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PHY 184 Spring 2007 Lecture 19 Title: Kirchoff’s Rules for Circuits 2/12/07 184 Lecture 19 1 Results of Midterm 1 (Sec 2) Pretty high average; will even be higher after the correction set! Midterm Section 1- average 63% (9 correct problems out of 10 gives 100%) 30 27 26 23 23 23 19 20 Result 19 30 27 15 25 9 10 6 5 5 2 0 0 1 2 3 4 5 6 7 8 9 10 Number of Students 24 23 24 23 22 20 18 15 9 10 6 Correct Answers 5 2/12/07 00 % 10 0. 9% 88 .8 8% 77 .7 7% 66 .6 6% 55 .5 4% 44 .4 3% 33 .3 2% 22 .2 1% 0 11 .1 3/9 or 33.3% --> 53% 5/9 or 55.6% --> 69% 2 00 % Correction set: Example 0. Number of Students 25 Percent Correct 184 Lecture 19 2 Announcements Corrections Set 1 will open soon • Corrections Set 1 is just Midterm 1 with different numbers You will get 30% credit for those problems you did incorrectly on Midterm 1 • However, you must do all the problems in Corrections Set 1 to get credit • After Corrections Set 1 is due, we will open Midterm 1 so that you can see your particular questions and answers This week we will continue with circuits and start magnetism. 2/12/07 184 Lecture 19 3 Resistivity of Wires in a Circuit The circuit on the right shows a resistor of resistance R=6 connected to an ideal battery providing 12V by means of two copper wires. Each wire has the length 20 cm and radius 1 mm. We generally neglect their resistance, the voltage drop across them and the energy dissipated. Justify by calculating the voltage across R accounting for the influence of the wire! Idea: For each wire Total resistance of the circuit is R+2Rw 2/12/07 184 Lecture 19 4 Resistivity of Wires in a Circuit (2) The circuit on the right shows a resistor of resistance R=6 connected to an ideal battery providing 12V by means of two copper wires. Each wire has the length 20.cm and radius 1 mm. We generally neglect their resistance, the voltage drop across them and the energy dissipated. Justify by calculating the voltage across R accounting for the influence of the wire! The voltage across R is V=Ri to be compared to 12 V if we neglect the wires. Neglecting the resistance of the wire introduces an error of only 0.03% 2/12/07 184 Lecture 19 5 Clicker Question The circuit on the right shows a resistor of resistance R=6 connected to an ideal battery providing 12V by means of two copper wires. Each wire has the length 20.cm and radius 1 mm. We generally neglect their resistance, the voltage drop across them and the energy dissipated. What is the total voltage drop across the two wires (i=1.9993 A, R=0.0011 each)? A) 5.5 10-4 V B) 0.01875 V C) 4.4 mV D) 2.2 mV 2/12/07 One wire: Vw = iR = 2.2 mV 184 Lecture 19 6 Circuits We have been working with simple circuits containing either capacitors or resistors. Capacitors wired in parallel n Ceq Ci i 1 Capacitors wired in series 2/12/07 184 Lecture 19 n 1 1 Ceq i 1 Ci 7 Circuits (2) Resistors wired in parallel Resistors wired in series n 1 1 Req i 1 Ri n Req Ri i 1 We dealt with circuits containing either capacitors or resistors that could be resolved in systems of parallel or series components. 2/12/07 184 Lecture 19 8 Complex Circuits Circuits can be constructed that cannot be resolved into series or parallel systems of capacitors or resistors 2/12/07 184 Lecture 19 9 Kirchhoff’s Rules for Multi-loop Circuits To handle these types of circuits, we must apply Kirchhoff’s Rules. Kirchhoff’s Rules can be stated as • Kirchhoff’s Junction Rule • The sum of the currents entering a junction must equal the sum of the currents leaving a junction. • Kirchhoff’s Loop Rule • The sum of voltage drops around a complete circuit loop must sum to zero. 2/12/07 184 Lecture 19 10 Kirchhoff’s Junction Rule Kirchhoff’s Junction Rule is a direct consequence of the conservation of charge. In a conductor, charge cannot be created or destroyed. At a junction: all charges streaming into the junction must also leave the junction i1 i2 i3 2/12/07 184 Lecture 19 i1=i2+i3 11 Kirchhoff’s Loop Rule Kirchhoff’s Loop Rule is a direct consequence of the conservation of electric potential energy. Suppose that this rule was not valid • we could construct a way around a loop in such a way that each turn would increase the potential of a charge traveling around the loop • we would always increase the energy of this charge, in obvious contradiction to energy conservation Kirchhoff’s Loop Rule is equivalent to the law of energy conservation. 2/12/07 184 Lecture 19 12 EMF Devices - Directions An emf device (e.g., a battery) keeps the positive terminal (labeled +) at a higher electrical potential than the negative terminal (labeled -). When a battery is connected in a circuit, its internal chemistry causes a net current inside the battery : positive charge carriers move from the negative to the positive terminal, in direction of the emf arrow. Or: Positive charge carriers move from a region of low electric potential (negative terminal) to a region of high electric potential (positive terminal) This flow is part of the current that is set up around the circuit in that same direction. 