2. Norah Ali Al- Moneef1

3.  Sources of emf The source that maintains the current in a closed circuit is called a source of emf – Any devices that increase the potential energy of charges circulating in circuits are sources of emf – Examples include batteries and generators SI units are Volts – The emf is the work done per unit charge Norah Ali Al- Moneef2 28.1 Electromotive Force All sources of emf have what is known as INTERNAL RESISTANCE (r) to the flow of electric current. The internal resistance of a fresh battery is usually small but increases with use. Thus the voltage across the terminals of a battery is less than the emf of the battery

4. Norah Ali Al- Moneef3 Direct Current When the current in a circuit has a constant direction, the current is called direct current – Most of the circuits analyzed will be assumed to be in steady state, with constant magnitude and direction Because the potential difference between the terminals of a battery is constant, the battery produces direct current The battery is known as a source of emf

5.  emf and Internal Resistance A real battery has some internal resistance Therefore, the terminal voltage is not equal to the emf Norah Ali Al- Moneef4 All sources of emf have what is known as INTERNAL RESISTANCE (r) to the flow of electric current. The internal resistance of a fresh battery is usually small but increases with use. Thus the voltage across the terminals of a battery is less than the emf of the battery  r + -

6. The internal resistance of the source of emf is always considered to be in a series with the external resistance present in the electric circuit The terminal voltage is ΔV = V b -V a ΔV = ε – Ir For the entire circuit, ε = IR + Ir Norah Ali Al- Moneef5 The schematic shows the internal resistance, r The internal resistance of the source of emf is always considered to be in a series with the external resistance present in the electric circuit The terminal voltage is ΔV = V b -V a ΔV = ε – Ir For the entire circuit, ε = IR + Ir

7. the relationship between Emf, resistance, current, and terminal voltage Circuit model looks like this : I r  R Terminal voltage = V V = IR  = I r + I R   = I (r + R) I =  / (r + R)  - Ir = V = IR The terminal voltage decrease =  - Ir as the internal resistance r increases or when I increases. If r =0 ( no internal resistance ) ( ideal battery )  = V = I R I = V / R I =  / R Norah Ali Al- Moneef6

8. –Power dissipated by a resistor, is amount of energy per time that goes into heat, light, etc. Power P = I Δ V Iε the total power output Iε of the battery I 2 R I 2 r is delivered to the external load resistance in the amount I 2 R and to the internal resistance in the amount I 2 r. Total power of emf Power dissipated as joule heat in: Internal resistorLoad resistor Norah Ali Al- Moneef7  Potential gain in the battery  Potential drop at all resistances  In an old, “used-up” battery emf is nearly the same, but internal resistance increases enormously

9. ε = ΔV + Ir = IR + Ir ε is equal to the terminal voltage when the current is zero – open-circuit voltage I = ε / (R + r) R is called the load resistance The current depends on both the resistance external to the battery and the internal resistance When R >> r, r can be ignored Power relationship: I ε = I 2 R + I 2 r When R >> r, most of the power delivered by the battery is transferred to the load resistor Norah Ali Al- Moneef8

10. 9 At V=10V someone measures a current of 1A through the below circuit. When she raises the voltage to 25V, the current becomes 2 A. Is the resistor Ohmic? a) YES b) NO

11. What are voltmeter and ammeter readings?  Example Norah Ali Al- Moneef10

12. the voltage across the internal resistor (lost voltage): v = Ir = 1.14 amps × 0.5 ohms = 0.57 volts From the figure find the current and the voltage across the internal resistor  example out the terminal voltage: Terminal voltage = emf - lost voltage = 12 - 0.57 = 11.43 volts Norah Ali Al- Moneef11

13. Example A battery has an emf of 12V and an internal resistance of 0.05Ω. Its terminals are connected to a load resistance of 3Ω. Find: a)The current in the circuit and the terminal voltage b)The power dissipated in the load, the internal resistance, and the total power delivered by the battery Norah Ali Al- Moneef12

14.  Example A car has a 12-V battery with internal resistance 0.020 . When the starter motor is cranking, it draws 125 A.What’s the voltage across the battery terminals while starting? Voltage across battery terminals = Battery terminals Norah Ali Al- Moneef13

15. ε=12 V R I = 2 A What does the energy balance look like in this circuit? When current leaves the battery the battery supplies power equal to: If current were forced to enter the battery, (as in charging it) then it absorbs the same power P = I ε = 24W Resistor: Electrical energy  heat  example Norah Ali Al- Moneef14

16. 0 Black 1 Brown 2 Red 3 Orange 4 Y ellow 5 Green 6 Blue 7 Violet 8 Gray 9 White Color Code Resistor Color Code 1 st digit 2 nd digit multiplier tolerance Norah Ali Al- Moneef15

17. Yellow = 4 Violet = 7 Red = 2 Gold = 5% 4700  5% of 4700 = 235 4700 + 235 = 4935 4700 - 235 = 4465 is the nominal value. The actual value can range from 4465 to 4935 . Applying the Color Code Norah Ali Al- Moneef16

18. What’s the nominal value and range? 1 k  (950 to 1050  ) 390  (370.5 to 409.5  ) 22 k  (20.9 to 23.1 k  ) 1 M  (950 k  to 1.05 M  ) Norah Ali Al- Moneef17

