Chapter 27: Current and Resistance Physics PDF

1  Section 27.1: Electric Current Assume charges are moving perpendicular to a surface of area A. If ∆𝑄 is the amount of charge that passes through 𝐴 in time ∆𝑡, then the average current is ∆𝑄...

1  Section 27.1: Electric Current Assume charges are moving perpendicular to a surface of area A. If ∆𝑄 is the amount of charge that passes through 𝐴 in time ∆𝑡, then the average current is ∆𝑄 𝐼𝑎𝑣𝑔 = ∆𝑡 If the rate at which the charge flows varies with time, the instantaneous current, 𝐼 , is defined as: ∆𝑄 𝑑𝑄 𝐼 = lim 𝐼𝑎𝑣𝑔 = lim → 𝐼= ∆𝑡→0 ∆𝑡→0 ∆𝑡 𝑑𝑡 The SI unit of current is the ampere (A): 1 A = 1 C / s. That is 1 A of current is equivalent to 1 C of charge passing through a surface in 1 s. The conventional direction of the current is the direction as the flow of positive charges. For a beam of protons in an accelerator, the direction of current is in the direction of protons’ motion. In conductors, current results from the motion of electrons thus the direction of current is opposite to the direction of electrons’ flow. In gasses and electrolytes, both positive and negative ions move  charge carriers. Exercise: Rank the current in the four regions shown. 𝑎>𝑏=𝑐>𝑑 1 2 3 Current results from the total flow of charge carriers. Dr. Suhad Sbeih 2 Microscopic Model of current To relate the motion of charge carriers to the current, consider the current in a cylindrical conductor of cross-sectional area 𝐴. The volume of the segment of length ∆𝑥 is 𝑉 = 𝐴 ∆𝑥. There are mobile charge carriers in the conductor, each of charge 𝑞. Let 𝑛 represents the number of mobile charge carriers per unit volume, charge carrier density, then the total charge ∆𝑄 in the wire segment is: 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑏𝑖𝑙𝑒 𝑐ℎ𝑎𝑟𝑔𝑒 𝑐𝑎𝑟𝑟𝑖𝑒𝑟𝑠 𝑛= 𝑢𝑛𝑖𝑡 𝑣𝑜𝑙𝑢𝑚𝑒 ∆𝑄 = (𝑛 𝐴 ∆𝑥) 𝑞 # of charge carriers = 𝑛 × 𝑉 Let ∆𝑡 be the time interval needed for the carriers to move across this segment 𝑣𝑑 : average drift speed with an average drift velocity v𝑑 in a direction parallel to the axis of cylinder. The magnitude of the displacement ∆𝑥 is given by: ∆𝑥 = 𝑣𝑑 ∆𝑡 The average current is: ∆𝑄 𝑛 𝐴 𝑣𝑑 ∆𝑡 𝑞 Microscopic model 𝐼𝑎𝑣𝑔 = = → 𝐼𝑎𝑣𝑔 = 𝑛 𝑞 𝑣𝑑 𝐴 ∆𝑡 ∆𝑡 of current In reality, free charge carriers (conduction electrons in metals) are randomly moving and colliding with atoms. In the presence of an electric field, in spite of all the collisions, the charge carriers slowly move along the conductor with an average drift velocity v𝑑. When a battery is connected to the ends of the conducting wire, the potential difference between its ends produces an electric field. The electric field exerts forces on the electrons in the wire causing them to move creating a current. Dr. Suhad Sbeih 3  Section 27.2: Resistance Consider a conductor of cross-sectional area 𝐴 carrying a current 𝐼 that is perpendicular to 𝐴. The uniform current density 𝐽 in the conductor is defined by the current per unit area, as: 𝐼 𝐽= = 𝑛 𝑞 𝑣𝑑 𝐴 SI units of current density is A/m2. A current density 𝐽 and an electric field 𝐸 are established in a conductor whenever a potential difference is maintained across the conductor. Ohm’s Law For many materials (including most metals), the ratio of the current density to the electric field is a constant 𝜎 that is independent of the electric field producing the current. 