2. 5.2.1Define electromotive force. 5.2.2Describe the concept of internal resistance. Topic 5: Electric currents 5.2 Electric circuits

3. Define electromotive force.  The electromotive force  (or emf) of a cell is the amount of chemical energy converted to electrical energy per unit charge.  Since energy per unit charge is volts, emf is measured in volts.  This battery has an emf of  = 1.6 V.  Because the battery is not connected to any circuit we say that it is unloaded. Topic 5: Electric currents 5.2 Electric circuits 01.6 00.0

4. EXAMPLE: How much chemical energy is converted to electrical energy by the cell if a charge of 15  C is drawn by the voltmeter? SOLUTION:  From ∆V = ∆E P /q we have  = ∆E P /q.  Thus ∆EP = q∆EP = q = (1.6)(15  10 -6 ) = 2.4  10 -5 J. Define electromotive force. Topic 5: Electric currents 5.2 Electric circuits 01.6

5. Describe the concept of internal resistance.  If we connect a resistor as part of an external circuit, we see that the voltage read by the meter goes down a little bit.  We say that the battery is loaded because there is a resistor con- nected externally in a circuit.  This battery has a loaded potential difference (p.d.) of V = 1.5 V. Topic 5: Electric currents 5.2 Electric circuits 01.5 01.6

6. Describe the concept of internal resistance.  All cells and batteries have limits as to how rapidly the chemical reactions can maintain the voltage.  There is an internal resistance r associated with a cell or a battery which causes the cell’s voltage to drop when there is an external demand for the cell’s electrical energy.  The best cells will have small internal resistances.  The internal resistance of a cell is why a battery becomes hot if there is a high demand for current from an external circuit.  Cell heat will be produced at the rate given by Topic 5: Electric currents 5.2 Electric circuits P = I 2 r cell heat rate r = internal resistance

7. Describe the concept of internal resistance. Topic 5: Electric currents 5.2 Electric circuits PRACTICE: A battery has an internal resistance of r = 1.25 . What is its rate of heat production if it is supplying an external circuit with a current of I = 2.00 A? SOLUTION:  Rate of heat production is power.  P = I 2 r  P = (2 2 )(1.25) = 5.00 J/s (5.00 W.) FYI  If you double the current, the rate of heat generation will quadruple because of the I 2 dependency.  If you accidentally “short circuit” a battery, the battery may even heat up enough to leak or explode!

8. Describe the concept of internal resistance.  If we wish to consider the internal resistance of a cell, we can use the cell and the resistor symbols together, like this:  And we may enclose the whole cell in a box of some sort to show that it is one unit.  Suppose we connect our cell with its internal resistance r to an external circuit consisting of a single resistor of value R.  All of the chemical energy from the battery is being consumed by the internal resistance r and the external resistance R. Topic 5: Electric currents 5.2 Electric circuits  r  r R

9. Describe the concept of internal resistance.  But from E P = qV we can write:  Electrical energy being created from chemical energy in the battery: E P = q .  Electrical energy being converted to heat energy by R: E P,R = qV R.  Electrical energy being converted to heat energy by r: E P,r = qV r.  From conservation of energy E P = E P,R + E P,r so that q  = qV R + qV r.  Note that the current I is the same everywhere.  From Ohm’s law V R = IR and V r = Ir so that q  = qIR + qIr. Topic 5: Electric currents 5.2 Electric circuits  = IR + Ir = I(R + r) emf relationship  r R

10. EXAMPLE: A cell has an unloaded voltage of 1.6 V and a loaded voltage of 1.5 V when a 330  resistor is connected as shown. (a) Find . (b) Then from the second circuit find I. (c) Finally, find the cell’s internal resistance. SOLUTION: (a) From the first schematic we see that  = 1.6 V. (Unloaded cell.) (b) From the second diagram we see that the voltage across the 330  resistor is 1.5 V.  V = IR so that 1.5 = I(330) and I = 0.0045 A. Describe the concept of internal resistance. Topic 5: Electric currents 5.2 Electric circuits  r 1.6 V  r R 1.5 V 330 

