Copyright © 2009 Pearson Education, Inc. Chapter 25 Electric Currents and Resistance

Copyright © 2009 Pearson Education, Inc. Recap

Copyright © 2009 Pearson Education, Inc. The wires used in homes to carry electricity have very low resistance. However, if the current is high enough, the power will increase and the wires can become hot enough to start a fire. To avoid this, we use fuses or circuit breakers, which disconnect when the current goes above a predetermined value Power in Household Circuits

Copyright © 2009 Pearson Education, Inc. Fuses are one-use items – if they blow, the fuse is destroyed and must be replaced Power in Household Circuits Very old type  Common in cars   Most common

Copyright © 2009 Pearson Education, Inc. Circuit breakers, which are now much more common in homes than they once were, are switches that will open if the current is too high; they can then be reset Power in Household Circuits

Copyright © 2009 Pearson Education, Inc. Current from a battery flows steadily in one direction (direct current, DC). Current from a power plant varies sinusoidally (alternating current, AC) Alternating Current

Copyright © 2009 Pearson Education, Inc. The voltage varies sinusoidally with time: as does the current: 25-7 Alternating Current,,

Copyright © 2009 Pearson Education, Inc. Multiplying the current and the voltage gives the power: 25-7 Alternating Current

Copyright © 2009 Pearson Education, Inc. Usually we are interested in the average power: 25-7 Alternating Current

Copyright © 2009 Pearson Education, Inc. The current and voltage both have average values of zero, so we square them, take the average, then take the square root, yielding the root-mean-square (rms) value: 25-7 Alternating Current

Copyright © 2009 Pearson Education, Inc Alternating Current Example 25-13: Hair dryer. (a) Calculate the resistance and the peak current in a 1000-W hair dryer connected to a 120-V line. (b) What happens if it is connected to a 240-V line in Britain?

Copyright © 2009 Pearson Education, Inc. Electrons in a conductor have large, random speeds just due to their temperature. When a potential difference is applied, the electrons also acquire an average drift velocity, which is generally considerably smaller than the thermal velocity Microscopic View of Electric Current: Current Density and Drift Velocity

Copyright © 2009 Pearson Education, Inc Microscopic View of Electric Current: Current Density and Drift Velocity We define the current density (current per unit area) – this is a convenient concept for relating the microscopic motions of electrons to the macroscopic current: If the current is not uniform:.

Copyright © 2009 Pearson Education, Inc. This drift speed is related to the current in the wire, and also to the number of electrons per unit volume: 25-8 Microscopic View of Electric Current: Current Density and Drift Velocity and

Copyright © 2009 Pearson Education, Inc Microscopic View of Electric Current: Current Density and Drift Velocity Example 25-14: Electron speeds in a wire. A copper wire 3.2 mm in diameter carries a 5.0- A current. Determine (a) the current density in the wire, and (b) the drift velocity of the free electrons. (c) Estimate the rms speed of electrons assuming they behave like an ideal gas at 20°C. Assume that one electron per Cu atom is free to move (the others remain bound to the atom).

Copyright © 2009 Pearson Education, Inc Microscopic View of Electric Current: Current Density and Drift Velocity The electric field inside a current-carrying wire can be found from the relationship between the current, voltage, and resistance. Writing R = ρ l /A, I = jA, and V = E l, and substituting in Ohm’s law gives:

Copyright © 2009 Pearson Education, Inc Microscopic View of Electric Current: Current Density and Drift Velocity Example 25-15: Electric field inside a wire. What is the electric field inside the wire of Example 25–14? (The current density was found to be 6.2 x 10 5 A/m 2.)

Copyright © 2009 Pearson Education, Inc. In general, resistivity decreases as temperature decreases. Some materials, however, have resistivity that falls abruptly to zero at a very low temperature, called the critical temperature, T C Superconductivity

Copyright © 2009 Pearson Education, Inc. Experiments have shown that currents, once started, can flow through these materials for years without decreasing even without a potential difference. Critical temperatures are low; for many years no material was found to be superconducting above 23 K. Since 1987, new materials have been found that are superconducting below 90 K, and work on higher temperature superconductors is continuing Superconductivity

Copyright © 2009 Pearson Education, Inc. A battery is a source of constant potential difference. Electric current is the rate of flow of electric charge. Conventional current is in the direction that positive charge would flow. Resistance is the ratio of voltage to current: Summary of Chapter 25

Copyright © 2009 Pearson Education, Inc. Ohmic materials have constant resistance, independent of voltage. Resistance is determined by shape and material: ρ is the resistivity. Summary of Chapter 25

