CourseNotes.28 - the resistivity of copper is ρ = 1 69 ×...

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7.3 Power in Electric Circuits From our everyday experiences, we should expect that powering electric circuits uses up some energy. Up until now in our discussion so circuits, we have not discussed this energy loss. We are now ready to do so because the energy lost in electrical circuits is mostly lost to the variance resistances in the circuit. The rate of electrical energy transfer in a circuit is given by P = iV Notice that for circuits which obey Ohm’s law we can rearrange this expression to get two different forms. P = V 2 R P = i 2 R 7.4 Problems Problem 26.24 Figure 22 gives the electric potential V ( x ) along a copper wire carrying uniform current, From a point of higher potential V s = 12 μV at x = 0 to a point of zero potential at x s = 3 m . The wire has a radius of 2 mm . What is the current in the wire? Figure 22: Plot for Problem 26.24 The problem tells us that the material is copper. From the table in the book we can see that
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Unformatted text preview: the resistivity of copper is ρ = 1 . 69 × 10-8 Ω · m Since we also know the length (from the plot) and the cross sectional area of the wire we can calculate the resistance. R = ρ L A = ρ x s πr 2 Hence, from Ohm’s law the current in the wire is i = V R = V s ρ x s πr 2 = V s πr 2 ρx s = 3 . × 10-3 A 8 Chapter 27: Circuits The last two chapters have, in essence, been a build up to this chapter. We have spent these last two chapters learning about three key concepts: capacitors, resistors, and energy in electric circuits. In this chapter we will bring these three ideas together and use them to talk about electric circuits as a whole. We will begin to see why electric circuits are useful and why they have become the cornerstone of the modern world. 28...
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