11. EXAMPLE: A cell has an unloaded voltage of 1.6 V and a loaded voltage of 1.5 V when a 330  resistor is connected as shown. (a) Find . (b) Then from the second circuit find I. (c) Finally, find the cell’s internal resistance. SOLUTION: (c) Now we can use the emf relationship  = IR + Ir. 1.6 = (0.0045)(330) + (0.0045)r 1.6 = 1.5 + 0.0045r 0.1 = 0.0045r r = 22 . Describe the concept of internal resistance. Topic 5: Electric currents 5.2 Electric circuits  r 1.6 V  r R 1.5 V 330 

12. 5.2.3Apply the equations for resistors in series and parallel. 5.2.4Draw circuit diagrams. 5.2.5Describe the use of ideal ammeters and ideal voltmeters. Topic 5: Electric currents 5.2 Electric circuits

13. Apply the equations for resistors in series and parallel.  Resistors can be connected to one another in series, which means one after the other.  Note there is only one current I and I is the same for all components of a series circuit.  Conservation of energy tells us  = V 1 + V 2 + V 3.  Thus  = IR 1 + IR 2 + IR 3 (from Ohm’s law)  = I(R 1 + R 2 + R 3 ) (factoring out I)  = I(R), where R = R 1 + R 2 + R 3. Topic 5: Electric currents 5.2 Electric circuits R1R1 R2R2 R3R3 R = R 1 + R 2 + R 3 equivalent resistance in series 

14. EXAMPLE: Three resistors of 330  each are connected to a 6.0 V bat- tery in series as shown. (a) What is the circuit’s equivalent resistance? (b) What is the current in the circuit? (c) What is the voltage on each resistor? SOLUTION: (a) In series, R = R 1 + R 2 + R 3 so that R = 330 + 330 + 330 = 990 . (b) Since the voltage on the entire circuit is 6 V, and since the total resistance is 990 , from Ohm’s law we have I = V/R = 6/990 = 0.0061 A. Apply the equations for resistors in series and parallel. Topic 5: Electric currents 5.2 Electric circuits R1R1 R2R2 R3R3 

15. EXAMPLE: Three resistors of 330  each are connected to a 6.0 V bat- tery in series as shown. (a) What is the circuit’s equivalent resistance? (b) What is the current in the circuit? (c) What is the voltage on each resistor? SOLUTION: (c) The current I we just found is the same everywhere. Thus each resis- tor has a current of I = 0.0061 A.  From Ohm’s law, each resistor has a voltage given by V = IR = (.0061)(330) = 2.0 V. Apply the equations for resistors in series and parallel. Topic 5: Electric currents 5.2 Electric circuits R1R1 R2R2 R3R3 FYI  In general the V’s are different if the R’s are.

16. Apply the equations for resistors in series and parallel.  Resistors can also be connected in parallel.  Each resistor is connected directly to the cell.  Thus each resistor has the same voltage V and V is the same for all components of a parallel circuit. We can then write  = V 1 = V 2 = V 3  V.  But there are three currents I 1, I 2, and I 3.  Since the total current I passes through the cell we see that I = I 1 + I 2 + I 3.  If R is the equivalent or total resistance of the three resistors, then I = I 1 + I 2 + I 3 becomes  /R = V 1 /R 1 + V 2 /R 2 + V 3 /R 3 (Ohm’s law I = V/R) Topic 5: Electric currents 5.2 Electric circuits R1R1 R2R2 R3R3 

17. Apply the equations for resistors in series and parallel.  Resistors can also be connected in parallel.  Each resistor is connected directly to the cell.  Thus each resistor has the same voltage V and V is the same for all components of a parallel circuit. We can then write  = V 1 = V 2 = V 3  V.  From  = V 1 = V 2 = V 3  V and  /R = V 1 /R 1 + V 2 /R 2 + V 3 /R 3 we get V/R = V/R 1 + V/R 2 + V/R 3.  Thus the equivalent resistance R is given by Topic 5: Electric currents 5.2 Electric circuits R1R1 R2R2 R3R3  1/R = 1/R 1 + 1/R 2 + 1/R 3 equivalent resistance in parallel

18. EXAMPLE: Three resistors of 330  each are connected to a 6.0 V cell in parallel as shown. (a) What is the circuit’s equivalent resistance? (b) What is the voltage on each resistor? (c) What is the current in each resistor? SOLUTION: (a) In parallel, 1/R = 1/R 1 + 1/R 2 + 1/R 3 so that 1/R = 1/330 + 1/330 + 1/330 = 0.00909. Thus R = 1/0.00909 = 110 . (b) The voltage on each resistor is 6 V, since the resistors are in parallel. (Or each resistor is clearly directly connected to the battery). Apply the equations for resistors in series and parallel. Topic 5: Electric currents 5.2 Electric circuits R1R1 R2R2 R3R3 

