HW 10 solution

HW 10 solution - song (shs546) – Homework 10 – weathers...

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Unformatted text preview: song (shs546) – Homework 10 – weathers – (22202) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider the combination of resistors shown in the figure. 4 . 3 Ω 1 . 3 Ω 3 . 9 Ω 9 . 1Ω 7Ω 2 . 1Ω 3 . 6 Ω a b What is the resistance between point a and point b ? Correct answer: 7 . 66849 Ω. Explanation: Let’s redraw the figure i R 1 i R 2 i R 5 i R 4 i R 3 i R 6 i R 7 a b Let : R 1 = 4 . 3 Ω , R 2 = 1 . 3 Ω , R 3 = 3 . 9 Ω , R 4 = 9 . 1 Ω , R 5 = 7 Ω , R 6 = 2 . 1 Ω , and R 7 = 3 . 6 Ω . Basic Concepts: • Equivalent resistance. • Ohm’s Law. There are two rules for adding up resis- tances. If the resistances are in series, then R series = R 1 + R 2 + R 3 + ··· + R n . If the resistances are parallel, then 1 R parallel = 1 R 1 + 1 R 2 + 1 R 3 + ··· + 1 R n . Solution: The key to a complex arrange- ments of resistors like this is to split the prob- lem up into smaller parts where either all the resistors are in series, or all of them are in parallel. It is easier to visualize the problem if you redraw the circuit each time you add them. i R 1 i R 2 i R 5 i R 4 i R 367 a b Step 1: The three resistors on the right are all in series, so R 367 = R 3 + R 6 + R 7 = (3 . 9 Ω) + (2 . 1 Ω) + (3 . 6 Ω) = 9 . 6 Ω . i R 1 i R 2 i R 3675 i R 4 a b Step 2: R 5 and R 367 are connected paral- lel, so R 3675 = parenleftbigg 1 R 5 + 1 R 367 parenrightbigg- 1 = R 5 R 367 R 5 + R 367 = (7 Ω) (9 . 6 Ω) 16 . 6 Ω = 4 . 04819 Ω . song (shs546) – Homework 10 – weathers – (22202) 2 i R 1 i R 36752 i R 4 a b Step 3: R 2 and R 3675 are in series, so R 23675 = R 2 + R 3675 = (1 . 3 Ω) + (4 . 04819 Ω) = 5 . 34819 Ω . Step 4: R 23675 and R 4 are parallel, so R 236754 = parenleftbigg 1 R 4 + 1 R 23675 parenrightbigg- 1 = R 4 R 23675 R 4 + R 23675 = (9 . 1 Ω) (5 . 34819 Ω) 14 . 4482 Ω = 3 . 36849 Ω . i R 1 i R 367524 a b Step 5: Finally, R 1 and R 236754 are in se- ries, so the equivalent resistance of the circuit is R eq = R 1 + R 236754 = 4 . 3 Ω + 3 . 36849 Ω = 7 . 66849 Ω . 002 10.0 points 3 Ω 4 Ω 7 Ω 9 Ω 11 Ω 2 Ω 2 Ω 2 Ω 12 V 25 V 36 V Find the magnitude of the current in the 12 V cell. Correct answer: 0 . 954802 A. Explanation: i 1 R 1 i 3 R 2 i 1 R 3 i 3 R 4 i 2 R 5 r 1 r 2 r 3 E 1 E 2 E 3 Let : E 1 = 12 V , E 2 = 25 V , E 3 = 36 V , R 1 = 3 Ω , R 2 = 4 Ω , R 3 = 7 Ω , R 4 = 9 Ω , R 5 = 11 Ω , r 1 = 2 Ω , r 2 = 2 Ω , and r 3 = 2 Ω . Basic Concepts: Kirchhoff’s Laws: summationdisplay V = 0 around a closed loop . summationdisplay I = 0 at a circuit junction . Solution: Applying Kirchhoff’s law to the outside loop and the lower loop we get 3 equa- tions in 3 unknowns; i.e. , E 1 − E 3 = ( R 3 + r 1 + R 1 ) i 1 + ( R 2 + r 3 + R 4 ) i 3 (1) E 2 − E 3 = ( R 5 + r 2 ) i 2 + ( R 2 + r 3 + R 4 ) i 3 (2) 0 = − i 1 − i 2 + i 3 . (3) Subtracting the first two equations, E 1 − E 2 = ( R...
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This note was uploaded on 04/22/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.

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HW 10 solution - song (shs546) – Homework 10 – weathers...

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