1. Introducing Current and Direct Current Circuits
Chapters 22 and 23
Introducing Current
and
Direct Current Circuits

2. Current
Current is defined as the flow of positive charge (conventional current).
I = Q/t
I: current in Amperes or Amps (A) 1A = 1C/s
Q: charge in Coulombs (C)
t: time in seconds

3. Charge carriers
Current flows because of a difference in electric potential across a circuit element.
Conventional current is the flow of positive charge.
Electron current is a flow of electrons. Flows opposite conventional current.

4. Charge Carriers
In a normal electrical circuit, it is the electrons that carry the charge.
So if the protons move this way, which way does the current move?

5. Sample problem
How many electrons per hour flow past a point in a circuit if it bears 11.4 mA of direct current?
If the electrons are moving north, in which direction is the current?

6. More Current Examples
A bolt of lightning discharges 1.7 C in 3.3 x 10-4 s. What is the average current during the discharge?
The amount of charge that moves through the heating coils of a hair dryer in 50.0 s is 37.0 C. What is the current in the hair dryer?

7. Cell Cells convert chemical energy into electrical energy.
The potential difference (voltage) provided by a cell is called its electromotive force (or emf).
The emf of a cell is constant, until near the end of the cell’s useful lifetime.
The emf is not really a force. It’s one of the biggest misnomers in physics!

8. Battery A battery is composed of more than one cell in series.
The emf of a battery is the sum of the emf’s of the cells.

9. Circuit components

10. Sample problem
If a typical AA cell has an emf of 1.5 V, how much emf do 4 AA cells provide?
Draw the battery composed of these 4 cells.

11. Circuit components
Light bulb
Wire
Switch
Resistor

12. Sample problem
Draw a single loop circuit that contains a cell, a light bulb, and a switch. Name the components.

13. Series arrangement of components
Series components are put together so that all the current must go through each one
Three bulbs or resistors in series all have the same current.

14. Sample problem
Draw a circuit with a cell and two bulbs in series.

15. Parallel arrangement of components
Parallel components are put together so that the current divides, and each component gets only a fraction of it.

16. Sample problem
Draw a circuit having a cell and four bulbs. Exactly two of the bulbs must be in parallel.

17. Circuit components continued
Voltmeter
Ohmmeter
Ammeter

18. Voltmeter A voltmeter measures potential difference.
Remember a voltage drop is the amount of energy lost as a charge moves through a resistor or circuit element.
It is placed in the circuit in a parallel connection.
A voltmeter has very high resistance, and does not contribute significantly to the total resistance of the circuit.

19. Ohmmeter Measures Resistance.
Placed across resistor when no current is flowing.

20. Ammeter An ammeter measures current.
It is placed in the circuit in a series connection.
An ammeter has very low resistance, and does not contribute significantly to the total resistance of the circuit.

21. Resistivity and Ohm’s Law
Electrical Resistance and Power

22. Conductors Conduct electricity easily. Have high “conductivity”.
Have low “resistivity”.
Metals are examples.
Wires are made of conductors.

23. Insulators Don’t conduct electricity easily. Have low “conductivity”.
Have high “resistivity”.
Rubber is an example.

24. Resistors
Resistors are devices put in circuits to reduce the current flow.
Resistors are built to provide a measured amount of “resistance” to electrical flow, and thus reduce the current.

25. Sample problem
Draw a single loop circuit containing two resistors and a cell. Draw voltmeters across each component.

26. Resistor Color Code
What is the resistance of this resistor?

28. Solution
Notice that the colors on this resistor are (in order) Red, Green, Orange, and Silver.
1. The first line is the first digit — Red = 2
2. The second line is the second digit —
Green = 5
3. The third line is the multiplier — Orange = 103
4. The last line (if any) is the tolerance —
Silver = ± 10%
So the final answer would be 2.5 X 104Ω ± 10%

29. Resistors Practice

30. Resistance, R
Resistance depends on resistivity and on geometry of the resistor.
R = ρL/A
ρ: resistivity (Ω m)
L: length of resistor (m)
A: cross sectional area of resistor (m2)
Unit of resistance: Ohms (Ω)
Analogy to flowing water

31. Resistivities (ρ) of common materials
Silver x Ω m
Copper x 10-8 Ω m
Gold x 10-8 Ω m
Aluminum x 10-8 Ω m
Tungsten x 10-8 Ω m
Iron x 10-8 Ω m
Nichrome x 10-8 Ω m
Carbon x 10-8 Ω m

32. Sample problem
What is the resistivity of a substance which has a resistance of 1000 Ω if the length of the material is 4.0 cm and its cross sectional area is 0.20 cm2?

33. Sample problem
What is the resistance of 1.61 X 103 m (a mile) of copper wire if the diameter is 5.0 mm?
(resistivity of copper is 1.72 x 10-8 Ω m)

34. Ohm’s Law
Resistance in a component in a circuit causes potential to drop according to the equation: ΔV = IR
ΔV: potential drop (Volts)
I: current (Amperes)
R: resistance (Ohms)
The drop in potential occurs as electrical energy is transformed to other forms (heat, light) and work is done.

35. Sample problem
Determine the current through a 333-Ω resistor if the voltage drop across the resistor is observed to be 1.5 V.

