HW5solutions

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

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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)....
View Full Document

## 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.

### Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online