2010 Chapter 25, Sections 6 7 Chapter 26, Sections 1 - 2 September 28, 2010 University Physics, Chapter 25 and 26 1
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September 28, 2010 University Physics, Chapter 25 2 Resistances in Parallel (1) Instead of connecting resistors in series so that all the current must pass through both resistors, we can connect the resistors in parallel such that the current is divided between the two resistors This type of circuit is shown below
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September 28, 2010 University Physics, Chapter 25 3 Resistance in Parallel (2) In this case, the voltage drop across each resistor is equal to the voltage provides by the source of emf Using Ohm’s Law we can write the current in each resistor The total current in the circuit must equal the sum of these currents Which we can rewrite as 1 1 emf V i R 2 2 emf V i R 12 i i i  1 2 1 2 11 emf emf emf VV i i i V R R R R     
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September 28, 2010 University Physics, Chapter 25 4 Resistance in Parallel (3) We can then rewrite Ohm’s Law for the complete circuit as .. where We can generalize this result for two parallel resistors to n parallel resistors 1 emf eq iV R 12 1 1 1 eq R R R  1 11 n i eq i RR = R eq
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September 28, 2010 University Physics, Chapter 25 5 Example: Network of Resistors (1) Consider the network of resistors shown below Calculate the current flowing in this circuit.
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September 28, 2010 University Physics, Chapter 25 6 Example: Network of Resistors (2) First look for pairs of resistors: R 3 and R 4 are in series Now note that R 34 and R 1 are in parallel 34 3 4 R R R  1 34 134 134 1 34 1 34 1 1 1 or RR R R R R R R
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September 28, 2010 University Physics, Chapter 25 7 Example: Network of Resistors (3) And now R 2 , R 5 , R 6 , and R 134 are in series   123456 2 5 6 134 1 34 123456 2 5 6 1 34 1 3 4 123456 2 5 6 1 3 4 123456 emf R R R R R RR R R R R R R R R R R R R R R V i R 
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September 28, 2010 University Physics, Chapter 25 8 Example 2: More Resistors (1) The figure shows a circuit containing one ideal 12-V battery (no internal resistance) and 4 resistors with R 1 =20 Ω , R 2 =20 Ω , R 3 =30 Ω , and R 4 =8 Ω . (Important note: Inputs are given by 2 significant numbers answers should be rounded off to 2 significant numbers !) Question: What is the current through the battery? Answer: Idea: Find the equivalent resistance and use Ohm’s Law. R 2 and R 3 are in parallel.
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September 28, 2010 University Physics, Chapter 25 9 Example 2: More Resistors (2) What is the current through the battery?
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241-25%2b26_Lec-11 (1) - Physics 241 Lecture 11 Y. E. Kim...

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