1. Current and Resistance
Chapter 26
Current and Resistance

3. Current and the Motion of Charges

4. Electric current I: steady flow of charge through a cross sectional area of a conductor.
I measured in Ampere (A)-fundamental unit in SI.
Differentially:

5. vd-drift velocity
Notations:
n is the number density (number of free charges per unit of volume)
q is the charge flowing through the conductor.
ΔV is the volume that includes the charges that will pass through A in the time Δt.
Definition: Current density J: electric current (constant) per unit of area.

6. Current Density General:
The magnitude of current density, J, is equal to the current per unit area through any element of cross section. It has the same direction as the velocity of the moving charges if they are positive and the opposite direction if they are negative.
If the current is uniform across the surface and parallel to dA, then J is also uniform
and parallel to dA.
Here, A is the total area of the surface.
The SI unit for current density is the ampere per square meter (A/m2).

7. Current Density:
Figure 26-4 shows how current density can be represented with a similar set of lines, which we can call streamlines.
The current, which is toward the right, makes a transition from the wider conductor at the left to the narrower conductor at the right. Since charge is conserved during the transition, the amount of charge and thus the amount of current cannot change.
However, the current density changes—it is greater in the narrower conductor.

8. Example Drift Speed I = nqvdA
A Cu wire has a radius of mm. Calculate vd for I=1A, assuming one free electron per atom. Density Cu = (8.83g/cm3), Molar mass = 63.5 g/mole.
I = nqvdA
n = N/V = #of moles x NA/V = [(m/M) x NA]/V = (m/V)x(NA/M)
Number Density n = number density of Cu atoms:
n = ρNA/M= [(8.83g/cm3)(6.02 x 1023atoms/mole)]/63.5 g/mole= =8.47 x 1022 atoms/cm3 = 8.47 x 1028 atoms/m3
q=e
A=πr2

9. In a certain particle accelerator, a current of 0
In a certain particle accelerator, a current of 0.5 mA is carried by a 5 MeV proton beam that has a radius of 1.5 mm (1eV = 1.6 x J, mp= 1.6 x 10-27kg).
a) Find the number density n of the protons in the beam.
1. I = nqvdA
2. Find the speed from kinetic energy:
3. Solve for speed:
4. Substitute to calculate n:

10. b) If the beam hits a target, how many protons hit the target in 1s?
1. The number N of protons that hit the target is related to the total charge that hits the target by the relationship: ΔQ = Ne
2. Use the current I to find the charge: ΔQ= It
3. The number of protons is then:

11. Example, Current Density, Uniform and Nonuniform:

12. Example, Current Density, Uniform and Nonuniform, cont.:

13. Example, In a current, the conduction electrons move very slowly.:

14. Resistance and Ohm’s Law

15. Va – Vb = EΔL
Electric charges (I) move in the direction of the drop in potential.
Resistance is a property of the conductor. Depends on material and geometry.

16. Resistance
Resistance (R): opposition to the flow of the electric current. Caused by collisions b/w electrons and the atoms of the metal.
Measured in ohms Ω.
Depends upon type of material (resistivity ρ), proportional to length L, inversely proportional to thickness (cross sectional area A). Measured in Ωm
R = ρL/A

17. Resistance, ResistivityL A E j
Increase the length, flow of electrons impeded (more collisions)
Increase the cross sectional area, flow facilitated (more space).
The structure of this relation is identical to heat flow through materials … think of a window for an intuitive example
How thick?
How big?
What’s it made of? or

18. Resistance Continuation
R (mainly because of resistivity ρ), is proportional to the temperature. At low temperatures conductors exhibit low resistance (superconductors).
Causes electric energy to turn into heat.

19. Ohm’s Law V=RI R I V Experimental: Vary applied voltage V.
Measure current I
Ratio remains constant-property of material V I R V I
slope = R or
V=RI

24. Since: Current Density and Resistivity:
If the streamlines representing the current density are uniform throughout the wire, the electric field, E, and the current density, J, will be constant for all points within the wire.

25. High conductivity is favorable to the current.
Conductivity and Resistivity:
The SI unit for r is W.m.
The conductivity s of a material is the reciprocal of its resistivity:
High conductivity is favorable to the current.