𝐽=𝜎𝐸 The constant of proportionality 𝜎 is a property of the Ohmic materials and is called the conductivity of the conductor. Some materials don’t have this property → nonohmic materials. Assume a conducting wire of length 𝑙 and cross-sectional area 𝐴. For a uniform 𝐸 in the wire, the potential difference is given by: ∆𝑉 = 𝐸 𝑙 From the definition of the uniform current density: 𝑙 𝐼 𝑙 ∆𝑉 𝑉𝑜𝑙𝑡 ∆𝑉 = 𝜎 𝐽 but 𝐽=𝐴 ∆𝑉 = 𝜎 𝐴 𝐼 = 𝑅 𝐼  𝑅 = 𝐼 , SI unit of 𝑅: Ohm (1 Ω = 1 𝐴𝑚𝑝 ) Resistivity of the material 𝝆: 𝑹 Resistance of the Conductor R 1 𝑙 It is the inverse of conductivity 𝜌 = 𝜎  𝑅=𝜌 𝑙 , 𝑅 ∝ 𝐴 , depends on conductor’s geometry 𝐴 SI unit of 𝜌 is: ohm-meters (Ω.m) 𝐴 𝜌=𝑅 Dr. Suhad Sbeih 4 𝑙  Every ohmic material has a characteristic resistivity that depends on the properties of the material and on temperature (sec. 27.4). An ideal conductor would have zero resistivity. An ideal insulator would have infinite resistivity. Ohmic material has a linear 𝐼−∆𝑉 graph and Non-ohmic material has a non-linear 𝐼−∆𝑉 graph and its slope is constant (constant resistance). its slope changes (e.g. 𝑅 decreases in a diode). Dr. Suhad Sbeih 5 a 𝑙 𝑅=𝜌 𝐴 𝑅 𝜌 𝜌 1×10−6 = = = 𝑙 𝐴 𝜋 𝑟2 𝜋 0.32×10−3 2 𝑅 = 3.1 Ω/m 𝑙 ∆𝑉 ∆𝑉 10 𝐼= = 𝑅 = = 3.2 A 𝑅 𝑙 3.1 ×1 𝑙 Dr. Suhad Sbeih 6  Section 27.4: Resistance and Temperature As Over a limited temperature range, the resistivity of a conductor varies approximately linearly with the temperature. The resistivity increases as the collisions between the electrons and atoms increase at high temperatures. 𝜌 = 𝜌0 1 + 𝛼 𝑇 − 𝑇0 𝜌0 is the resistivity at some reference temperature 𝑇0 (usually taken to be 20 ℃). 𝛼 is the temperature coefficient of resistivity ( SI units are ℃−1 ), see the previous table. A nonlinear region exists at very low temperatures in metals, due to the impurities and imperfections. Thus, the resistivity reaches some finite value as the temperature approaches absolute zero. For some materials, semiconductors, 𝛼 has negative values. This means that the resistivity decreases with temperature. The conductivity of semiconductors depends mainly on the type and concentration of impurities. Temperature variations of resistance: 𝑙 𝑙 Since 𝑅 = 𝜌 , multiply both sides of the equation above by : 𝐴 𝐴 𝑅 = 𝑅0 1 + 𝛼 𝑇 − 𝑇0 We use this relation to determine the resistance of a conductor, that is made of a certain material, as a function of temperature. Dr. Suhad Sbeih 7  Section 27.6: Electrical Power Power is the rate of energy transfer. For a circuits consisting of a source (battery) and a resistor (light bulb), energy is delivered from battery (that is stored as a chemical energy) to the resistor (as an electrical potential energy). Let us assume what happens as a positive charge 𝑄 moves through the battery (external agent) to the resistor (system), neglecting the