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PE Unreasonable Results (a) What current is needed to

Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli ISBN: 9780130606204 3

Solution for problem 69PE Chapter 20

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Physics: Principles with Applications | 6th Edition

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Problem 69PE

PE Unreasonable Results (a) What current is needed to transmit 1.00×10? MW of 2? power at 480 V? (b) What power is dissipated by the transmission lines if they have a 1.00 - ? resistance? (c) What is unreasonable about this result? (d) Which assumptions are unreasonable, or which premises are inconsistent?

Step-by-Step Solution:

Step-by-step solution Given: Power = 1.00×10 MW Voltage = 480 V Resistance = 1.00 - 6 1 MW = 10 Watt Calculate: Power is dissipated by the transmission lines= Comment on unreasonable,assumptions = Step 1 of 5 Power is defined as the rate at which energy is consumed. The relation for power P, electric current I and voltage V is given by equation (1) P = VI …………………..(1) Step 2 of 5 Rearranging equation (1) , we can write as below in equation (2) I = P/V…………..(2) 2 Substitute 1.00×10 MW for P and 480 V for V in equation (2) we get, I = 1.00 x 10 MW / 480 V 2 6 I = 1.00 x 10 x 10 W/ 480 V 5 = 2.08 x 10 A So, the current needed is 2.08 x 10 A 5 . Step 3 of 5 The power dissipated in presence of resistance R in the transmission line will be given by equation ( 3) as below P= I R……………….(3) 5 Substitute 2.08 x 10 A for I and 1.00 - for R:in equation (3), we get 5 2 P = (2.08 x 10 A) (1.00 - ) = 4.33 x 10 W10 = 43.3 G W Therefore, the power dissipated is 43.3 G W.

Step 4 of 5

Chapter 20, Problem 69PE is Solved
Step 5 of 5

Textbook: Physics: Principles with Applications
Edition: 6
Author: Douglas C. Giancoli
ISBN: 9780130606204

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PE Unreasonable Results (a) What current is needed to

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