19. EXAMPLE: Three resistors of 330  each are connected to a 6.0 V cell in parallel as shown. (a) What is the circuit’s equivalent resistance? (b) What is the voltage on each resistor? (c) What is the current in each resistor? SOLUTION: (c) Using Ohm’s law (I = V/R): I 1 = V 1 /R 1 = 6/330 = 0.018 A. I 2 = V 2 /R 2 = 6/330 = 0.018 A. I 3 = V 3 /R 3 = 6/330 = 0.018 A. Apply the equations for resistors in series and parallel. Topic 5: Electric currents 5.2 Electric circuits R1R1 R2R2 R3R3  FYI  In general the I’s are different if the R’s are different.

20. Draw circuit diagrams.  A complete circuit will always contain a cell or a battery.  The schematic diagram of a cell is this:  A battery is just a group of cells connected in series:  If each cell is 1.5 V, then the battery shown above is 3(1.5) = 4.5 V.  Often a battery is shown in a more compact form:  Of course a fixed-value resistor looks like this: Topic 5: Electric currents 5.2 Electric circuits cell battery resistor

21. Draw circuit diagrams. Topic 5: Electric currents 5.2 Electric circuits EXAMPLE: Draw schematic diagrams of each of the following circuits: SOLUTION: For A: For B: A B solder joints

22. PRACTICE: Draw a schematic diagram for this circuit: SOLUTION: FYI  Be sure to position the voltmeter across the correct resistor in parallel. Describe the use of ideal ammeters and ideal voltmeters. Topic 5: Electric currents 5.2 Electric circuits 1.06

23. PRACTICE: Draw a schematic diagram for this circuit: SOLUTION: FYI  Be sure to position the ammeter between the correct resistors in series. Describe the use of ideal ammeters and ideal voltmeters. Topic 5: Electric currents 5.2 Electric circuits.003

24. PRACTICE: Draw a schematic diagram for this circuit: SOLUTION: FYI  This circuit is a combination series-parallel. In a later slide you will learn how to find the equivalent resistance of the combo circuit. Draw circuit diagrams. Topic 5: Electric currents 5.2 Electric circuits

25. Describe the use of ideal ammeters and ideal voltmeters.  We have seen that voltmeters are always connected in parallel.  The voltmeter reads the voltage of only the component it is in parallel with.  The green current represents the amount of current the battery needs to supply to the voltmeter in order to make it register.  The red current is the amount of current the battery supplies to the original circuit.  In order to NOT ALTER the original properties of the circuit, ideal voltmeters have extremely high resistance (  ) to minimize the green current. Topic 5: Electric currents 5.2 Electric circuits

26. Describe the use of ideal ammeters and ideal voltmeters.  We have seen that ammeters are always connected in series.  The ammeter reads the current of the original circuit.  In order to NOT ALTER the original properties of the circuit, ideal ammeters have extremely low resistance (0) to minimize the effect on the red current. Topic 5: Electric currents 5.2 Electric circuits

27. 5.2.6Describe a potential divider. 5.2.7Explain the use of sensors in potential divider circuits. 5.2.8Solve problems involving electric circuits. Topic 5: Electric currents 5.2 Electric circuits

28. Describe a potential divider.  Consider a battery of  = 6 V. Suppose we have a light bulb that can only use three volts. How do we obtain 3 V from a 6 V battery?  A potential divider is a circuit made of two (or more) series resistors that allows us to tap off any voltage we want that is less than the battery voltage.  The input voltage is the emf of the battery.  The output voltage is the voltage drop across R 2.  Since the resistors are in series R = R 1 + R 2. Topic 5: Electric currents 5.2 Electric circuits OUT IN potential divider R1R1 R2R2

29. Describe a potential divider.  Consider a battery of  = 6 V. Suppose we have a light bulb that can only use three volts. How do we obtain 3 V from a 6 V battery?  A potential divider is a circuit made of two (or more) series resistors that allows us to tap off any voltage we want that is less than the battery voltage.  From Ohm’s law the current I of the divider is given by I = V IN /R = V IN /(R 1 + R 2 ).  But V OUT = V 2 = IR 2 so that V OUT = IR 2 = R 2  V IN /(R 1 + R 2 ). Topic 5: Electric currents 5.2 Electric circuits V OUT = V IN [ R 2 /(R 1 + R 2 ) ] potential divider OUT IN potential divider R1R1 R2R2

