HW5solutions

HW5solutions - Physics 213 HW #5 – Solutions Spring 2008...

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Unformatted text preview: Physics 213 HW #5 – Solutions Spring 2008 26.50. IDENTIFY: P = VI = I 2 R SET UP: Problem 25.77 says that for 12-gauge wire the maximum safe current is 2.5 A. EXECUTE: (a) I = P V = 4100 W 240 V = 17.1 A. So we need at least 14-gauge wire (good up to 18 A). 12 gauge is also ok (good up to 25 A). (b) P = V 2 R and 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)(4.1 kW) = 45 / c . EVALUATE: The cost to operate the device is proportional to its power consumption. 26.61. IDENTIFY: Apply the junction rule to express the currents through the 5.00 ¡ and 8.00 ¡ resistors in terms of I 1 , I 2 and I 3 . Apply the loop rule to three loops to get three equations in the three unknown currents. SET UP: See the circuit sketched at right. The current in each branch has been written in terms of I 1 , I 2 and I 3 such that the junction rule is satisfied at each junction point. EXECUTE: Apply the loop rule to loop (1). ¡ 12.0 V + I 2 1.00 ¢ ( ) + I 2 ¡ I 3 ( ) 5.00 ¢ ( ) = 0 I 2 6.00 ¡ ( ) ¢ I 3 5.00 ¡ ( ) = 12.0 V eq.(1) Apply the loop rule to loop (2)....
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This note was uploaded on 08/27/2008 for the course PHYS 2213 taught by Professor Perelstein,m during the Spring '07 term at Cornell.

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HW5solutions - Physics 213 HW #5 – Solutions Spring 2008...

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