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Chapter 17. Current and Resistance. Example 17.3 – Page 578. A) Calculate the resistance per unit length of a 22-gauge nichrome wire of radius 0.321mm. B) If a potential difference of 10V is maintained across a 1m length of the nichrome wire, what is the current in the wire?

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## Chapter 17

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**Chapter 17**Current and Resistance**Example 17.3 – Page 578**• A) Calculate the resistance per unit length of a 22-gauge nichrome wire of radius 0.321mm. • B) If a potential difference of 10V is maintained across a 1m length of the nichrome wire, what is the current in the wire? • C) The wire is melted down and recast with twice its original length. Find the new resistance RN as a multiple of the old resistance R0.**Example 17.3 – continue**• A) • B) • C)**Temperature Variation of Resistivity**• For most metals, resistivity increases with increasing temperature • With a higher temperature, the metal’s constituent atoms vibrate with increasing amplitude • The electrons find it more difficult to pass through the atoms**Temperature Variation of Resistivity, cont**• For most metals, resistivity increases approximately linearlywith temperature over a limited temperature range • ρ is the resistivity at some temperature T • ρo is the resistivity at some reference temperature To • To is usually taken to be 20° C • is the temperature coefficient of resistivity**Temperature Variation of Resistance**• Since the resistance of a conductor with uniform cross sectional area is proportional to the resistivity, you can find the effect of temperature on resistance**Example 17.4 – Page 580**• A resistance thermometer, which measures temperature by measuring the change in resistance of a conductor, is made of platinum and has a resistance of 50Ω at 200C. • A) when the device is immersed in a vessel containing melting indium, its resistance increases to 76.8Ω. From this information, find the melting point of indium. • B) the indium is heated further until it reaches a temperature of 2350C. What is the ratio of the new current in the platinum to the current Imp at the melting point?**Example 17.4 – continue**• A) • B)**Electrical Energy and Power**• In a circuit, as a charge moves through the battery, the electrical potential energy of the system is increased by ΔQΔV • The chemical potential energy of the battery decreases by the same amount • As the charge moves through a resistor, it loses this potential energy during collisions with atoms in the resistor • The temperature of the resistor will increase**Energy Transfer in the Circuit**• Consider the circuit shown • Imagine a quantity of positive charge, DQ, moving around the circuit from point A back to point A**Energy Transfer in the Circuit, cont**• Point A is the reference point • It is grounded and its potential is taken to be zero • As the charge moves through the battery from A to B, the potential energy of the system increases by DQDV • The chemical energy of the battery decreases by the same amount**Energy Transfer in the Circuit, final**• As the charge moves through the resistor, from C to D, it loses energy in collisions with the atoms of the resistor • The energy is transferred to internal energy • When the charge returns to A, the net result is that some chemical energyof the battery has been delivered to the resistor and caused its temperature to rise**Electrical Energy and Power, cont**• The rate at which the energy is lost is the power • From Ohm’s Law, alternate forms of power are**Electrical Energy and Power, final**• The SI unit of power is Watt (W) • I must be in Amperes, R in ohms and DV in Volts • The unit of energy used by electric companies is the kilowatt-hour • This is defined in terms of the unit of power and the amount of time it is supplied • 1 kWh = 3.60 x 106 J

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