19. Resistors Under 10  The multiplier band is either gold or silver. For gold, multiply by 0.1. For silver, multiply by 0.01. 56  5.6  0.56  10  Norah Ali Al- Moneef18

20. Five-Band Code The first three stripes are digits. The fourth stripe is the multiplier. The tolerance is given by the fifth stripe: Brown = 1% Red = 2 % Green = 0.5% Blue = 0.25% Violet = 0.1% 1.58 k  Norah Ali Al- Moneef19

21. Zero-Ohm Resistor No resistance Used for connecting together two points on a printed-circuit board Body has a single black band around it Wattage ratings are typically 1/8 or 1/4- Watt Norah Ali Al- Moneef20

22. Resistors in Series When two or more resistors are connected end-to- end, they are said to be in series The current is the same in all resistors because any charge that flows through one resistor flows through the other The sum of the potential differences across the resistors is equal to the total potential difference across the combination Norah Ali Al- Moneef21 28.2 Resistors in Series and Parallel

23. Norah Ali Al- Moneef22 Potentials add – ΔV = IR 1 + IR 2 = I (R 1 +R 2 ) – Consequence of Conservation of Energy The equivalent resistance has the effect on the circuit as the original combination of resistors

24. Equivalent Resistance – Series R eq = R 1 + R 2 + R 3 + … always greater than any of the individual resistors The equivalent resistance of a series combination of resistors is the algebraic sum of the individual resistances and is always greater than any of the individual resistors Norah Ali Al- Moneef23 Current has one pathway - same in every part of the circuit Total resistance is sum of individual resistances along path Current in circuit equal to voltage supplied divided by total resistance Sum of voltages across each lamp equal to total voltage One bulb burns out - circuit broken - other lamps will not light

25. Resistors in Parallel The potential difference across each resistor is the same because each is connected directly across the battery terminals A junction is a point where the current can split The current, I, that enters a point must be equal to the total current leaving that point – I = I 1 + I 2 – The currents are generally not the same – Consequence of Conservation of Charge Norah Ali Al- Moneef24

26. Equivalent Resistance – Parallel Equivalent Resistance The inverse of the equivalent resistance of two or more resistors connected in parallel is the algebraic sum of the inverses of the individual resistance – The equivalent is always less than the smallest resistor in the group Norah Ali Al- Moneef25

27. Equivalent Resistance – Parallel, Example Equivalent resistance replaces the two original resistances Household circuits are wired so the electrical devices are connected in parallel – Circuit breakers may be used in series with other circuit elements for safety purposes Norah Ali Al- Moneef26

28. Norah Ali Al- Moneef27 Resistors in Parallel, Final In parallel, each device operates independently of the others so that if one is switched off, the others remain on In parallel, all of the devices operate on the same voltage The current takes all the paths – The lower resistance will have higher currents – Even very high resistances will have some currents Household circuits are wired so that electrical devices are connected in parallel

29. Equivalent Resistance – Series: An Example Four resistors are replaced with their equivalent resistance Norah Ali Al- Moneef28

30. Problem-Solving Strategy, 1 Combine all resistors in series – They carry the same current – The potential differences across them are not the same – The resistors add directly to give the equivalent resistance of the series combination: R eq = R 1 + R 2 + … Norah Ali Al- Moneef29

31. Problem-Solving Strategy, 2 Combine all resistors in parallel – The potential differences across them are the same – The currents through them are not the same – The equivalent resistance of a parallel combination is found through reciprocal addition: Norah Ali Al- Moneef30

32. Problem-Solving Strategy, 3 A complicated circuit consisting of several resistors and batteries can often be reduced to a simple circuit with only one resistor – Replace any resistors in series or in parallel using steps 1 or 2. – Sketch the new circuit after these changes have been made – Continue to replace any series or parallel combinations – Continue until one equivalent resistance is found Norah Ali Al- Moneef31

33. Problem-Solving Strategy, 4 If the current in or the potential difference across a resistor in the complicated circuit is to be identified, start with the final circuit found in step 3 and gradually work back through the circuits – Use ΔV = I R and the procedures in steps 1 and 2 – The greater the resistance in a circuit, the lower the current. Norah Ali Al- Moneef32

34. With the switch in this circuit (figure a) open, there is no current in R 2. There is current in R 1 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 R 2. When the switch is closed, the reading on the ammeter (a) increases, (b) decreases, or (c) remains the same. Norah Ali Al- Moneef33 (a). When the switch is closed, resistors R 1 and R 2 are in parallel, so that the total circuit resistance is smaller than when the switch was open. As a result, the total current increases.  Example

35. You have a large supply of lightbulbs and a battery. You start with one light bulb connected to the battery and notice its brightness. You then add one light bulb at a time, each new bulb being added in parallel to the previous bulbs. As the light bulbs 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. Norah Ali Al- Moneef34 (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 decreases  Example

36. what is the equivalent resistance of all resistors as placed in the below circuit? If V=12V, what is the current I? Norah Ali Al- Moneef35 I R1R1 R2R2 V R3R3 R 1 =3 Ohm R 2 =3 Ohm R 3 =3 Ohm V=12V R 2 & R 3 are in parallel 1/R 23 =1/R 2 +1/R 3 =1/3+1/3=2/3 R 23 =3/2 Ohm R 1 is in series with R 23 R 123 =R 1 +R 23 =3+3/2=9/2 Ohm I=V/R=12/(9/2)=24/9=8/3 A  Example