Copyright © 2009 Pearson Education, Inc. Power in an electric circuit: Direct current is constant. Alternating current varies sinusoidally: Summary of Chapter 25

Copyright © 2009 Pearson Education, Inc. The average (rms) current and voltage: Relation between drift speed and current: Summary of Chapter 25

Copyright © 2009 Pearson Education, Inc. Chapter 26 DC Circuits

Copyright © 2009 Pearson Education, Inc. EMF and Terminal Voltage Resistors in Series and in Parallel Kirchhoff’s Rules Series and Parallel EMFs; Battery Charging Circuits Containing Resistor and Capacitor (RC Circuits) Ammeters and Voltmeters Units of Chapter 26

Copyright © 2009 Pearson Education, Inc. A series connection has a single path from the battery, through each circuit element in turn, then back to the battery Resistors in Series and in Parallel Series = absolutely, positively, no-question-about-it same current through resistors!

Copyright © 2009 Pearson Education, Inc. The current through each resistor is the same; the voltage depends on the resistance. The sum of the voltage drops across the resistors equals the battery voltage: 26-2 Resistors in Series and in Parallel

Copyright © 2009 Pearson Education, Inc. From this we get the equivalent resistance (that single resistance that gives the same current in the circuit): 26-2 Resistors in Series and in Parallel

Copyright © 2009 Pearson Education, Inc. Electric circuit needs battery or generator to produce current – these are called sources of emf (archaic term: electromotive force). Battery is a nearly constant voltage source, but does have a small internal resistance, which reduces the actual voltage from the ideal emf: 26-1 EMF and Terminal Voltage

Copyright © 2009 Pearson Education, Inc. This resistance behaves as though it were in series with the emf EMF and Terminal Voltage

Copyright © 2009 Pearson Education, Inc EMF and Terminal Voltage Example 26-1: Battery with internal resistance. A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calculate (a) the current in the circuit, (b) the terminal voltage of the battery, V ab, and (c) the power dissipated in the resistor R and in the battery’s internal resistance r.

ConcepTest 26.1bSeries Resistors II 12 V R 1 = 4  R 2 = 2  In the circuit below, what is the voltage across ? In the circuit below, what is the voltage across R 1 ? 1) 12 V 2) zero 3) 6 V 4) 8 V 5) 4 V

ConcepTest 26.1bSeries Resistors II 12 V R 1 = 4  R 2 = 2  In the circuit below, what is the voltage across ? In the circuit below, what is the voltage across R 1 ? 1) 12 V 2) zero 3) 6 V 4) 8 V 5) 4 V The voltage drop across R 1 has to be twice as big as the drop across R 2.V 1 = 8 V The voltage drop across R 1 has to be twice as big as the drop across R 2. This means that V 1 = 8 V and V 2 = 4 V. Or else you could find the current I = V/R = (12 V)/(6  = 2 A, and then use Ohm’s law to get voltages.

Copyright © 2009 Pearson Education, Inc. A parallel connection splits the current; the voltage across each resistor is the same: 26-2 Resistors in Series and in Parallel Parallel = absolutely, positively, no-question-about-it same potential drop across resistors!

Copyright © 2009 Pearson Education, Inc. The total current is the sum of the currents across each resistor: 26-2 Resistors in Series and in Parallel

Copyright © 2009 Pearson Education, Inc. An analogy using water may be helpful in visualizing parallel circuits. The water (current) splits into two streams; each falls the same height, and the total current is the sum of the two currents. With two pipes open, the resistance to water flow is half what it is with one pipe open Resistors in Series and in Parallel

Copyright © 2009 Pearson Education, Inc Resistors in Series and in Parallel Conceptual Example 26-2: Series or parallel? (a) The lightbulbs in the figure are identical. Which configuration produces more light? (b) Which way do you think the headlights of a car are wired? Ignore change of filament resistance R with current.

ConcepTest 26.2bParallel Resistors II 1) increases 2) remains the same 3) decreases 4) drops to zero Points P and Q are connected to a battery of fixed voltage. As more resistors R are added to the parallel circuit, what happens to the total current in the circuit?

ConcepTest 26.2bParallel Resistors II 1) increases 2) remains the same 3) decreases 4) drops to zero resistance of the circuit drops resistance decreasescurrent must increase As we add parallel resistors, the overall resistance of the circuit drops. Since V = IR, and V is held constant by the battery, when resistance decreases, the current must increase. Points P and Q are connected to a battery of fixed voltage. As more resistors R are added to the parallel circuit, what happens to the total current in the circuit? Follow-up: What happens to the current through each resistor?

