Factors that govern RSlide 2The work that must be done to take Q through VSlide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Chapter 18 DC CircuitsSlide 13Slide 14Slide 15When a piece of wire is used to connect points b and c in this figure, the brightness of bulb R1 (a) increases, (b) decreases, or (c) stays the same. The brightness of bulb R2 (a) increases, (b) decreases, or (c) stays the same.R1 becomes brighter. Connecting a wire from b to c provides a nearly zero resistance path from b to c and decreases the total resistance of the circuit from R1 + R2 to just R1. Ignoring internal resistance, the potential difference maintained by the battery is unchanged while the resistance of the circuit has decreased. The current passing through bulb R1 increases, causing this bulb to glow brighter. Bulb R2 goes out because essentially all of the current now passes through the wire connecting b and c and bypasses the filament of Bulb R2.With the switch in this circuit (figure a) closed, no current exists in R2 because the current has an alternate zero-resistance path through the switch. Current does exist in R1 and this current is measured with the ammeter at the right side of the circuit. If the switch is opened (figure b), current exists in R2. After the switch is opened, the reading on the ammeter (a) increases, (b) decreases, (c) does not change.(b). When the switch is opened, resistors R1 and R2 are in series, so that the total circuit resistance is larger than when the switch was closed. As a result, the current decreases.With the switch in this circuit (figure a) open, there is no current in R2. There is current in R1 and this current is measured with the ammeter at the right side of the circuit. If the switch is closed (figure b), there is current in R2. When the switch is closed, the reading on the ammeter (a) increases, (b) decreases, or (c) remains the same.(a). When the switch is closed, resistors R1 and R2 are in parallel, so that the total circuit resistance is smaller than when the switch was open. As a result, the total current increases.Slide 22Slide 23Five resistors are connected as shown in the diagram. The potential difference between points A and B is 15 V.Slide 25Slide 26Slide 27Slide 28Slide 29Electromotive Force or EmfSlide 31Slide 32Slide 33Kirchhoff’s RulesSlide 35Slide 36You have a large supply of lightbulbs and a battery. You start with one lightbulb connected to the battery and notice its brightness. You then add one lightbulb at a time, each new bulb being added in parallel to the previous bulbs. As the lightbulbs are added, what happens (a) to the brightness of the bulbs? (b) to the current in the bulbs? (c) to the power delivered by the battery? (d) to the lifetime of the battery? (e) to the terminal voltage of the battery? Hint: Do not ignore the internal resistance of the battery.(a) The brightness of the bulbs decreases (b) The current in the bulbs decreases (c) The power delivered by the battery increases (d) The lifetime of the battery decreases (e) The terminal voltage of the battery decreasesFactors that governFactors that govern R•Material  The ability to carry electric current varies drastically •Length L The longer the conductor, the greater its resistance•Cross sectional area A The thicker the conductor, the less its resistanceR=L/Ais the resistivity, and depends on the material used  reflects intrinsic properties of the material is a function of temperatureThe unit of  is -mThe work that must be done to The work that must be done to take Q through Vtake Q through V W=QV Since I=Q/t or Q=It, W=IVt (Electric work) Recall P=work done/time interval=W/t, P=IV (Electric Power)Using Ohm’s law I=V/R, P=I2R=V2/RThe unit for Power is Watt, 1W=1J/sSuperconductorT0ConductorTcThe highest Tc is about –150 C or –300 F found in “high temperature” superconductorsI=Q/t Ohm’s law V=IR R=L/A P=IV=I2R=V2/RQuestion:A 10-A current is maintained in a simple circuit with a total resistance of 200   What net charge passes through any point in the circuit during a 1-minute interval?(a) 200 C (b) 500 C (c) 1200 C (d) 400 C (e) 600 CXQuestion: A current in a 12  toaster operated at 120 V is (a) 0.1 A(b) 10 A(c) 12 A(d) 1440 AAnswer: bQuestion: The power rating of an electric motor that draws a current of 3 A when operated at 120 V is (a) 40 W(b) 360 W(c) 540 W(d) 1080 WAnswer: bQuestion: The temperature of a copper wire is raised. Its resistivity (a) decreases(b) remains the same (c) increases(d) any of the above, depending on the temperatures involvedAnswer: cQuestion: Which draws more current, a 100-W light bulb or a 75-W bulb? Which has the higher resistance? Answer: Let P1=100 W and P2=75 W. Since P=IV, V=constant P1/P2=I1/I2=100/75=4/3 The 100-W bulb draws more current Also, P=V2/R, where V=constant P1/P2=R2/R1=100/75=4/3 The 75-W has higher resistanceQuestion: Electric power is transferred over large distances at very high voltages. Explain how the high voltage reduces power losses in the transmission lines. Answer: The power transmitted P=IV. So the higher V, the smaller I. Since the power lost P=I2R, the smaller I leads to smaller I2R loss for a given transmission line of resistance R.Chapter 18 DC CircuitsChapter 18 DC CircuitsResistors in SeriesResistors in ParallelElectromotive Force (emf)Kirchhoff’s RulesR2R3Resistors in SeriesResistors in SeriesV=V1+V2+V3I=I1=I2=I3 (Resistors in series have the same current)V=IR=IR1+IR2+IR3=I(R1+R2+R3) R=R1+R2+R3IIR1II RVResistors in ParallelResistors in Parallel I=I1+I2+I3 V=V1=V2=V3 (Resistors in parallel have the same voltage) I=V/R=V/R1+V/R2+V/R3=V/(R1+R2+R3) 1/R=1/R1+1/R2+1/R3R1R2IIR3I RVI3I2I1R=L/AR2R3R1Increasing LR=R1+R2+R3+…>>R1R2IIR3Increasing A1/R=1/R1+1/R2+1/R3+…QUICK QUIZ 18.1When a piece of wire is used to connect points b and c in this figure, the brightness of bulb R1 (a) increases, (b) decreases, or (c) stays the same. The brightness of bulb R2 (a) increases, (b) decreases, or (c) stays the same.QUICK QUIZ 18.1 ANSWERR1 becomes brighter. Connecting a wire from b to c provides a nearly zero resistance path from b to c and decreases the total resistance of the circuit from R1 + R2 to just R1. Ignoring internal resistance, the potential difference maintained by the battery is unchanged while the resistance of the circuit has decreased. The