1. Resistor and Resistance
Shatin Tsung Tsin Secondary School
Mr. C.K. Yu

2. How do electrons (charges) flow within a conductor
Direction of electron flows
Direction of current
Current : flow of positive charges

3. How do electrons (charges) flow within a conductor
When electrons travel through, they bump onto the outer electrons or the atoms and energies are given to the conductor.

4. Ohm’s Law
(i) Conductors (導電體) are material which can conduct electricity (傳電).
(ii) Electrons (negative charges) or positive charges can go through conductor.
(iii) Negative charges give energy to the conductor when they travel through the conductor.

5. Ohm’s Law
(iv) If there is a potential difference across a conductor, a current will flow through the conductor.
(v) Ohm discovered (發現) that, in some conductors, the current (I) and the potential difference or voltage (V) across the conductor is directly proportional.

6. Ohm’s Law (vi) So, his discovery is called Ohm's law.
The proportional constant (the slope) is
called the resistance of that material

7. Ohm’s Law-summary
Ohm’s law states that the p.d. across an ohmic conductor is directly proportional to the current through it, provided that the temperature and other physical conditions are constant (the same). The proportional constant is called resistance .
A material is called ohmic material if it follows the ohm’s law

8. Experiment – Ohm’s Law Objective : To verify the ohm’s law Apparatus:
1 power supply 1 resistor (red)
1 voltmeter 1 ammeter
1 circuit board connecting wires
1 switch

9. Experiment – Ohm’s Law
Your power supply consists of 4 electric cells. In this experiment, you are to measure the relationship between the current passing through and the potential difference across a resistor.

10. Experiment – Ohm’s Law
Connect a red resistor, an ammeter, a switch and a power supply (4 cells) in series as shown in Figure 1.
Figure 1

11. Experiment – Ohm’s Law
2. Close the switch and measure the current shown in the ammeter, record the reading in Table 1.

12. Table 1 No. Reading of Ammeter (I/A) Reading of Voltmeter (p.d./V)
Remark 1
1 cell 2
2 cells 3
3 cells 4
4 cells ==
========

13. Experiment – Ohm’s Law
Keep the switch closed. Use the voltmeter to measure the potential difference across the red resistor as shown in figure 2. Record the reading in Table 1.
Figure 2

14. Experiment – Ohm’s Law
4. Repeat steps 1 to 3 with different numbers of electric cells in the power supply. Also record the results in Table 1.

15. Experiment – Ohm’s Law
5.In the graph below , plot the graph of V against I.
V/V
I/A

16. Experiment – Conclusion
The proportional constant (resistance) of the resistor is : ___________ Ω
Ohm’s law states that the p.d. across a conductor is __________ proportional to the current through it, provided that the ___________ and other physical conditions are constant.
The ratio of p.d./current is the resistance of the conductor. It is measured in ___________.
directly
temperature
Ohm, or Ω

17. HOT questions
If the battery of a circuit is exhausted, what will happen to the current in the circuit?
__________________________________
What are the possible reasons if the current in a circuit increases?
____________________________________________________________________
The current will decrease and become zero.
Either the resistance is lower or the supply
source is of higher emf.

18. Resistors
(i)Any conductors which work under Ohm's Law are called resistor.
(ii)The proportional constant (正比常數) of voltage and current is called the resistance (電阻值) of that conductor.
(iii)Unit of resistance is "Ohm", or W .

19. Resistors (iv)The symbol of a resistor in a circuit :
(v)The mathematical equation for Ohm’s law is:
Where R : the resistance of the resistor,
V : the voltage across the resistor,
I : the current flow through the resistor.
12W
e.g.

20. (c) Simple resistor circuit
15 Ω
The values of R, V, I follows the ohm’s law.

21. Example
5 C of charge passes a resistor in a circuit in 2 s, the total electrical energy dissipated by the charge is 8 J.
(i) What is the resistance of the resistor ?
V = E / Q = 8 J / 5C = 1.6 V
I = Q / t = 5 C /2 s = 2.5 A
R = V / I = 1.6 V / 2.5 A = 0.64 W

22. Example
5 C of charge passes a resistor in a circuit in 2 s, the total electrical energy dissipated by the charge is 8 J.
(ii) What is the power dissipation of the resistor ?
P = E / t = 8 J / 2 s = 4 W

23. Revisit of Electric Power
( by definition) P = t P =
V x I
(V=E/Q,I=Q/t, V x I=E/ t) P =
(I = V/R)
V2 / R P =
(V = I R)
I2 R

24. Summary Table Quantities Symbol Unit Unit symbol Time t Second s
Charge Q
Coulomb C
Current I
Ampere A
Electromotive force
e.m.f
Volt V
Voltage
Potential difference
p.d.
Power P
Watt W
Resistance R
Ohm

25. Resistors in series Resistors can be connected together at each end.
(ii) One of the basic methods is to connect the resistors one by one in a line, called in series.

