Hw_ch21 - Instructor Tomoyuki Nakayama Tuesday June 1 2010 PHY 2005 Applied Physics 2 Summer C 2010 Solutions for Suggested Homework

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Instructor: Tomoyuki Nakayama Tuesday, June 1, 2010 PHY 2005 Applied Physics 2 - Summer C 2010 Solutions for Suggested Homework Problems (Chapter 21) ________________________________________________________________________________ 1. (a) Kirchhoff’s loop rule yields 6 + 9 – 45 I = 0 I = 0.333 A. (b) Again we apply the loop rule, we get 9 -6 -45 I = 0 I = 0.0667 A. Note that the potential drops if you across a battery from its positive to negative terminal. 3. (a) I 2 is flowing into junction A, whereas I 1 and I 3 are flowing out. The junction rule gives I 2 = I 1 + I 3. (b) We the loop rule to obtain +7 I 2 + 12 +5 I 1 +3 I 1 -6 = 0 Note if you traverse a register in the opposite direction to that of the current flowing through it, the potential rises across the resistor. 7. Two resistors are connected in parallel. The equivalent resistance is R eq = (1/15 + 1/6) -1 = 4.29 . The current through the battery is I 3 = 12/ R eq = 2.8 A. The voltage across each resistor is the same as the voltage provided by the battery, we get I 1 = 12 / R eq = 0.80 A & I 2 = 12 / R 2 = 2.0 A. 10. The current through the 6- resistor is 0.8/6 = 0.133 A. Applying the loop rule to loop ABDA, we get -6 I 1 + 8 I 3 +2 = 0 I 3 = (6 I 1 – 2)/8 = -0.150 A. The junction rule at point D yields 0 =
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This note was uploaded on 07/16/2011 for the course PHY 2005 taught by Professor Lee during the Summer '08 term at University of Florida.

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Hw_ch21 - Instructor Tomoyuki Nakayama Tuesday June 1 2010 PHY 2005 Applied Physics 2 Summer C 2010 Solutions for Suggested Homework

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