PHY 184

1. PHY 184 Spring 2007 Lecture 19 Title: Kirchoff’s Rules for Circuits 184 Lecture 19

2. Results of Midterm 1 (Sec 2) • Pretty high average; will even be higher after the correction set! Correction set: Example 3/9 or 33.3% --> 53% 5/9 or 55.6% --> 69% 184 Lecture 19

3. 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. 184 Lecture 19

4. 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 184 Lecture 19

5. 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% 184 Lecture 19

6. Clicker Question One wire: Vw = iR = 2.2 mV • 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 184 Lecture 19

7. Circuits • We have been working with simple circuits containing either capacitors or resistors. • Capacitors wired in parallel • Capacitors wired in series 184 Lecture 19

8. 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. 184 Lecture 19

9. Complex Circuits • Circuits can be constructed that cannot be resolved into series or parallel systems of capacitors or resistors 184 Lecture 19

10. 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. 184 Lecture 19

11. Kirchhoff’s Junction Rule i2 i1 i3 • 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 184 Lecture 19

12. 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. 184 Lecture 19

13. 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. 184 Lecture 19

15. The half-reactions are: • At the cathode… • 2 MnO2 + H2O + 2 e- —>Mn2O3 + 2 OH- • At 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.

16. Alkaline battery Al Kaline batter

17. 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. 184 Lecture 19

18. 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. 184 Lecture 19

19. 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. 184 Lecture 19

20. Circuit Analysis Conventions i is the magnitude of the assumed current 184 Lecture 19

21. 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. 184 Lecture 19

22. 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 184 Lecture 19

23. 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 dropV1 given by -iR1 • Continuing around the circuit we find resistor R2, which produces a voltage dropV2 given by -iR2 • Next we meet a second source of emf, Vemf,2 • This source of emf is wired into the circuitwith a polarity opposite that of Vemf,1 • We treat this component as a voltage dropof -Vemf,2 rather than a voltage gain • We now have completed the circuitand we are back at point a. 184 Lecture 19

24. Single Loop Circuits (4) • 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. 184 Lecture 19

25. Single Loop Circuits (5) • Now let’s analyze the same circuit in the counter-clockwise direction starting at point a • The first circuit element is Vemf,2 which is a positive voltage gain • The next element is R2 184 Lecture 19

26. Single Loop Circuits (6) • Because we have assumed that thecurrent is in the clockwise directionand we are analyzing the loop in thecounter-clockwise direction, thisvoltage 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,1 184 Lecture 19

27. 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. 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 B) pointing to the left C) doesn’t matter 184 Lecture 19

29. 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 184 Lecture 19