1. 2/12/07184 Lecture 191 PHY 184 Spring 2007 Lecture 19 Title: Kirchoff’s Rules for Circuits

2. 2/12/07184 Lecture 192 Results of Midterm 1 (Sec 2)  Pretty high average; will even be higher after the correction set! 3/9 or 33.3% --> 53% 5/9 or 55.6% --> 69% Correction set: Example

3. 2/12/07184 Lecture 193AnnouncementsAnnouncements  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.

4. 2/12/07184 Lecture 194 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+2R w

5. 2/12/07184 Lecture 195 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%

6. 2/12/07184 Lecture 196 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 One wire: V w = iR = 2.2 mV

7. 2/12/07184 Lecture 197CircuitsCircuits  We have been working with simple circuits containing either capacitors or resistors.  Capacitors wired in parallel  Capacitors wired in series

8. 2/12/07184 Lecture 198 Circuits (2)  Resistors wired in parallel  Resistors wired in series  We dealt with circuits containing either capacitors or resistors that could be resolved in systems of parallel or series components.

9. 2/12/07184 Lecture 199 Complex Circuits  Circuits can be constructed that cannot be resolved into series or parallel systems of capacitors or resistors

10. 2/12/07184 Lecture 1910 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.

11. 2/12/07184 Lecture 1911 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 i1i1 i2i2 i3i3 i 1 =i 2 +i 3

12. 2/12/07184 Lecture 1912 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.

13. 2/12/07184 Lecture 1913 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.

14. 2/12/0714

15. 2/12/0715 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. Anode (negative terminal): Zinc powder Cathode (positive terminal): Manganese dioxide (MnO 2 ) powder Electrolyte: Potassium hydroxide (KOH)  The half-reactions are: At the cathode… 2 MnO 2 + H 2 O + 2 e- —>Mn 2 O 3 + 2 OH- At the anode… Zn + 2 OH- —> ZnO + H 2 O + 2 e-  The overall reaction is: Zn + 2MnO 2 —> ZnO + Mn 2 O 3 + [E=1.5 V]

16. 2/12/0716 Alkaline battery Al Kaline batter

17. 2/12/07184 Lecture 1917 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.

18. 2/12/07184 Lecture 1918 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.

19. 2/12/07184 Lecture 1919 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.

20. 2/12/07184 Lecture 1920 Circuit Analysis Conventions i is the magnitude of the assumed current

21. 2/12/07184 Lecture 1921 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 (V emf,1 and V emf,2 ) and two resistors ( R 1 and R 2 ) connected in series in a single loop  We will assume that the two sources of emf have opposite polarity.

22. 2/12/07184 Lecture 1922 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

23. 2/12/07184 Lecture 1923  Start at point a.  The first circuit component is a source of emf V emf,1, which produces a positive voltage gain of +V emf,1  Next we find resistor R 1, which produces a voltage drop V 1 given by -iR 1  Continuing around the circuit we find resistor R 2, which produces a voltage drop V 2 given by -iR 2  Next we meet a second source of emf, V emf,2  This source of emf is wired into the circuit with a polarity opposite that of V emf,1  We treat this component as a voltage drop of -V emf,2 rather than a voltage gain  We now have completed the circuit and we are back at point a. Single Loop Circuits (3)

24. 2/12/07184 Lecture 1924  Kirchoff’s loop rule for the voltage drops states  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. Single Loop Circuits (4)

25. 2/12/07184 Lecture 1925  Now let’s analyze the same circuit in the counter- clockwise direction starting at point a  The first circuit element is V emf,2 which is a positive voltage gain  The next element is R 2 Single Loop Circuits (5)

26. 2/12/07184 Lecture 1926 Single Loop Circuits (6)  Because we have assumed that the current is in the clockwise direction and we are analyzing the loop in the counter-clockwise direction, this voltage drop is +iR 2  Proceeding to the next element in the loop, R 1, we use a similar argument to designate the voltage drop as +iR 1  The final element in the circuit is V emf,1, which is aligned in a direction opposite to our analysis direction, so the voltage drop across this element is -V emf,1

27. 2/12/07184 Lecture 1927 Single Loop Circuits (7)  Kirchhoff’s Loop Rule then gives us  Comparing this result with the result we obtained by analyzing the circuit in the clockwise direction  … we see that they are equivalent.  The direction that we choose to analyze the circuit does not matter.

28. 2/12/07184 Lecture 1928 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

29. 2/12/07184 Lecture 1929 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