26. Resistors in series
(iii) The following diagram shows three resistors connected in series :
(iv) they can be considered as a single resistor, with a total resistance RT. RT
All charges flow through R1 will also flow through R2 and R3
RT follows Ohm’s law

27. Resistors in series In calculation : RT = R1 + R2 + R3 + ….. Example :
R1 = 5 W, R2 = 3 W , R3 = 5 W
If the above three resistors are connected in series, the total resistance is
______________________
5 W + 3 W + 5 W = 13 W

28. Resistors in Parallel
(i) Another method is to connect the resistors one over another at two ends, called in parallel.
(ii) The diagram shows
three resistors
connected in parallel : A B
Charges flowing from A to B will only go through one of the resistors in parallel.

29. Resistors in Parallel
(iii) These resistors can be considered as a single resistor, with a total resistance RT

30. Resistors in Parallel RT
(iii) These resistors can be considered as a single resistor, with a total resistance RT
RT follows Ohm’s law

31. Resistors in Parallel In calculation : + … … … Example :
R1 = 5 W, R2 = 10 W , R3 = 5 W
If the above three resistors are connected in parallel, the total resistance is __ W . 2

32. Examples
Two resistors of 5 ohms and 15 ohms are connected in series. The total resistance is ______ ohms. 20

33. Examples
Three resistors of the same resistance are connected in series and the total resistance is 15 ohms. What is the resistance of each resistor?
5 ohms

34. Examples
Two resistors of the same resistance are connected in parallel and the total resistance is 5 ohms. What is the resistance of each resistor?
10 ohms

35. Examples
Two resistors are connected in parallel and the total resistance is 4 ohms. One of them is 5 ohms, what is the resistance of the other resistor?
20 ohms

36. Examples
Three resistors are connected in parallel and the total resistance is 2 ohms. Two of them are 5 ohms and 10 ohms, what is the resistance of the third resistor?
5 ohms

37. Class Example
(i) If 2 resistors of 2 Ω and 5 Ω are connected in series, the total resistance is :
R1 = 2 W, R2 = 5 W
By RT = R1 + R2
= = 7 Ω

38. Class practices
(ii) If 4 resistors of 2W, 3W , 8W and 9 W, are connected in series, the total resistance is :
The total resistance is : = 22W

39. Class practices
(iii) If 2 resistors of 2W and 5W, are connected in parallel, the total resistance is :
The total resistance :
1/R = 1/2+1/5 = 7/10
R = 10/7 or 1 3/7 W

40. Class practices
(iv) If 4 resistors of 2W, 3W, 8Wand 9W, are connected in parallel, the total resistance is :
The total resistance :
1/R = 1/2+1/3+1/8+1/9
= ( )/72
=77/72
R = 72/77 W

41. Class Practices
If 3 resistors of 2W, 3W, 8W are connected in series and then in parallel with a resistor of 9W, the total resistance of this connection is :
The total resistance :
1/R = 1/(2+3+8)+1/9
= (9+13)/117
=22/117
R = 117/22 or 5 7/22 W

42. HOT – Home Practice
Only resistors of 5 ohms are given. How do we connect the resistors to form a total resistance of 4 ohms?

43. Summary
(i) Resistors connected either in series or in parallel can be considered as a single resistor of resistance RT.
(ii) The current and the voltage across the resistors or the equivalent resistor can also be calculated using the equations :
RT = VT / IT

44. Simple measurement experiments
Objectives:
Measure current and voltage with ammeter and voltmeter
Find out the characteristics of resistors connected in parallel

45. Construct the circuit as shown by connecting a 6V battery and two identical resistors, R1 and R2 , in parallel. Show your connection to your teacher.

46. Connect an Ammeter in series with each resistor to measure current I1 and I2. (I1 and I2 can be called branch currents)
I1=_______ A
I2=_______ A

47. Then connect an ammeter in series with the battery as shown in the diagram. Measure the current I.
Compare the three reading, I1, I2, and I
I=_______ + ________ = ________ A
I I2

48. Finally, connect a Voltmeter, in turns, in parallel across each resistor to measure p.d. across each of them respectively
p.d. across R1 =____V,
p.d. across R2 =____V.
Conclusion : The p.d. across each parallel branch is __________.
the same

49. Summary of resistors in series or in parallel
Resistors in parallel :
Voltages across all branches in parallel are the same. V = V1 = V2 = V3 =….
The current before and after the resistors in parallel is the sum of the currents of all branch. I = I1+I2+I3+ ….

50. Summary of resistors in series or in parallel
Current of each resistor in a series is the same. I = I1 = I2 = I3 =….
The p.d. (voltage) across the beginning and the end of the resistors in series is the sum of the p.d. across each resistors.
V = V1+V2+V3+ ….

