Class Examples for Chapter 27

Class Examples for Chapter 27 - Class Examples for Chapter...

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1 Class Examples for Chapter 27 12. (a) For each wire, R wire = L/A where A = r 2 . Consequently, we have R wire = (1.69 10 8 m  )(0.200 m)/ (0.00100 m) 2 = 0.0011 . The total resistive load on the battery is therefore tot R = 2 R wire + R =  0.0011   6.00    . Dividing this into the battery emf gives the current tot 12.0 V 1.9993 A 6.0022 i R . The voltage across the R = 6.00 resistor is therefore V iR  (1.9993 A)(6.00 ) = 11.996 V 12.0 V. (b) Similarly, we find the voltage-drop across each wire to be wire wire V iR (1.9993 A)(0.0011 ) = 2.15 mV. (c) P = i 2 R = (1.9993 A)(6.00 ) 2 = 23.98 W 24.0 W. (d) Similarly, we find the power dissipated in each wire to be 4.30 mW. 13. (a) We denote L = 10 km and = 13 /km. Measured from the east end we have R 1 = 100 = 2 ( L – x ) + R , and measured from the west end R 2 = 200 = 2 x + R . Thus, x R R L 2 1 4 2 200 100 4 13 10 2 69 km km km. b g . (b) Also, we obtain R R R L 1 2 2 100 200 2 13 10 20 km km b gb g .
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2 30. Using the junction rule ( i 3 = i 1 + i 2 ) we write two loop rule equations: 10.0 V – i 1 R 1 – ( i 1 + i 2 ) R 3 = 0 5.00 V – i 2 R 2 – ( i 1 + i 2 ) R 3 = 0. (a) Solving, we find i 2 = 0, and (b) i 3 = i 1 + i 2 = 1.25 A (downward, as was assumed in writing the equations as we did).
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This document was uploaded on 11/04/2011 for the course PHY 108 at SUNY Buffalo.

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Class Examples for Chapter 27 - Class Examples for Chapter...

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