2/12/07 184 Lecture 19 13 2/12/07 14 The half-reactions are: At the cathode… 2 MnO2 + H2O + 2 e- —>Mn2O3 + 2 OHAt the anode… Zn + 2 OH- —> ZnO + H2O + 2 e The overall reaction is: Zn + 2MnO2 —> ZnO + Mn2O3 + [E=1.5 V] Anode (negative terminal): Zinc powder Cathode (positive terminal): Manganese dioxide (MnO2) powder Electrolyte: Potassium hydroxide (KOH) The flow of electrons is always from anode—to--cathode outside of the cell (i.e., in the circuit) and from cathode—to--anode inside the cell. Inside a chemical cell, ions are carrying the electrons from cathode—to--anode inside the cell. 2/12/07 15 Alkaline battery 2/12/07 Al Kaline batter 16 Single Loop Circuits We begin our study of more complicated circuits by analyzing circuits with several sources of emf and resistors connected in series in a single loop. We will apply Kirchhoff’s Rules to these circuits. To apply these rules: establish conventions for determining the voltage drop across each element of the circuit depending on the assumed direction of current and the direction of the analysis of the circuit. Because we do not know the direction of the current in the circuit before we start, we must choose an arbitrary direction for the current. 2/12/07 184 Lecture 19 17 Single Loop Circuits (2) We can determine if our assumption for the direction of the current is correct after the analysis is complete. If our assumed current is negative, then the current is flowing in the direction opposite to the direction we chose. We can also choose the direction in which we analyze the circuit arbitrarily. Any direction we choose will give us the same information. 2/12/07 184 Lecture 19 18 Single Loop Circuits (3) If we move around the circuit in the same direction as the current, the voltage drop across a resistor will be negative. If we move around the circuit in the direction opposite to the current, the voltage drop across the resistors will be positive. If we move around the circuit and encounter a source of emf pointing in the same direction, we assume that this source of emf contributes a positive voltage. If we encounter a source of emf pointing in the opposite direction, we consider that component to contribute a negative voltage. 2/12/07 184 Lecture 19 19 Circuit Analysis Conventions Element 2/12/07 Analysis Direction Current Direction Voltage Drop iR iR iR iR Vemf Vemf Vemf Vemf 184 Lecture 19 i is the magnitude of the assumed current 20 Single Loop Circuits We have studied circuits with various networks of resistors but only one source of emf. Circuits can contain multiple sources of emf as well as multiple resistors. We begin our study of more complicated circuits by analyzing a circuit with two sources of emf (Vemf,1 and Vemf,2 ) and two resistors ( R1 and R2 ) connected in series in a single loop We will assume that the two sources of emf have opposite polarity. 2/12/07 184 Lecture 19 21 Single Loop Circuits (2) We assume that the current is flowing around the circuit in a clockwise direction. Starting at point a with V=0, we analyze around the circuit in a clockwise direction. Because the components of the circuit are in series, all components have the same current, i i i i 2/12/07 184 Lecture 19 22 Single Loop Circuits (3) Start at point a. The first circuit component is a source of emf Vemf,1, which produces a positive voltage gain of +Vemf,1 Next we find resistor R1, which produces a voltage drop V1 given by -iR1 Continuing around the circuit we find resistor R2, which produces a voltage drop V2 given by -iR2 Next we meet a second source of emf, Vemf,2 This source of emf is wired into the circuit with a polarity opposite that of Vemf,1 We treat this component as a voltage drop of -Vemf,2 rather than a voltage gain We now have completed the circuit and we are back at point a. 2/12/07 184 Lecture 19 23 Single Loop Circuits (4) Kirchoff’s loop rule for the voltage drops states Vemf ,1 V1 V2 Vemf ,2 Vemf ,1 iR1 iR2 Vemf ,2 0 Generalization: the voltage drops across components in a single loop circuit must sum to zero. This statement must be qualified with conventions for assigning the sign of the voltage drops around the circuit. 2/12/07 184 Lecture 19 24 Single Loop Circuits (5) Now let’s analyze the same circuit in the counterclockwise direction starting at point a i The first circuit element is Vemf,2 which is a positive i voltage gain The next element is R2 2/12/07 184 Lecture 19 25 Single Loop Circuits (6) i Because we have assumed that the current is in the clockwise direction and we are analyzing the loop in the counter-clockwise direction, this i voltage drop is +iR2 Proceeding to the next element in the loop, R1, we use a similar argument to designate the voltage drop as +iR1 The final element in the circuit is Vemf,1, which is aligned in a direction opposite to our analysis direction, so the voltage drop across this element is -Vemf,1 2/12/07 184 Lecture 19 26 Single Loop Circuits (7) Kirchhoff’s Loop Rule then gives us Vemf ,2 iR2 iR1 Vemf ,1 0 Comparing this result with the result we obtained by analyzing the circuit in the clockwise direction Vemf ,1 iR1 iR2 Vemf ,2 0 … we see that they are equivalent. The direction that we choose to analyze the circuit does not matter. 2/12/07 184 Lecture 19 27 Clicker Question The figure shows the current i in a single-loop circuit with a battery B and a resistance R. How should the emf arrow be drawn? A) pointing to the right B) pointing to the left C) doesn’t matter 2/12/07 184 Lecture 19 28 Clicker Question The figure shows the current i in a single-loop circuit with a battery B and a resistance R. How should the emf arrow be drawn? A) pointing to the right in direction of the current 2/12/07 184 Lecture 19 29