37. A tree is decorated with a string of many equal lights placed in parallel. If one burns out (no current flow through it), what happens to the others? a) They all stop shining b) the others get a bit dimmer c) the others get a bit brighter d) the brightness of the others remains the same Norah Ali Al- Moneef36 R R V I Before the one light fails: 1/R eq =1/R 1 +1/R 2 +…+1/R n if there are 3 lights of 1 Ohm: R eq =1/3 I=V/R eq I j =V/R j (if 3 lights: I=3V I j =V/1 After one fails: 1/R eq =1/R 1 +1/R 2 +….+1/R n-1 if there are 2 lights left: R eq =1/2 I=V/R eq I j =V/R j (if 2 lights: I=2V I j =V/1) The total resistance increases, so the current drops. The two effects cancel each other  Example

38. A tree is decorated with a string of many equal lights placed in parallel. If one burns out (no current flow through it), what happens to the others? – a) They all stop shining – b) the others get a bit dimmer – c) the others get a bit brighter – d) the brightness of the others remains the same Norah Ali Al- Moneef37 R R V I Before the one light fails: 1/R eq =1/R 1 +1/R 2 +…+1/R n if there are 3 lights of 1 Ohm: R eq =1/3 I=V/R eq I j =V/R j (if 3 lights: I=3V I j =V/1 After one fails: 1/R eq =1/R 1 +1/R 2 +….+1/R n-1 if there are 2 lights left: R eq =1/2 I=V/R eq I j =V/R j (if 2 lights: I=2V I j =V/1) The total resistance increases, so the current drops. The two effects cancel each other  Example

39. Norah Ali Al- Moneef 38 a person designs a new string of lights which are placed in series. One fails, what happens to the others? – a) They all stop shining – b) the others get a bit dimmer – c) the others get a bit brighter – d) the brightness of the others remains the same Assume: If one fails, the wire inside it is broken and Current cannot flow through it any more.  Example

40. Find the current through the 2-  resistor in the circuit. Equivalent of parallel 2.0-  & 4.0-  resistors: Total current is Equivalent of series 1.0- , 1.33-  & 3.0-  resistors:  Voltage across of parallel 2.0-  & 4.0-  resistors: Current through the 2-  resistor: Norah Ali Al- Moneef39

41. Equivalent Resistance Norah Ali Al- Moneef40 Three resistors are connected in parallel as shown in the Figure a potential difference of 18.0 V is maintained between points a and b. (A) Find the current in each resistor,total current ( I )  Example

42. Norah Ali Al- Moneef41 (B) Calculate the power delivered to each resistor and the total power delivered to the combination of resistors. P = I 2 R We apply the relationship P = I 2 R to each resistor and obtain This shows that the smallest resistor receives the most power. Summing the three quantities gives a total power of 198 W.

43. Equivalent Resistance – Complex Circuit Norah Ali Al- Moneef42  Example

44. inthe circuit (a) below to get the equivalent resistance. b ) calculate the power P = I 2 R dissipated in each resistor. Norah Ali Al- Moneef43  Example

45.  Example : Four equivalent light bulbs R 1 = R 2 = R 3 = R 4 = 4.50 W, emf = 9.00 Volts. Find current and power in each light bulb. Which bulb is brightest? Later, if bulb 4 is removed which bulbs get brighter? Dimmer? Still Bulb 1 is the brightest Norah Ali Al- Moneef44 I total =I 1 =1.5A,, I 2 =I3=I4+0.5A., P 1 =10.1W, P 2 =P 3 =P 4 =1.125W., V2=V3=V4=2.25V P 1 > P 2, P 3, P 4 Bulb 1 is the brightest The brightness (intensity) of the bulb is basically the power dissipated in the resistor. –For example A 200 W bulb is brighter than a 75 W bulb, all other things equal

46. What is the current through a? 1 /R eq = (1 /5) + (1 /10) (1 /30) R eq =30 / (6+3+1) = 3  What is the current through e? What is the current each branch b-d? 55 10  30  30v a e I tot = 10A b c d Same voltage is across each path b: V= IR30 = Ix5 I= 6A c: 30= Ix10 I= 3A d: 30= Ix30 I= 1A Norah Ali Al- Moneef45  Example

47. If all N branches have the same resistance, total resistance is equal to the resistance of one branch divided by the number of branches Total resistance= 1 /R eq = (1 /30) + (1 /30) (1 /30) R eq = 30 /3 = 10  Total current= V / R eq I total = 30 /10 = 3A Current in b= 30 /3 0 = 1 A Norah Ali Al- Moneef46  Example

48. If there are only two branches, the total resistance is equal to the product of the resistances divided by the sum of the resistances Total resistance= R eq = 12 X 4 /( 12 + 4 ) = 3  Norah Ali Al- Moneef47  Example

49. What is the: oTotal resistance? oTotal current flow? oCurrent flow through b oCurrent flow through c oCurrent flow through d oVoltage between b and d oVoltage between c and d oVoltage between d and e  Example Norah Ali Al- Moneef48

