HW5solutions - Physics 213 HW #5 Solutions Spring 2009...

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Physics 213 HW #5 – Solutions Spring 2009 26.23. [Kirchoff’s Circuit Rules] We have to use Kirchoff’s rules to find the unknown currents, emfs, and resistances. Apply the junction rule at points a, b, c and d to calculate the unknown currents. Then apply the loop rule to three loops to calculate 12 , and . R  The circuit is sketched to the right. Apply the junction rule to point a : 3 3.00 A 5.00 A 0 I 3 8.00 A I Apply the junction rule to point b : 4 2.00 A 3.00 A 0 I 4 1.00 A I Apply the junction rule to point c : 3 4 5 0 I I I 5 3 4 8.00 A 1.00 A 7.00 A I I I As a check, apply the junction rule to point d : 5 2.00 A 5.00 A 0 I 5 7.00 A I (b) Apply the loop rule to loop (1):      13 3.00 A 4.00 3.00 0 I        1 12.0 V 8.00 A 3.00 36.0 V   Apply the loop rule to loop (2):      23 5.00 A 6.00 3.00 0 I        2 30.0 V 8.00 A 3.00 54.0 V   (c) Apply the loop rule to loop (3):   2.00 A 0 R 21 54.0 V 36.0 V 9.00 2.00 A 2.00 A R  Finally, we can apply the loop rule to loop (4) as a check of our calculations:         2.00 A 3.00 A 4.00 5.00 A 6.00 0 R        2.00 A 9.00 12.0 V 30.0 V 0   18.0 V 18.0 V 0 26.50. [Electric Dryer] (a) P VI I 2 R , so I P V 4100 W 240 V 17.1 A. Recall from page 201 that 12-gauge wire can carry up to 20A safely. Thus, we can use 12-gauge wire safely for this electric dryer. (b) 2 V P VI R  . Solving for the resistance, R V 2 P (240 V) 2 4100 W 14 . (c) At 11 c per kWH, for 1 hour the cost is (11 c/kWh)(1 h)(4.1kW) 45 c . (d) It makes sense that the cost to operate the device is proportional to its power consumption. The number also sounds about right. I am charged about one dollar to run a load of clothes through a dryer for an hour. 23.82. [Alpha Particle Collision] As the alpha particle moves from a to b, its total energy, kinetic plus potential, is conserved: K a U a K b U b . Assume the particles initially are far apart, so U a 0 , The alpha particle has zero speed at the distance of closest approach, so K b 0 . Conservation of energy then gives K a U b =11MeV.
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This note was uploaded on 04/14/2009 for the course PHYS 2213 taught by Professor Perelstein,m during the Spring '07 term at Cornell University (Engineering School).

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HW5solutions - Physics 213 HW #5 Solutions Spring 2009...

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