30. PRACTICE: Find the output voltage if the battery has an emf of 9.0 V, R 1 is a 2200  resistor, and R 2 is a 330  resistor. SOLUTION:  Use V OUT = V IN [ R 2 /(R 1 + R 2 ) ] V OUT = 9[ 330/(2200 + 330) ] V OUT = 9[ 330/2530 ] V OUT = 1.2 V. Solve problems involving electric circuits. Topic 5: Electric currents 5.2 Electric circuits OUT IN R1R1 R2R2 FYI  The bigger R 2 is in comparison to R 1, the bigger V OUT is in proportion to the total voltage.

31. PRACTICE: Find the value of R 2 if the battery has an emf of 9.0 V, R 1 is a 2200  resistor, and we want an output voltage of 6 V. SOLUTION:  Use the formula V OUT = V IN [ R 2 /(R 1 + R 2 ) ]. Thus 6 = 9[ R 2 /(2200 + R 2 ) ] 6(2200 + R 2 ) = 9R 2 13200 + 6R 2 = 9R 2 13200 = 3R 2 R 2 = 4400  Solve problems involving electric circuits. Topic 5: Electric currents 5.2 Electric circuits OUT IN R1R1 R2R2

32. PRACTICE: A light-dependent resistor (LDR) has a resistance of 25  in bright light and a resistance of 22000  in low light. An electronic switch will turn on a light when its p.d. is above 7.0 V. What should the value of R 1 be? SOLUTION: Use V OUT = V IN [ R 2 /(R 1 + R 2 ) ]. Thus 7 = 9[ 22000/(R 1 + 22000) ] 7(R 1 + 22000) = 9(22000) 7R 1 + 154000 = 198000 7R 1 = 44000 R 1 = 6300  (6286) Explain the use of sensors in potential divider circuits. Topic 5: Electric currents 5.2 Electric circuits R1R1 R2R2 electronic switch 9.0 V

33. PRACTICE: A thermistor has a resistance of 250  when it is in the heat of a fire and a resistance of 65000  when at room temperature. An electronic switch will turn on a sprinkler system when its p.d. is above 7.0 V. (a) Should the thermistor be R 1 or R 2 ? SOLUTION:  Because we want a high voltage at a high temperature, and because the thermistor’s resistance decreases with temperature, it should be placed at the R 1 position. Explain the use of sensors in potential divider circuits. Topic 5: Electric currents 5.2 Electric circuits R1R1 R2R2 electronic switch 9.0 V

34. PRACTICE: A thermistor has a resistance of 250  when it is in the heat of a fire and a resistance of 65000  when at room temperature. An electronic switch will turn on a sprinkler system when its p.d. is above 7.0 V. (b) What should the value of R 2 be? SOLUTION: In fire the thermistor is R 1 = 250 . 7 = 9[ R 2 /(250 + R 2 ) ] 7(250 + R 2 ) = 9R 2 1750 + 7R 2 = 9R 2 R 2 = 880  (875) Explain the use of sensors in potential divider circuits. Topic 5: Electric currents 5.2 Electric circuits R1R1 R2R2 electronic switch 9.0 V V2V2 V1V1

35. PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W” on its package. The potentiometer has a resistance from X to Z of 24  and has linear variation. (a) Sketch the variation of the p.d. V vs. the current I for a typical filament lamp. Is it ohmic? SOLUTION:  Since the temperature increases with the current, so does the resistance.  But from V = IR we see that R = V/I, which is the slope.  Thus the slope should increase with I. Solve problems involving electric circuits. Topic 5: Electric currents 5.2 Electric circuits 7.0 V X Y Z V I ohmic means linear non-ohmic

36. PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W” on its package. The potentiometer has a resistance from X to Z of 24  and has linear variation. (b) The potentiometer is adjusted so that the meter shows 4.0 V. Will it’s contact be above Y, below Y, or exactly on Y? SOLUTION:  The circuit is acting like a potential divider with R 1 being the resistance between X and Y and R 2 being the resistance between Y and Z.  Since we need V OUT = 4 V, and since V IN = 6 V, the contact must be adjusted above the Y. Solve problems involving electric circuits. Topic 5: Electric currents 5.2 Electric circuits 7.0 V X Y Z

37. PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W” on its package. The potentiometer has a resistance from X to Z of 24  and has linear variation. (c) The potentiometer is adjusted so that the meter shows 4.0 V. What are the current and the resistance of the lamp at this instant? SOLUTION:  P = 0.80 and V = 4.0.  From P = IV we get 0.8 = I(4) so that I = 0.20 A.  From V = IR we get 4 = 0.2R so that R = 20. .  You could also use P = I 2 R for this last one. Solve problems involving electric circuits. Topic 5: Electric currents 5.2 Electric circuits 7.0 V X Y Z

38. PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W” on its package. The potentiometer has a resistance from X to Z of 24  and has linear variation. (d) The potentiometer is adjusted so that the meter shows 4.0 V. What is the resistance of the Y-Z portion of the potentiometer? SOLUTION:  Let R 1 = X-Y resistance, R 2 = Y-Z resistance.  Then R 1 + R 2 = 24 so that R 1 = 24 – R 2.  From V OUT = V IN [ R 2 /(R 1 + R 2 ) ] we get 4 = 7[ R 2 /(24 – R 2 + R 2 ) ]. R 2 = 14 . (13.71) Solve problems involving electric circuits. Topic 5: Electric currents 5.2 Electric circuits 7.0 V X Y Z R1R1 R2R2

39. PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W” on its package. The potentiometer has a resistance from X to Z of 24  and has linear variation. (e) The potentiometer is adjusted so that the meter shows 4.0 V. What is the current in the Y-Z portion of the potentiometer? SOLUTION:  V 2 = 4.0 V because it is in parallel with the lamp.  Then I 2 = V 2 /R 2 = 4/13.71 (use unrounded value) = 0.29 A Solve problems involving electric circuits. Topic 5: Electric currents 5.2 Electric circuits 7.0 V X Y Z R1R1 R2R2

40. PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W” on its package. The potentiometer has a resistance from X to Z of 24  and has linear variation. (f) The potentiometer is adjusted so that the meter shows 4.0 V. What is the current in the ammeter? SOLUTION:  There are two currents supplied by the battery.  The red current is 0.29 A because it is the I 2 we just calculated in (e).  The green current is 0.20 A found in (c).  The ammeter has both so I = 0.29 + 0.20 = 0.49 A. Solve problems involving electric circuits. Topic 5: Electric currents 5.2 Electric circuits 7.0 V X Y Z R1R1 R2R2

41. PRACTICE: Which circuit shows the correct setup to find the V-I characteristics of a filament lamp? SOLUTION:  The voltmeter must be in paral- lel with the lamp.  It IS, in ALL cases.  The ammeter must be in series with the lamp and must read only the lamp’s current.  The correct response is B. Solve problems involving electric circuits. Topic 5: Electric currents 5.2 Electric circuits two currents no currents short circuit! lamp current

42. PRACTICE: A non-ideal voltmeter is used to measure the p.d. of the 20 k  resistor as shown. What will its reading be? SOLUTION:  There are two currents passing through the circuit because the voltmeter does not have a high enough resistance to prevent the green one from flowing.  The 20 k  resistor is in parallel with the 20 k  voltmeter. Thus their equivalent resistance is 1/R = 1/20000 + 1/20000 = 2/20000. R = 20000/2 = 10 k .  But then we have two 10 k  resistors in series and each takes half the battery voltage, or 3 V. Solve problems involving electric circuits. Topic 5: Electric currents 5.2 Electric circuits equivalent ckt

43. PRACTICE: All three circuits use the same resistors and the same cells. Which one of the following shows the correct ranking for the currents passing through the cells? SOLUTION:  The bigger the resistance the lower the current. Solve problems involving electric circuits. Topic 5: Electric currents 5.2 Electric circuits 2R2R 0.5R R 1.5R parallel series combo Highest I Lowest I Middle I

44. PRACTICE: The battery has negligible internal resis- tance and the voltmeter has infinite resistance. What are the readings on the voltmeter when the switch is open and closed? SOLUTION:  With the switch open the green R is not part of the circuit. Red and orange split the battery emf.  With the switch closed the red and green are in paral- lel and are (1/2)R. Solve problems involving electric circuits. Topic 5: Electric currents 5.2 Electric circuits E