50. Total resistance: oIn compound circuits, reduce all parallel parts to a single resistance until you have a simpler series circuit oThe resistance between a and b is 2  oTherefore, total resistance is 4  (2 + 2)  Example Norah Ali Al- Moneef49

51. Total current: V = I total R eq 20 v = I X 4  I tot = 5 A Norah Ali Al- Moneef50 R 1eq = 3 X 6 / ( 3 + 6 ) = 2  the total resistance of the series portion of the circuit R eq = 2 + 2 = 4  Find the Total current. othe total resistance of the parallel portion of the circuit  Example

52. Current flow through b oWe need to know the voltage drop across b-d oVoltage drop across e-d will be (V= 5A X 2  = 10 v ) oTherefore, voltage drop across each parallel branch (c and b) must be ( V da =20 V -10 V = 10 V) oCurrent flow in b: 10 = I X 3  ; I = 3.33A oCurrent flow in c: 10 = I X 6  ; I= 1.67A  Example Norah Ali Al- Moneef51

53. Norah Ali Al- Moneef52 22 66 12  3. 20  4.  Example

54.  In the following circuit with source voltage V and Total current I, which resistor will have the greatest voltage across it?  The resistor with the largest resistance (30  )  Which resistor has the greatest current flow through it?  Same for all because series circuit  If we re-ordered the resistors, what if any of this would change?  Nothing would change Norah Ali Al- Moneef53  Example

55. If we added a resistor in series with these, what would happen to the total resistance, total current, voltage across each resistor, and current through each resistor?  Total resistance would increase  Total current would decrease  Voltage across each resistor would decrease (All voltage drops must still sum to total in series circuit; Kirchhoff’s law of voltages)  Current through each resistor would be lower (b/c total current decreased, but same through each one) Norah Ali Al- Moneef54  Example

56.  In the following circuit with source voltage V and Total current I, which resistor will have the greatest voltage across it?  All the same in parallel branches  Which resistor has the greatest current flow through it?  The “path of least resistance” (10  )  What else can you tell me about the current through each branch  They will sum to the total I (currents sum in parallel circuits; Kirchhoff’s law of current) Norah Ali Al- Moneef55  Example

57. If we added a resistor in parallel with these, what would happen to the total resistance, total current, voltage across each resistor, and current through each resistor? Total resistance would decrease Total current would increase Voltage across each resistor would still be V Current through each resistor would be higher and would sum to new total I  Example Norah Ali Al- Moneef56

58. Norah Ali Al- Moneef57 R 1 =100 ohms R 2 =100 ohms R 3 =100 ohms R eq = ? Ohms 1 / R eq = 1 / R1 + 1 / R2 + 1 / R3 1 / R eq = 1/100 + 1/100 + 1/100 1 / R eq =.01 + 01 +.01 =.03 R eq = 1/.03 = 33.33 ohms  Example

59. Norah Ali Al- Moneef58 A parallel circuit is shown in the diagram above. In this case the current supplied by the battery splits up, and the amount going through each resistor depends on the resistance. If the values of the three resistors are: R 1 = 8 Ω R 2 = 8 Ω R 3 = 4 Ω 1 / R eq = 1 / 8 + 1 / 8 + 1 / 4 = 4 / 8 R eq = 2 Ω With a 10 V battery, by V = I R the total current in the circuit is: I = V / R = 10 / 2 = 5 A. The individual currents can also be found using I = V / R. The voltage across each resistor is 10 V, so: I 1 = 10 / 8 = 1.25 A I 2 = 10 / 8 = 1.25 A I 3 =10 / 4 = 2.5 A Note that the currents add together to 5A, the total current.  Example

60. What would V 1 read in the illustration? Ohm’s Law states: Therefore: Norah Ali Al- Moneef59 At this point there is insufficient data because I (amp) is unknown. Using Ohm’s Law to solve for the current in the circuit: Knowing the amount of current we can calculate the voltage drop.  Example

61. Determine the readings for A1 and A2 in the illustration. In a series circuit the electricity has no alternative paths, therefore the amperage is the same at every point in the circuit. Norah Ali Al- Moneef60 The current in the circuit is determined by dividing the voltage by the circuit resistance.  Example

62. Determine the readings for amp meters A 1 and A 2 in the circuit. Norah Ali Al- Moneef61 In a parallel circuit the amperage varies with the resistance. In the illustration, A 1 will measure the total circuit amperage, but A 2 will only measure the amperage flowing through the 6.3 Ohm resistor. To determine the total resistance of the circuit must be  Example

63. When the total resistance is known, the circuit current (Amp’s) can be calculated. Norah Ali Al- Moneef62 Branch current is: A 2 = 1.9 A A 1 = 12.76 A Total current is: When the circuit current (Amp’s) is known, the current for each branch circuit can be calculated.  Example

64. To calculate the circuit current the total circuit resistance must be known. Norah Ali Al- Moneef63 Determine the readings for the two volt meters in the illustration. Volt meter one (V 1 ) will read source voltage: Volt meter two (V 2 ) will read the voltage in the circuit after the 2.3  resistor. To determine this reading, the voltage drop across the resistor must be calculated. Before the voltage drop across the resistor can be calculated, the circuit current must be determined. 12 V  Example

