241-25-26_Lec-11 - 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 ² 7 Chapter 26, Sections 1 - 2 September 27, 2010 University Physics, Chapter 25 and 26 1
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September 27, 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 27, 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 ± 8VLQJ 2KP·V /DZ ZH FDQ ZULWH WKH FXUUHQW LQ HDFK UHVLVWRU ± 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 V i R 12 ii i ± 1 2 11 VV i V RR R R §· ± ± ± ¨¸ ©¹
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September 27, 2010 University Physics, Chapter 25 4 Resistance in Parallel (3) ± :H FDQ WKHQ UHZULWH 2KP·V /DZ IRU WKH FRPSOHWH FLUFXLW DV ± .. where ± We can generalize this result for two parallel resistors to n parallel resistors 1 emf eq iV R 12 11 1 RR R ± 1 n i i ¦ = R
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September 27, 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 27, 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 RR R ± 13 4 134 134 1 34 1 34 11 1 or R R R R ± ±
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September 27, 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 ± ² 123456 2 5 6 134 1 34 123456 2 5 6 1 34 13 4 123456 2 5 6 134 123456 emf RR R R R R R R R R R V i R ³³³ ³ ³ ³³
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September 27, 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 ȍ , R 2 =20 ȍ , R 3 =30 ȍ , and R 4 =8 ȍ . (Important note: Inputs are given by 2 significant numbers Æ answers should be rounded off to the same 2 significant numbers !) Question: What is the current through the battery? Answer: Idea: Find the equivalent resistance DQG XVH 2KP·V /DZ± R 2 and R 3 are in parallel.
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September 27, 2010 University Physics, Chapter 25 9 Example 2: More Resistors (2) What is the current through the battery?
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This note was uploaded on 11/28/2010 for the course PHY 222 taught by Professor Yu during the Spring '10 term at Beaufort County Community College.

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241-25-26_Lec-11 - Physics 241 Lecture 11 Y. E. Kim...

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