chpt 6

# chpt 6 - PHYS 133 WINTER 2008 WEEK 6 CHAPTER 31 31.4 Solve...

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PHYS 133 WINTER 2008 WEEK 6, CHAPTER 31 31.4. Solve: The potential difference between the ends of a copper wire that carries a current can be obtained from Ohm’s law and the relationship R = ! L A . Finding the resistivity of copper from Table 30.1, ! V = IR = I " L A = 3.0 A ( ) 1.7 # 10 \$ 8 % m ( ) 0.20 m ( ) & 0.5 # 10 \$ 3 m ( ) 2 = 13 mV 31.10. Model: The batteries and the connecting wires are ideal. Visualize: Please refer to Figure Ex31.10. Solve: (a) Choose the current I to be in the clockwise direction. If I ends up being a positive number, then the current really does flow in this direction. If I is negative, the current really flows counterclockwise. There are no junctions, so I is the same for all elements in the circuit. With the 9 V battery being labeled 1 and the 6 V battery being labeled 2. Kirchhoff’s loop law is ! V i " = ! V bat 1 + ! V R + ! V bat 2 = + E 1 # IR # E 2 = 0 ! I = E 1 ! E 2 R = 9 V ! 6 V 30 " = 0.100 A Note the signs: Potential is gained in battery 1, but potential is lost both in the resistor and in battery 2. Because I is positive, we can say that I = 0.100 A flows from left to right through the resistor.

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