65. Circuit current is: Norah Ali Al- Moneef64 Volt meter 1 will read source voltage: 12 V Volt meter 2 will read 3.6 V. The voltage drop caused by the 2.3 Ohm resistance is: The voltage remaining in the circuit is:  Example

66. Determine the readings for the amp meters in the illustration. The first amp meter will read circuit amps and the second one will measure the current in that branch of the circuit. To determine circuit amperage, the total circuit resistance must be known. This was calculated in the previous slide as 3.29 . Norah Ali Al- Moneef65 Total circuit current was also calculated as 3.65 A. A 1 = 3.65 A To determine the reading for A 2, the current flowing in that part of the circuit must be calculated. A 2 = 0.62 A In the previous slide the voltage left after the 2.3  resistor was 3.61 V.  Example

67. Norah Ali Al- Moneef66 When a piece of wire is used to connect points b and c in this figure, the brightness of bulb R 1 (a) increases, (b) decreases, or (c) stays the same. The brightness of bulb R 2 (a) increases, (b) decreases, or (c) stays the same. R 1 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 R 1 + R 2 to just R 1. 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 R 1 increases, causing this bulb to glow brighter. Bulb R 2 goes out because essentially all of the current now passes through the wire connecting b and c and bypasses the filament of Bulb R 2.  Example

68. Norah Ali Al- Moneef67 With the switch in this circuit (figure a) closed, no current exists in R 2 because the current has an alternate zero-resistance path through the switch. Current does exist in R 1 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 R 2. 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 R 1 and R 2 are in series, so that the total circuit resistance is larger than when the switch was closed. As a result, the current decreases.  Example

69. Norah Ali Al- Moneef68 What is the value of R? I R = V / R =. 8 / 12 =.667A 1 / R eq = 1 / 10 + 1 / 15 + 1 / R =1 50+25 R / 150 R R eq = 150 R / 150+ 25R R eq = 150 R / 150+ 25R R eq = 8 / 2 = 4 Ω R eq = 8 / 2 = 4 Ω 4= 150 R /1 50+25 R 4 (150+ 25R ) = 150 R 4 (150+ 25R ) = 150 R 100 R + 600 = 150R R = 12 Ω  Example

70. What is the Value of R? Norah Ali Al- Moneef69 R = Δ V / I = 5.0 V / ( 100 X 10 -3 )A R = 25Ω  Example

71. What is the current through the circuit? What are the values of potential at points a-f? Norah Ali Al- Moneef70 I: 1A a:0V b:12V c:9V d: 6V e: 0V f: 0V  Example

72. 28.3 Kirchhoff’s Rules There are ways in which resistors can be connected so that the circuits formed cannot be reduced to a single equivalent resistor Two rules, called Kirchhoff’s Rules can be used instead Norah Ali Al- Moneef71  Statement of Kirchhoff’s Rules Junction Rule – The sum of the currents entering any junction must equal the sum of the currents leaving that junction A statement of Conservation of Charge Loop Rule – The sum of the potential differences across all the elements around any closed circuit loop must be zero A statement of Conservation of Energy

73.  the Junction Rule I 1 = I 2 + I 3 From Conservation of Charge Diagram b shows a mechanical analog Norah Ali Al- Moneef72 I1I1 I3I3 I2I2 I4I4 I5I5 I 1 +I 2 +I 3 =I 4 +I 5

74.  Setting Up Kirchhoff’s Rules Norah Ali Al- Moneef73 Assign symbols and directions to the currents in all branches of the circuit – If a direction is chosen incorrectly, the resulting answer will be negative, but the magnitude will be correct When applying the loop rule, choose a direction for transversing the loop – Record voltage drops and rises as they occur

75. Electrical Work and Power II+- Higher V 1 Lower V 2 Resistance R Current I flows through a potential difference  V Follow a charge Q : at positive end, U 1 = QV 1 at negative end, U 2 = QV 2 P.E. Decreases: The speed of the charges is constant in the wires and resistor. What is electrical potential energy converted to? Norah Ali Al- Moneef74

76. Electrical resistance converts electrical potential energy to thermal energy (heat), just as friction in mechanical systems converts mechanical energy to heat. This thermal energy means the atoms in the conductor move faster and so the conductor gets hotter. The average kinetic energy of the electrons doesn’t increase once the current reaches a steady state; the electrons lose energy in collisions with the atoms as fast as it is supplied by the field. Norah Ali Al- Moneef75

77. More About the Loop Rule Traveling around the loop from a to b In a In a, the resistor is transversed in the direction of the current, the potential across the resistor is –IR In b In b, the resistor is transversed in the direction opposite of the current, the potential across the resistor is +IR Norah Ali Al- Moneef76 In c In c, the source of emf is transversed in the direction of the emf (from – to +), the change in the electric potential is +ε In d In d, the source of emf is transversed in the direction opposite of the emf (from + to -), the change in the electric potential is -ε

78. Junction Equations from Kirchhoff’s Rules Use the junction rule as often as needed, so long as, each time you write an equation, you include in it a current that has not been used in a previous junction rule equation – In general, the number of times the junction rule can be used is one fewer than the number of junction points in the circuit Norah Ali Al- Moneef77

