Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Current Conservation of current Batteries Resistance and resistivity Simple circuits Chapter 22 Current and Resistance Topics: Sample question: How can the measurement of an electric current passed through a person’s body allow a determination of the percentage body fat? Slide 22-1

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Series and Parallel Slide Series Circuit elements in a chain between 2 points Same current flows through circuit elements I 1 = I 2 Electric Potentials add => Delta V total = Delta V 1 + Delta V 2 Parallel Circuit elements on multiple paths connecting the same points Since paths connect the same points, Delta V’s are the same Currents Add => I total = I 1 + I 2

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Show equipotentials in a circuit Slide 22-35

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. The filament of a 100-W bulb carries a current of 0.83 A at the normal operating voltage of 120 V. A.What is the resistance of the filament? B.If the filament is made of tungsten wire of diameter mm, how long is the filament? Example Problem Slide 22-28

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Why is one bulb brighter? Which has the greatest resistance? In parallel, which carries the greatest current? Why is one bulb Brighter, 40 W vs. 100 W Slide 22-35

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Why is one bulb brighter? Which has the greatest resistance? In parallel, which carries the greatest current? How are the plugs in your home wired? A.In series B.In parallel C.In combination How can you tell? Which bulb is brighter in series, 40 W vs. 100 W Slide 22-35

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Conductivity and Resistivity This woman is measuring her percentage body fat by gripping a device that sends a small electric current through her body. Because muscle and fat have different resistivities, the amount of current allows the fat-to-muscle ratio to be determined. Slide 30-66

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Example Problem: Measuring Body Fat Slide 22-35

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Analyzing a Simple Circuit In a circuit using a battery, a lightbulb, and wires (a flashlight), the lightbulb has a resistance of ~3 Ω. The wires typically have a much lower resistance. We use an ideal wire where its resistance is 0. The potential difference in the wire is 0, even if there is current in it.

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Analyzing a Simple Circuit For the ideal-wire model, two wires are connected to a resistor. Current flows through all three, but the current only requires a potential difference across the resistor.

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Analyzing a Simple Circuit Current moves in the direction of decreasing potential, so there is a voltage drop when the current passes through the resistor left to right.

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Analyzing a Simple Circuit The electric field in a resistor carrying a current in a circuit is uniform. The strength of the electric field is

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Charge Carriers  The outer electrons of metal atoms are only weakly bound to the nuclei.  In a metal, the outer electrons become detached from their parent nuclei to form a fluid-like sea of electrons that can move through the solid.  Electrons are the charge carriers in metals. Slide 30-22

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.  The wire is already full of electrons!  We don’t have to wait for electrons to move all the way through the wire from one plate to another.  We just need to slightly rearrange the charges on the plates and in the wire. Discharging a Capacitor Slide 30-30

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Establishing the Electric Field in a Wire  The figure shows two metal wires attached to the plates of a charged capacitor.  This is an electrostatic situation.  What will happen if we connect the bottom ends of the wires together? Slide 30-32

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.  Within a very brief interval of time (  10  9 s ) of connecting the wires, the sea of electrons shifts slightly.  The surface charge is rearranged into a nonuniform distribution, as shown in the figure. Establishing the Electric Field in a Wire Slide 30-33

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.  The nonuniform distribution of surface charges along a wire creates a net electric field inside the wire that points from the more positive end toward the more negative end of the wire.  This is the internal electric field that pushes the electron current through the wire. Establishing the Electric Field in a Wire Slide 30-34

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Surface charge is distributed on a wire as shown. Electrons in the wire QuickCheck 30.2 A. Drift to the right. B. Drift to the left. C. Move upward. D. Move downward. E. On average, remain at rest. Slide 30-35

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Surface charge is distributed on a wire as shown. Electrons in the wire QuickCheck 30.2 A. Drift to the right. B. Drift to the left. C. Move upward. D. Move downward. E. On average, remain at rest. Slide Electric field from nonuniform surface charges is to the right. Force on negative electrons is to the left.

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. A Model of Conduction  Within a conductor in electrostatic equilibrium, there is no electric field.  In this case, an electron bounces back and forth between collisions, but its average velocity is zero. Slide 30-37

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.  In the presence of an electric field, the electric force causes electrons to move along parabolic trajectories between collisions.  Because of the curvature of the trajectories, there is a slow net motion in the “downhill” direction. A Model of Conduction Slide 30-38

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.  The graph shows the speed of an electron during multiple collisions.  The average velocity is the electron drift speed A Model of Conduction Slide 30-39

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Circuits containing multiple elements Series and parallel combinations RC circuits Electricity in the nervous system Chapter 23 Circuits Topics: Sample question: An electric eel can develop a potential difference of over 600 V. How do the cells of the electric eel’s body generate such a large potential difference? Slide 23-1

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Kirchhoff’s Laws Kirchhoff’s junction law, as we learned in Chapter 22, states that the total current into a junction must equal the total current leaving the junction. This is a result of charge and current conservation:

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Kirchhoff’s Laws The gravitational potential energy of an object depends only on its position, not on the path it took to get to that position. The same is true of electric energy. If a charged particle moves around a closed loop and returns to its starting point, there is no net change in its electric energy: Δu elec = 0. Because V = U elec /q, the net change in the electric potential around any loop or closed path must be zero as well.

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Kirchhoff’s Laws

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Kirchhoff’s Laws For any circuit, if we add all of the potential differences around the loop formed by the circuit, the sum must be zero. This result is Kirchhoff’s loop law: ΔV i is the potential difference of the ith component of the loop.

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Kirchhoff’s Laws Slide 23-11

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Kirchhoff’s Laws Text: pp. 729–730

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Kirchhoff’s Laws Text: p. 730

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Kirchhoff’s Laws ΔV bat can be positive or negative for a battery, but ΔV R for a resistor is always negative because the potential in a resistor decreases along the direction of the current. Because the potential across a resistor always decreases, we often speak of the voltage drop across the resistor.

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Circuit Analysis using Kirchoff’s rules (Loop + Junction) Slide Circuit from End of Circuit Activity 1

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. QuickCheck 23.6 The diagram below shows a segment of a circuit. What is the current in the 200 Ω resistor? 0.5 A 1.0 A 1.5 A 2.0 A There is not enough information to decide.

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. QuickCheck 23.6 The diagram below shows a segment of a circuit. What is the current in the 200 Ω resistor? 0.5 A 1.0 A 1.5 A 2.0 A There is not enough information to decide.

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Series and Parallel Circuits There are two possible ways that you can connect the circuit. Series and parallel circuits have very different properties. We say two bulbs are connected in series if they are connected directly to each other with no junction in between.

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Calculating Equivalent Resistance Slide 22-35

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Parallel Resistors The current I bat from the battery splits into currents I 1 and I 2 at the top of the junction. According to the junction law, Applying Ohm’s law to each resistor, we find that the battery current is

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Parallel Resistors Can we replace a group of parallel resistors with a single equivalent resistor? To be equivalent, ΔV must equal ℇ and I must equal I bat : This is the equivalent resistance, so a single R eq acts exactly the same as multiple resistors.

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Parallel Resistors

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. 1.By using only two resistors singly, in series, or in parallel you are able to obtain resistances of 18.0, 24.0, 72, and 96. What are the two resistances? (Make sure your answer is consistent.) 2.Find the equivalent resistance of the following circuit: Be sure to show each step. 3.Find the heat energy generated by the 3 Ohm and the 12 Ohm resistors in the circuit in problem 2 Equivalent Resistance Examples Slide 22-35