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Chapter 17. Current and Resistance. André-Marie Ampère 1775 – 1836. Electric Current. Let us look at the charges flowing perpendicularly to a surface of area A The electric current is the rate at which the charge flows through this surface

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## Chapter 17

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**Chapter 17**Current and Resistance**André-Marie Ampère**1775 – 1836 Electric Current • Let us look at the charges flowing perpendicularly to a surface of area A • The electric current is the rate at which the charge flows through this surface • The SI unit of current is Ampere (A): 1 A = 1 C/s**Electric Current**• The conventional direction of the current is the direction positive charge would flow • In a common conductor (e.g., copper), the current is due to the motion of the negatively charged electrons • It is common to refer to a moving charge as a mobile charge carrier • A charge carrier can be positive or negative**Current and Drift Speed**• Charged particles move through a conductor of cross-sectional area A and a charge carrier density n • The total number of charge carriers: n A Δx • The total charge is the number of carriers times the charge per carrier, q: ΔQ = (n A Δx) q • The drift speed, vd, is the speed at which the carriers move: vd = Δx / Δt ΔQ = (n A vd Δt) q I = ΔQ / Δt = n q vd A**Current and Drift Speed**• If the conductor is isolated, the electrons undergo random motion (due to collisions with the atoms) • When an electric field is set up in the conductor, it creates an electric force on the electrons and hence a current • The zigzag line represents the motion of charge carrier in a conductor**Current and Drift Speed**• The drift speed is much smaller than the average speed between collisions • When a circuit is completed, the electric field travels with a speed close to the speed of light • Therefore, although the drift speed is on the order of 10-4 m/s the effect of the electric field is felt on the order of 108 m/s**Meters in a Circuit – Ammeter, Voltmeter**• An ammeter is used to measure current in line with the bulb – all the charge passing through the bulb also must pass through the meter • A voltmeter is used to measure voltage (potential difference) – connects to the two ends of the bulb**Georg Simon Ohm**1787 – 1854 Resistance • In a conductor, the voltage applied across the ends of the conductor is proportional to the current through the conductor • The constant of proportionality is the resistance of the conductor – it arises due to collisions between the electrons carrying the current with the fixed atoms inside the conductor • SI unit of resistance is ohm (Ω): 1 Ω = 1 V / A**Non-ohmic**Ohm’s Law • For certain materials, including most metals, the resistance remains constant over a wide range of applied voltages or currents – this statement has become known as Ohm’s Law ΔV = I R • Materials that obey Ohm’s Law are said to be ohmic (the relationship between current and voltage is linear)**L**R = r A Resistivity • The resistance of an ohmic conductor is proportional to its length, L, and inversely proportional to its cross-sectional area, A • ρ is the constant of proportionality and is called the resistivity of the material (See table 17.1)**Chapter 17Problem 12**Suppose that you wish to fabricate a uniform wire out of 1.00 g of copper. If the wire is to have a resistance R = 0.500 Ω, and if all of the copper is to be used, what will be (a) the length and (b) the diameter of the wire?**Temperature Variation of Resistivity**• For most metals, resistivity increases with increasing temperature – the atoms vibrate with increasing amplitude so the electrons find it more difficult to pass through the atoms • For most metals, resistivity increases approximately linearly with temperature over a limited temperature range • ρ0 – resistivity at some reference temperature T0 (usually taken to be 20° C); α – 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, the effect of temperature on resistance is similar**Chapter 17Problem 22**A metal wire has a resistance of 10.00 Ω at a temperature of 20°C. If the same wire has a resistance of 10.55 Ω at 90°C, what is the resistance of the wire when its temperature is −20°C?**Superconductors**• Superconductors – a class of materials whose resistances fall to virtually zero below a certain temperature, TC (critical temperature) • The value of TC is sensitive to chemical composition, pressure, and crystalline structure • Once a current is set up in a superconductor, it persists without any applied voltage (since R = 0) • One application is superconducting magnets**Electrical Energy and Power**• In a circuit, as a charge moves through the battery, the electrical potential energy of the system is increased by ΔQ ΔV (the chemical potential energy of the battery decreases by the same amount) • The charge moving through a resistor loses this potential energy during collisions with atoms in the resistor (the temperature of the resistor increases) • When the charge returns to A, the net result is that some chemical energy of the battery has been delivered to the resistor and caused its temperature to rise**Electrical Energy and Power**• The rate at which the energy is lost is the power • From Ohm’s Law, alternate forms of power are • The SI unit of power is Watt (W) (I must be in Amperes, R in ohms and ΔV in Volts) • The unit of energy used by electric companies is the kilowatt-hour (defined in terms of the unit of power and the amount of time it is supplied): 1 kWh = 3.60 x 106 J**Chapter 17Problem 60**In a certain stereo system, each speaker has a resistance of 4.00 Ω. The system is rated at 60.0 W in each channel. Each speaker circuit includes a fuse rated at a maximum current of 4.00 A. Is this system adequately protected against overload?**Chapter 17Problem 28**A toaster rated at 1 050 W operates on a 120-V household circuit and a 4.00-m length of nichrome wire as its heating element. The operating temperature of this element is 320°C. What is the cross-sectional area of the wire?**Answers to Even Numbered Problems**Chapter 17: Problem 2 5.21 × 10−5 m / s**Answers to Even Numbered Problems**Chapter 17: Problem 10 500 mA**Answers to Even Numbered Problems**• Chapter 17: • Problem 18 • 2.8 × 108 A • 1.8 × 107 A**Answers to Even Numbered Problems**Chapter 17: Problem 20 1.4 × 103 °C**Answers to Even Numbered Problems**• Chapter 17: • Problem 32 • $0.29 • $2.6

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