79. Loop Equations from Kirchhoff’s Rules The loop rule can be used as often as needed so long as a new circuit element (resistor or battery) or a new current appears in each new equation You need as many independent equations as you have unknowns Norah Ali Al- Moneef78

80. Problem-Solving Strategy – Kirchhoff’s Rules Draw the circuit diagram and assign labels and symbols to all known and unknown quantities Assign directions to the currents. Apply the junction rule to any junction in the circuit Apply the loop rule to as many loops as are needed to solve for the unknowns Solve the equations simultaneously for the unknown quantities Check your answers Norah Ali Al- Moneef79

81. The emfs and resistances in the circuit have the following values Starting at point a lets apply the loop rule counter-clockwise and tally the voltages along the way Norah Ali Al- Moneef80  Example

82. What is the potential difference between the terminals of battery 1 ?? This is the equivalent of asking what the potential difference is between points b and a. So lets travel from point b to point a clockwise V b – i r 1 + ε 1 = V a V a - V b = - i r 1 + ε 1 = -(0.24A) (2.3  ) + 4.4V = 3.8V Norah Ali Al- Moneef81

83. Example. Multiloop Circuit Find the current in R 3 in the figure below. Loop 1:  Loop 2:  Norah Ali Al- Moneef82

84. Label the 3 branch currents I 2, I 3, and I 4. ε Since V AB across all 3 branches is the same and is known: V 4 = I 4 R 4 = 5A X (4Ω) = 20 Volts, the currents and ε can be readily solved. Norah Ali Al- Moneef83 V 4 = I 4 R 4 = 5A (4  ) = 20 Volts I 3 = V 3 / R 3 = 4V / 3  = 1.33 A At junction B,  I in =  I out I 4 = I 2 + I 3 ; I 2 = I 4 – I 3 = 5A - 1.3A I 2 = 3.7A Loop #1:  V i =0 -  + I 2 R 2 + I 4 R 4  = 3.7A (2  ) + 5A(4  ) = 27.4 V  Example

85. Norah Ali Al- Moneef84 Find r and ε. I in = I out I = 1 A + 2A = 3 A Loop 1: - ε +1 A X 1Ω + 2A X 3 Ω = 0 ε = 7 V Loop 3 : 12 – 3 A Xr – 2 A X 3 Ω = 0 r = 6 V / 3 A = 2 Ω  Example

86. Norah Ali Al- Moneef85 1) I 1 = 12 / ( 9 /2 ) = 8/3 A Kirchhof 2 2) V-I 1 R 1 -I 2 R 2 =0 12-3 I 1 -3 I 2 =0 Kirchhof 2 3) V-I 1 R 1 -I 3 R 3 =0 12-3I 1 -3I 3 =0 Kirchhof 2 4) 0-I 3 R 3 +I 2 R 2 =0 -3I 3 +3I 2 =0 Kirchhoff I 5) I 1 -I 2 -I 3 =0 I 1 =I 2 +I 3 1)& 2) 12-8-3I 2 =0 so 4=3I 2 and I 2 =4/3 A 1) & 3) 12-8-3I 3 =0 so 4=3I 3 and I 3 =4/3 A Use V=IR for R 1 V 1 =8/3X3= 8 V for R 2 V 2 =4/3X3= 4 V for R 3 V 3 =4/3X3= 4 V  Given V=12V, what is the current through and voltage over each resistor R eq = 3 +( 3X 3 / 3 + 3 ) = 3 + ( 9 / 6 ) = 4.5 Ω I=I 1 R1R1 R2R2 V R3R3 R 1 =3 Ω R 2 =3 Ω R 3 =3 Ω V=12V I3I3 I1I1 I2I2 I=I 1  Example

87. Norah Ali Al- Moneef86 which of the following cannot be correct? a) V-I 1 R 1 -I 3 R 3 -I 2 R 2 =0 b) I 1 -I 3 -I 4 =0 c) I 3 R 3 -I 6 R 6 -I 4 R 4 =0 d) I 1 R 1 -I 3 R 3 -I 6 R 6 -I 4 R 4 =0 e) I 3 +I 6 +I 2 =0 NOT a LOOP I 4 R 4 I 3 R 3 I 2 R 2 I 1,R 1 I 5 R 5 V I 6 R 6  Example

88. Norah Ali Al- Moneef87 consider the circuit. Which of the following is/are not true? 1.If R 2 =R 3 =2R 1 the potential drops over R 1 and R 2 are the same 2.for any value of R 1,R 2 and R 3 the potential drop over R 1 must be equal to the potential drop over R 2 3.The current through R 1 is equal to the current through R 2 plus the current through R 3 (I 1 =I 2 +I 3 ) I R1R1 R2R2 V R3R3 THIS IS TRUE 1)if R 2 =R 3 =2R 1 then 1/R 23 =1/R 2 +1/R 3 =1/R 1 so R 23 =R 1 and I 1 =I 23 and potential of R 1 equals the potential over R 23 and thus R 2 and R 3. THIS IS TRUE 2)no 2)no, this is only TRUE in the case of 1) 3)true: 3)true: conservation of current.  Example