resistance in the connecting wires. As 𝑄 moves from 𝑎 to 𝑏, through the potential difference ∆𝑉, the battery losses chemical energy of the amount 𝑄 ∆𝑉. Then as the charge moves from 𝑐 to 𝑑, the resistor gains energy at the same amount and stores it as an internal energy (vibrations of atoms) and heats up. The air in contact with the bulb gets warmer. After some time, the non-isolated system reaches a steady state (bulb reaches constant temperature). The rate at which the energy is transferred (delivered) to the resistor is the power: An electron takes hours 𝑑𝑈 𝑑 𝑑𝑄 to move all the way 𝑃= = 𝑄 ∆𝑉 = ∆𝑉 = 𝐼∆𝑉 around the circuit at the 𝑑𝑡 𝑑𝑡 𝑑𝑡 drift velocity. Using the formula ∆𝑉 = 𝐼𝑅, the power delivered in a resistor can be expressed in Current is the same different forms: everywhere in the circuit. ∆𝑉 2 2 𝑃 = 𝐼∆𝑉 = 𝐼 𝑅 = 𝑅 SI units of power is Watt: 1 W = 1 J/s. Power is delivered to our homes at high voltages and low currents to minimize power losses. Since power lines have high resistance (copper wires are very expensive). Dr. Suhad Sbeih 8  𝑄 The current is the electron’s flow rate: 𝐼 = 𝑡 The time needed for the electron to complete one period in its circular motion: 2𝜋𝑅 𝑇= 𝑣 So: 𝑒 𝑣 1.6×10−9 ×2.19×106 𝐼= 2𝜋𝑅 = 2𝜋 5.29×10−11 = 1.05 mA Dr. Suhad Sbeih 9  Electron’s density in copper: 𝑛 = 8.46 × 1028 electrons/m3 (a) Microscopic definition of the current: 𝐼 = 𝑛𝑞𝑣𝑑 𝐴 Where the cross-sectional area is: 𝐴 = 𝜋𝑟 2 𝐼 𝑣𝑑 = 𝑛𝑞𝐴 3.7 = 8.46×1028 ×1.6×10−19 ×𝜋× 1.25×10−3 2 𝑣𝑑 = 5.57 × 10−5 m/s (b) The drift speed is smaller because more electrons are being conducted. To create the same current, the drift speed doesn’t need to be as great. Dr. Suhad Sbeih 10 (a) The instantaneous current at some instant 𝑡 is given by: 𝑑𝑞 𝐼= 𝑑𝑡 = 12𝑡 2 + 5 At 𝑡 = 1 s: 2 𝐼 = 12 1 + 5 = 17 A (b) The current density: 𝐼 17 𝐽= = = 85 kA/m2 𝐴 2×10−4 Dr. Suhad Sbeih 11  120 𝜋𝑡 v v Dr. Suhad Sbeih 12 Dr. Suhad Sbeih 13  Resistivity of tungsten: 𝜌 = 5.6 × 10−8 Ω.m From Ohm’s law: ∆𝑉 𝐼= 𝑅 𝜌𝑙 also: 𝑅= 𝐴 5.6×10−8 ×1.5 𝑅= = 1.4 Ω 0.6×10−6 So the current is: 0.9 𝐼= = 6.43 A 1.4 Dr. Suhad Sbeih 14  𝜌𝐴𝑙 = 2.82 × 10−8 Ω.m and 𝜌𝐶𝑢 = 1.7 × 10−8 Ω.m 𝜌𝑙 𝑅= 𝐴 and 𝑅𝐴𝑙 = 𝑅𝐶𝑢 𝜌𝐴𝑙 𝑙 𝜌𝐶𝑢 𝑙 𝜌𝐴𝑙 𝜌𝐶𝑢 𝜌𝐴𝑙 𝜌𝐶𝑢 so: = → 2 = 2 → 2 = 2 𝐴𝐴𝑙 𝐴𝐶𝑢 𝜋𝑟𝐴𝑙 𝜋𝑟𝐶𝑢 𝑟𝐴𝑙 𝑟𝐶𝑢 2 𝑟𝐴𝑙 𝜌𝐴𝑙 𝑟𝐴𝑙 𝜌𝐴𝑙 → 2 = → = 𝑟𝐶𝑢 𝜌𝐶𝑢 𝑟𝐶𝑢 𝜌𝐶𝑢 𝑟𝐴𝑙 2.82×10−8 → = = 1.29 𝑟𝐶𝑢 1.7×10−8 Dr. Suhad Sbeih 15  𝛼𝑖𝑟𝑜𝑛 = 5 × 10−3 ℃−1 If we ignore the thermal expansion, the resistance will change from 𝑅0 to 𝑅, that is given by: 𝑅 = 𝑅0 1 + 𝛼 𝑇 − 𝑇0 𝑅−𝑅0 The fractional change in resistance is defined as: 𝑓 = 𝑅0 We can arrange the first equation to find 𝑓: 𝑅 = 𝑅0 + 𝛼𝑅0 𝑇 − 𝑇0 𝑅 − 𝑅0 = 𝛼𝑅0 𝑇 − 𝑇0 𝑅−𝑅0 = 𝛼 𝑇 − 𝑇0 𝑅0 𝑓 = 5 × 10−3 50 − 25 = 0.12 Dr. Suhad Sbeih 16 𝛼𝑠𝑖𝑙𝑣𝑒𝑟 = 3.8 × 10−3 ℃−1 𝑅 = 𝑅0 1 + 𝛼 𝑇 − 𝑇0 = 6 1 + 3.8 × 10−3 34 − 20 𝑅 = 6.32 Ω Dr. Suhad Sbeih 17 Dr. Suhad Sbeih 18 (a) Using the equation of the power in a resistor: 𝑃 1000 𝑃 = 𝐼 ∆𝑉 → 𝐼 = ∆𝑉 = 120 = 8.33 A (b) To find the resistance, use one of the other forms for the power equation: ∆𝑉 2 ∆𝑉 2 120 2 𝑃= →𝑅= = = 14.4 Ω 𝑅 𝑃 1000 Dr. Suhad Sbeih 19 W Dr. Suhad Sbeih 20 Dr. Suhad Sbeih 21