51. HOT Questions
Which of the following is true in the following description about a wire.
The longer the wire, the higher the resistance of the wire.
The larger the diameter of the wire, the higher the resistance of the wire 

52. Comparison table In series In parallel Current I = I1=I2=I3=…
IT=I1+I2+I3+…
Voltage, p.d.
VT=V1+V2+V3+…
V=V1=V2=V3+…
Total Resistance
RT=R1+R2+R3+…

53. Advanced Example (i) Below shows a series circuit, calculate
(1)the total resistance of the circuit,
(2)the current, I of the circuit,
(3)the current I1 passed through the resistor R1,
(4)the current I2 passed through the resistor R2,
(5)V1 , the potential difference across the resistor R1, and
(6)V2 , the potential difference across the resistor R2.

54. Advanced Example (i) Below shows a series circuit, calculate
(1)the total resistance of the circuit,
RT = R1 + R2 + R3 + ..
RT = 15 W + 9 W = 24 W

55. Advanced Example (i) Below shows a series circuit, calculate R = V/ I
(2)the current, I of the circuit,
R = V/ I
24W = 6V / I
I = 0.25 A

56. Advanced Example (i) Below shows a series circuit, calculate I = I1=I2
(3)the current I1 passed through the resistor R1,
I = I1=I2
I1 = 0.25A

57. Advanced Example (i) Below shows a series circuit, calculate I = I1=I2
(4)the current I2 passed through the resistor R2,
I = I1=I2
I2 = 0.25A

58. Advanced Example (i) Below shows a series circuit, calculate
(5)V1 , the potential difference across the resistor R1, and
By R1 = V1/I1
15W = V1 / 0.25 A
V1 = 3.75 V

59. Advanced Example (i) Below shows a series circuit, calculate
(6)V2 , the potential difference across the resistor R2.
By R2 = V2/I2
9W = V2 / 0.25 A
V2 =2.25 V

60. Advanced Example (ii) Below shows a parallel circuit, calculate
(1)the total resistance of the circuit,
(2)the current, I, passed through the circuit,
(3)the p.d. across the resistor R1,
(4) the p.d. across the resistor R2.
(5)the current I1 passed through the resistor, and
(6)the current I2 passed through the resistor R2 .

61. ` (ii) Below shows a parallel circuit, calculate By
(1)the total resistance of the circuit, By
1/RT = 1/15 + 1/9 = 8/45
RT = W
Exact value

62. Advanced Example (ii) Below shows a parallel circuit, calculate
(2)the current, I, passed through the circuit,
By RT = VT/IT
5.625 W = 6V / I
I = A
I = 1.07 A

63. Advanced Example (ii) Below shows a parallel circuit, calculate
(3)the p.d. across the resistor R1,
VT = V1=V2=V3= ….
Then p.d. = 6V

64. Advanced Example (i) Below shows a parallel circuit, calculate
(4) the p.d. across the resistor R2.
VT = V1=V2=V3= ….
Then p.d. = 6V

65. Advanced Example (i) Below shows a parallel circuit, calculate
(5)the current I1 passed through the resistor, and
By R1 = V1/I1
15W = 6V / I1
I1 = 0.4 A

66. Advanced Example (i) Below shows a parallel circuit, calculate
(6)the current I2 passed through the resistor R2 .
By R2 = V2/I2
9W = 6V / I2
I2 = A

67. Mixed Problem

68. Mixed Problem The total resistance of the circuit
The current, I, passed through the circuit
The current I3 passed through the resistor R3
The p.d. across the resistor R1
The p.d. across the resistor R2
The current I1 passed through the resistor R1 and
The current I2 passed through the resistor R2

69. The total resistance of the circuit

70. The total resistance of the circuit
R = 1/(1/R1 + 1/R2)

71. 1) The total resistance of the circuit
RT = 1/(1/R1 + 1/R2) + R3
RT = 1/(1/12 + 1/4) + 9= = 12 W

72. 2) The current, I By RT = VT/IT , 12W = 6V / I I = 0.5 A
RT = 1/(1/R1 + 1/R2) + R3
By RT = VT/IT , 12W = 6V / I
I = 0.5 A

73. 3) The current, I3
By I3 = I
I3 = 0.5 A

74. The p.d. across R3
By R3 = V3/I3
9W = V3 /0.5A V3 = 4.5 V

75. 4) The p.d. across R1 By 6V = V + V3 , V = 6V - 4.5 V = 1.5 V
R = 1/(1/R1 + 1/R2) V V3
By 6V = V + V3 , V = 6V V = 1.5 V

76. 5) The p.d. across R2 The p.d. across R2 is also 1.5 V
R = 1/(1/R1 + 1/R2) V V3
The p.d. across R2 is also 1.5 V

77. 6) The Current I1
By R1 = V1/I1
12W = 1.5V/I1
I1 = A

78. 7) The Current I2
By R2 = V2/I2
4W = 1.5V/I2 I2 = A

79. Thank you. What should you do?
The End
Thank you. What should you do?