89. Norah Ali Al- Moneef88  Example 12V A BC D At which point (A,B,C,D) is the potential highest and at which point lowest? All resistors are equal. a)highest B, lowest A b)highest C, lowest D c)highest B, lowest D d)highest C, lowest A e)highest A, lowest B circuit 1: I=12/(2R)=6/R V A =12-6/R*R=6V circuit 2: V B =12V circuit 3: I=12/(3R)=4/R V C =12-4/R*R=8V V D =12-4/R*R-4/R*R=4V 1 2 3

90. Norah Ali Al- Moneef89 R1R1 R3R3 R2R2 V1V1 V2V2 R 1 =R 2 =R 3 = 3 Ohm V 1 =V 2 = 12 V  what is the current through and voltage over each R? I1I1 I3I3 I2I2 apply kirchhoff’s rules 1) I 1 +I 2 -I 3 =0 (kirchhoff I) 2) left loop: V 1 -I 1 R 1 +I 2 R 2 =0 so 12-3I 1 +3I 2 =0 3) right loop: V 2 +I 3 R 3 +I 2 R 2 =0 so 12+3I 3 +3I 2 =0 4) outside loop: V 1 -I 1 R 1 -I 3 R 3 -V 2 =0 so –3I 1 -3I 3 =0 so I 1 =-I 3 combine 1) and 4) I 2 =2I 3 and put into 3) 12+9I 3 =0 so I 3 = 4/3 A and I 1 = - 4/3 A and I 2 = 8/3 A  Example

91. Norah Ali Al- Moneef90 what is the power dissipated by R 3 ? P=VI=V 2 /R=I 2 R I=I 1 R1R1 R2R2 V R3R3 R 1 = 1 Ω, R 2 = 2 Ω, R 3 = 3 Ω, R 4 = 4 Ω V= 5V I3I3 I1I1 I2I2 I=I 1 R4R4 I4I4 We need to know V 3 and/or I 3. Find equivalent R of whole circuit. 1/R 23 =1/R 2 +1/R 3 = 1/2+1/3= 5/6 R 23 = 6/5 Ω R 1234 = R 1 +R 23 +R 4 = 1+6/5+4=31/5 Ω I=I 1 =I 4 = V / R 1234 = 5 / (31/5) I = 25 / 31 A Kirchhoff 1: I 1 =I 2 +I 3 = 25 / 31 Kirchhoff 2: I 3 R 3 -I 2 R 2 =0 so 3I 3 -2I 2 =0 I 2 = 3 / 2 I 3 Combine: 3/2 I 3 +I 3 = 25 / 31 so 5 / 2 I 3 =25 / 31 I 3 =10/31 A P=I 2 R so P=(10 / 31) 2 X 3=(100/961)X3= 0.31 W  Example

92. Norah Ali Al- Moneef91 What is Kirchhoff II for the right small loop (with R 2 and R 3 )? a)I 3 R 3 +I 2 R 2 =0 b) I 3 R 3 -I 2 R 2 =0 c) I 3 R 3 -I 2 R 2 +V=0 What is Kirchhoff II for the loop (with V,R 4 and R 3 )? a) V-I 4 R 4 +I 3 R 3 =0 b) V+I 4 R 4 -I 3 R 3 =0 c) V-I 4 R 4 -I 3 R 3 =0 R1R1 R2R2 V R3R3 I3I3 I1I1 I2I2 I4I4 R4R4 What is Kirchhoff I for ? a)I 1 +I 2 -I 3 -I 4 =0 b) I 1 +I 2 +I 3 +I 4 =0 c) I 1 -I 2 -I 3 -I 4 =0 What is Kirchhoff II for the left small loop ( with R 4 and R 1 ?) a) I 4 R 4 +I 1 R 1 =0 b) I 4 R 4 - I 1 R 1 =0 c) I 4 R 4 +I 1 R 1 -V=0  Example

93. The Capacitance of a Capacitor Q = [( ε 0 A)/d]∆V c Capacitance (C) = ( ε 0 A)/d. For a given charge, a capacitor with a larger capacitance will have a greater potential difference. The SI unit of capacitance is the farad: 1 farad = 1 F = 1 Coulomb/Volt. Typical capacitors have values in the microfarad to pico farad range. Norah Ali Al- Moneef92 28.4 RC Circuits

94. RC Circuits If a direct current circuit contains capacitors and resistors, the current will vary with time At the instant the switch is closed, the charge on the capacitor is zero Norah Ali Al- Moneef93 When the circuit is completed, the capacitor starts to charge, and the potential difference across the capacitor increases, until the charge reaches its maximum ( Q = C ε ) After the switch is closed, the battery will move charge from the bottom plate to the top plate of the capacitor. As the charge on the capacitor increases, the current through the circuit will decrease.

95. Norah Ali Al- Moneef94 RC Circuits

96. RC Circuit  Unlike a battery, a capacitor cannot provide a constant source of potential difference.  This value is constantly changing as the charge leaves the plate.  Current due to a discharging capacitor is finite and changes over time.  The capacitor continues to charge until it reaches its maximum charge (Q = C ε)  Once the capacitor is fully charged, the current in the circuit is zero Norah Ali Al- Moneef95

97. Charging Capacitor in an RC Circuit Once the capacitor is fully charged, the current in the circuit is zero Once the maximum charge is reached, the current in the circuit is zero, and the potential difference across the capacitor matches that supplied by the battery Norah Ali Al- Moneef96 After the switch is closed, the battery will move charge from the bottom plate to the top plate of the capacitor. As the charge on the capacitor increases, the current through the circuit will decrease.