Copyright © 2009 Pearson Education, Inc Resistors in Series and in Parallel Conceptual Example 26-3: An illuminating surprise. A 100-W, 120-V lightbulb and a 60-W, 120-V lightbulb are connected in two different ways as shown. In each case, which bulb glows more brightly? Ignore change of filament resistance with current (and temperature).

Copyright © 2009 Pearson Education, Inc Resistors in Series and in Parallel Example: What is the current through each resistor shown?

Copyright © 2009 Pearson Education, Inc Resistors in Series and in Parallel Conceptual Example 26-6: Bulb brightness in a circuit. The circuit shown has three identical lightbulbs, each of resistance R. (a) When switch S is closed, how will the brightness of bulbs A and B compare with that of bulb C ? (b) What happens when switch S is opened? Use a minimum of mathematics in your answers.

Copyright © 2009 Pearson Education, Inc Resistors in Series and Parallel Example 26-8: Analyzing a circuit. A 9.0-V battery whose internal resistance r is 0.50 Ω is connected in the circuit shown. (a) How much current is drawn from the battery? (b) What is the terminal voltage of the battery? (c) What is the current in the 6.0-Ω resistor?

Copyright © 2009 Pearson Education, Inc. Multiple configurations Assume that in each circuit the battery gives 12 V and each resistor has a resistance of 4 ohms. In which circuit does the largest current flow through the battery? What is that current?

Copyright © 2009 Pearson Education, Inc. Multiple configurations Assume that in each circuit the battery gives 12 V and each resistor has a resistance of 4 ohms. In which circuit does the largest current flow through the battery? What is that current?

Copyright © 2009 Pearson Education, Inc. Circuit Maze If each resistor has a resistance of 4  and each battery is a 4 V battery, what is the current flowing through the resistor labelled “R”?

Copyright © 2009 Pearson Education, Inc. Circuit Maze If each resistor has a resistance of 4  and each battery is a 4 V battery, what is the current flowing through the resistor labelled “R”?

Copyright © 2009 Pearson Education, Inc. Some circuits cannot be broken down into series and parallel connections. For these circuits we use Kirchhoff’s rules Kirchhoff’s Rules

Copyright © 2009 Pearson Education, Inc. Junction rule: The sum of currents entering a junction equals the sum of the currents leaving it (i.e., charge does not pile up) Kirchhoff’s Rules

Copyright © 2009 Pearson Education, Inc. Loop rule: The sum of the changes in potential around a closed loop is zero Kirchhoff’s Rules

Copyright © 2009 Pearson Education, Inc. Problem Solving: Kirchhoff’s Rules 1. Label each current, including its direction. 2. Identify unknowns. 3. Apply junction and loop rules; you will need as many independent equations as there are unknowns. Each new equation MUST include a new variable. 4.Solve the equations, being careful with signs. If the solution for a current is negative, that current is in the opposite direction from the one you have chosen Kirchhoff’s Rules

Copyright © 2009 Pearson Education, Inc Kirchhoff’s Rules Example: Using Kirchhoff’s rules. a)Calculate the currents (call them I 1, I 2, and I 3 ) through the three batteries of the circuit in the figure. b)What is V a -V b ?

ConcepTest 26.12Kirchhoff’s Rules ConcepTest More Kirchhoff’s Rules 2 V 2  2 V 6 V 4 V 3  1  I1I1 I3I3 I2I2 Which of the equations is valid for the circuit below? 1) 2 – I 1 – 2I 2 = 0 2) 2 – 2I 1 – 2I 2 – 4I 3 = 0 3) 2 – I 1 – 4 – 2I 2 = 0 4) I 3 – 4 – 2I = 0 5) 2 – I 1 – 3I 3 – 6 = 0

ConcepTest 26.12Kirchhoff’s Rules ConcepTest More Kirchhoff’s Rules 2 V 2  2 V 6 V 4 V 3  1  I1I1 I3I3 I2I2 Eq. 3 is valid for the left loop Eq. 3 is valid for the left loop: The left battery gives +2 V, then there is a drop through a 1  resistor with current I 1 flowing. Then we go through the middle battery (but from + to – !), which gives –4 V. Finally, there is a drop through a 2  resistor with current I 2. Which of the equations is valid for the circuit below? 1) 2 – I 1 – 2I 2 = 0 2) 2 – 2I 1 – 2I 2 – 4I 3 = 0 3) 2 – I 1 – 4 – 2I 2 = 0 4) I 3 – 4 – 2I = 0 5) 2 – I 1 – 3I 3 – 6 = 0

Copyright © 2009 Pearson Education, Inc. Questions?