36. Sample problem
Draw a circuit with a AA cell attached to a light bulb of resistance 4 Ω.
Determine the current through the bulb. (Calculate)

37. Power in General
Power is energy per unit time converted by electric circuit into another form of energy.
Energy is changed from electrical to thermal energy
P = W/t
P = ΔE/Δt
SI Unit:
Joules/second = Watts

38. Power in Electrical Circuits
P = I ΔV
P: power (W)
I: current (A)
ΔV: potential difference (V)
P = I2R
P = (ΔV)2/R

39. Sample problem
How much current flows through a 100-W light bulb connected to a 120 V DC power supply?
What is the resistance of the bulb?

40. The Kilowatt Hour A kilowatt hour is a unit of energy.
It is the rate of energy used (power) multiplied by a time in hours.
1 kWh = 3.6 X 106 J
Cost = Rate (cost/kWh) X Energy
Energy E = Pt
Electric companies sell energy not power.

41. Sample problem
If electrical power is 5.54 cents per kilowatt hour, how much does it cost to run a 100-W light bulb for 24 hours?

42. Sample Problem
A 9.0 V battery costs $3.00 and will deliver A for 26.0 h before it must be replaced. Calculate the cost per kWh.

43. Series and Parallel Circuit
Chapters 23
Series and Parallel Circuit

44. Resistors in circuits
Resistors can be placed in circuits in a variety of arrangements in order to control the current.
Arranging resistors in series increases the resistance and causes the current to be reduced.
Arranging resistors in parallel reduces the resistance and causes the current to increase.
The overall resistance of a specific grouping of resistors is referred to as the equivalent resistance.

45. Resistors in series
Req = R1 + R2 + R3
Req = ΣRi

46. Characteristics of a Series Circuit
Current is the same in each element of a series circuit.
IT = I1 = I2 = I3 = …
The total potential drop in a series circuit is equal to the sum of the individual potential drops.
VT = V1 + V2 + V3 + …...
The equivalent (total) resistance in a series circuit is the sum of the individual resistances.
RT = R1 + R2 + R3 + ….

47. Resistors in parallel
1/Req = 1/R1 + 1/R2 + 1/R3
1/Req = Σ(1/Ri )

48. Characteristic of a Parallel Circuit
The total current in a parallel circuit is the sum of the currents in each path or branch.
IT = I1 + I2 + I3 + ….
In a parallel circuit all the potential drops are the same.
VT = V1 = V2 = V3 = …...
The equivalent resistance in a parallel circuit is found by adding the inverses of the resistances in the circuit and taking the inverse of the sum.
Req = (R1-1 + R2-1 + R3-1)-1

49. Sample Problem
What is the equivalent resistance of a 100Ω, a 330Ω and a 530Ω resistor when these are in a series arrangement? (Draw and calculate)

50. Sample Problem
What is the equivalent resistance of a 100Ω, a 330Ω and a 530Ω resistor when these are in a parallel arrangement? (Draw and calculate)

51. Sample Problem
Draw a circuit containing, in order (1) a 3.3 V battery, (2) a 100Ω resistor, (3) a 330Ω resistor in parallel with a 100Ω resistor, (4) a 560Ω resistor, and (5) a switch.
Calculate the equivalent resistance.
Calculate the current through the cell.
Calculate the current through the 330Ω resistor.

53. Usefulness of Series and Parallel Circuits
Series circuit examples:
A voltage divider – a series circuit used to produce a voltage source of desired magnitude from a higher voltage battery.
A lightmeter (photoresistor) is a good example.
In household wiring a fuse is connected in series to protect all devices. Schematic:
For a parallel circuit, household wiring is connected in parallel.

54. Equation for a Voltage Divider

55. General rules
How does the voltage from a cell or battery get dispersed in a circuit…
when there is one component?
when there are two components in series?
when there are two components in parallel?
when there is one component in series with two components in parallel?

56. Kirchhoff’s Rules

57. Kirchhoff’s 1st Rule
Kirchhoff’s 1st rule is also called the “junction rule”.
The sum of the currents entering a junction equals the sum of the currents leaving the junction.
This rule is based upon the conservation of charge.

58. Sample problem
Find the current I4 (magnitude and direction).

59. Kirchhoff’s 2nd Rule
Kirchoff’s 2nd rule is also referred to as the “loop rule”.
The net change in electrical potential in going around one complete loop in a circuit is equal to zero.
This rule is based upon conservation of energy.

60. Sample problem
Use the loop rule to determine the potential drop across the light bulb.

61. Series – Parallel Combination Circuits
A combination circuit can be understood by analyzing them in steps.
If any resistors are connected in parallel calculate the equivalent resistance that can replace them first.
If any equivalent resistances are now connected in series, calculate a new single equivalent resistance that can replace them second.
By repeating the above steps you can reduce the circuit to a single resistance. The total current can now be calculated. The voltage drops across each resistor and the current through individual resistors can be calculated.

62. Sample Problem
Redraw the following circuit to reflect a simplified, manageable drawing.

64. Terminal Voltage and EMF
When a current is drawn from a battery, the voltage across its terminals drops below its rated EMF.
The chemical reactions in the battery cannot supply charge fast enough to maintain the full EMF.
Thus the battery is said to have an internal resistance, designated r.
Ex: Starting a car with the headlights on, the lights dim. The starter draws a large current and the battery voltage drops as a result.

65. Terminal Voltage and EMF
A real battery is then modeled as if it were a perfect emf E. in series with a resistor r.
Terminal voltage Vab
When no current is drawn from the battery, the terminal voltage equals the emf.
When a current I flows from the battery, there is an internal drop in voltage equal to Ir, thus the terminal voltage (actual voltage delivered) is
Vab = E - Ir