98. Charging Capacitor in an RC Circuit Norah Ali Al- Moneef97 Integrating this expression, using the fact that q = 0 at t =0, we obtain Charge as a function of time for a capacitor being charged

99. Charging Capacitor in an RC Circuit The charge on the capacitor varies with time q = Q (1 – e -t/RC ) The time constant,  = RC, represents the time required for the charge to increase from zero to 63.2% of its maximum Norah Ali Al- Moneef98 In a circuit with a large (small) time constant, the capacitor charges very slowly (quickly) After t = 10 , the capacitor is over 99.99% charged We can find an expression for the charging current with respect to time. Using I = dq/dt, we find that Current as a function of time for a capacitor being charged

100. Norah Ali Al- Moneef99 The following dimensional analysis shows that  has the units of time: Because  = RC has units of time, the combination  = RC is dimensionless, as it must be in order to be an exponent of e The energy output of the battery as the capacitor is fully charged is Q ε = C ε 2. After the capacitor is fully charged, the energy stored in the capacitor is ½ Q ε = (1 / 2 )C ε 2 which is just half the energy output of the battery. to show that the remaining half of the energy supplied by the battery appears as internal energy in the resistor.

101. Norah Ali Al- Moneef100

102. Use Kirchoff’s loop law to analyze the circuit, starting at the capacitor: ∆V c + ∆V R = 0 Since charge is being removed from the plates: Norah Ali Al- Moneef101 Discharging Capacitor in an RC Circuit When a charged capacitor is placed in the circuit, it can be discharged the circuit shown, which consists of a capacitor carrying an initial charge Q, a resistor, and a switch. the circuit shown, which consists of a capacitor carrying an initial charge Q, a resistor, and a switch. When the switch is open, a potential difference Q /C exists across the capacitor and there is zero potential difference across the resistor because I = 0. If the switch is closed at t =0, the capacitor begins to discharge through the resistor. At some time t during the discharge, the current in the circuit is I and the charge on the capacitor is q a current that decreases in magnitude with time is set up in the direction shown, and the charge on the capacitor decreases exponentially with time

103. Discharging Capacitor in an RC Circuit The initial conditions were Q = Q 0 at t=0 (switch closes) Norah Ali Al- Moneef102 This is a differential equation, but is relatively easy to solve by rearranging terms: And then by integrating…. = -1/RC ln Q – ln Q 0 = -t/RC ln (Q/Q 0 ) = - t/RC

104. Discharging Capacitor in an RC Circuit Q e – t /RC The charge on the capacitor varies with time q = Q e – t /RC The charge decreases exponentially t =  = RC At t =  = RC, the charge decreases to 0.368 Q max ; i.e., in one time constant, the capacitor loses 63.2% of its initial charge The current that decreases in magnitude with time is set up in the direction shown, and the charge on the Capacitor decreases exponentially with time Norah Ali Al- Moneef103 Differentiating this expression with respect to time gives the instantaneous current as a function of time: The negative sign indicates that as the capacitor discharges, the current direction is opposite its direction when the capacitor was being charged.

105. Notes on Time Constant In a circuit with a large time constant, the capacitor charges very slowly The capacitor charges very quickly if there is a small time constant After t = 10 , the capacitor is over 99.99% charged Norah Ali Al- Moneef104

106. Norah Ali Al- Moneef105 A 10 μF capacitor initially charged to 20 μC is discharged through a 1 kΩ resistor. How long does it take to reduce the capacitor’s charge to 10 μC? 10 μC = (20 μC) e t/0.010 s Take the natural logarithm of both sides: ln ( 10 μC/20 μC ) = - t/0.010 s t = 6.93 s  Example

107. Norah Ali Al- Moneef106  Example given: V=10 V R=100 Ohm C=10x10 -6 F V The emf source is switched on at t=0. a) After how much time is the capacitor C charged to 75% of its full capacity? b) what is the maximum current through the system? a) if (1-e -t/RC )=0.75 then charged for 75%, so e -t/RC = 0.25 t / RC= - ln(0.25) ln: natural logarithm t= -RC x ln (0.25)=-100x10 -5 x(-1.39)= 1.39x10 -3 seconds. b) maximum current: at t=0 it is as if the capacitor C is not present so I= V/ R= 0.1 A

108. Norah Ali Al- Moneef107  Summary

109. Norah Ali Al- Moneef108 RC circuit for the charge on the capacitor for the voltage over the capacitor for the voltage over the resistor for the current e = 2.718…

110. Norah Ali Al- Moneef109  = I r + I R   = I (r + R) I =  / (r + R)  - Ir = V = IR Terminal voltage = V = IR Electromotive Force (  ) The terminal voltage decrease =  - Ir as the internal resistance r increases or when I increases.If r =0 ( no internal resistance ) ( ideal battery )  = V = I R I = V / R I =  / R

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112. Series connection Parallel connection Resistors connected in a circuit in series or parallel can be simplified using the following: Norah Ali Al- Moneef111