26. ConcepTest 17.1 Connect the Battery
Which is the correct way to light the lightbulb with the battery?
4) all are correct
5) none are correct 1) 2) 3)

27. ConcepTest 17.1 Connect the Battery
Which is the correct way to light the lightbulb with the battery?
4) all are correct
5) none are correct 1) 2) 3)
Current can only flow if there is a continuous connection from the negative terminal through the bulb to the positive terminal. This is only the case for Fig. (3).

28. ConcepTest 17.2 Ohm’s Law
You double the voltage across a certain conductor and you observe the current increases three times. What can you conclude?
1) Ohm’s Law is obeyed since the current still increases when V increases
2) Ohm’s Law is not obeyed
3) this has nothing to do with Ohm’s Law

29. Follow-up: Where could this situation occur?
ConcepTest 17.2 Ohm’s Law
You double the voltage across a certain conductor and you observe the current increases three times. What can you conclude?
1) Ohm’s Law is obeyed since the current still increases when V increases
2) Ohm’s Law is not obeyed
3) this has nothing to do with Ohm’s Law
Ohm’s Law, V = I R, states that the relationship between voltage and current is linear. Thus for a conductor that obeys Ohm’s Law, the current must double when you double the voltage.
Follow-up: Where could this situation occur?

30. ConcepTest 17.3a Wires I 1) dA = 4 dB
Two wires, A and B, are made of the same metal and have equal length, but the resistance of wire A is four times the resistance of wire B. How do their diameters compare?

31. ConcepTest 17.3a Wires I 1) dA = 4 dB
Two wires, A and B, are made of the same metal and have equal length, but the resistance of wire A is four times the resistance of wire B. How do their diameters compare?
The resistance of wire A is greater because its area is less than wire B. Since area is related to radius (or diameter) squared, the diameter of A must be two times less than B.

32. ConcepTest 17.3b Wires II 1) it decreases by a factor 4
3) it stays the same
4) it increases by a factor 2
5) it increases by a factor 4
A wire of resistance R is stretched uniformly (keeping its volume constant) until it is twice its original length. What happens to the resistance?

33. ConcepTest 17.3b Wires II
1) it decreases by a factor 4
2) it decreases by a factor 2
3) it stays the same
4) it increases by a factor 2
5) it increases by a factor 4
A wire of resistance R is stretched uniformly (keeping its volume constant) until it is twice its original length. What happens to the resistance?
Keeping the volume (= area x length) constant means that if the length is doubled, the area is halved.
Since , this increases the resistance by four.

34. Energy in Electric Circuits

35. Electric Energy
The equivalent effect is of charge ΔQ moving from higher potential Va to lower potential Vb
We will make the following notation: potential difference Va-Vb =V

36. Electric Energy Also: Since: V=IR, we can also write:
The work done by the electric field to move the charge ΔQ through the potential difference V is a measure of the electric energy expanded in this case.
Therefore the electric energy is given by the formula:
Since: V=IR, we can also write:
W= I2Rt (thermal or Joule effect)
Also:

37. Electric Power
Electric Power: is the time rate at which electricity does work or provides electric energy.
P=W/t, therefore:
Power = Voltage x Current
Power is measured in Watts; Watts = Volts x Amperes
1 kilowatt = 1000 watts.
1kwh=1000 W x 3600 s= 3,600,000 J
Watts; Watts = Volts x Amperes

38. Example, Rate of Energy Dissipation in a Wire Carrying Current:

39. The electric ray has two large organs on each side of his head (each like a battery), where current passes from the lower to the upper surface of the body.

40. EMF and Batteries
Batteries (Voltage sources, sources of emf):
Purpose is to provide a constant potential difference between two points. + - V OR
The potential difference across the terminals in an open circuit is called electromotive force or “emf” ξ.

41. Simple Electric Circuit: Mechanical Analogy

42. Work, Energy, and Emf:
In any time interval dt, a charge dq passes through any cross section of the circuit shown, such as aa’. This same amount of charge must enter the emf
device at its low-potential end and leave at its high-potential end.
The emf device must do an amount of work dW on the charge dq to force it to move in this way.
We define the emf of the emf device in terms of this work:
An ideal emf device is one that has no internal resistance to the internal movement of charge from terminal to terminal. The potential difference between the terminals of an ideal emf device is exactly equal to the emf of the device.
A real emf device, such as any real battery, has internal resistance to the internal movement of charge. When a real emf device is not connected to a circuit, and thus does not have current through it, the potential difference between its terminals is equal to its emf. However, when that device has current
through it, the potential difference between its terminals differs from its emf.

43. Calculating the Current in a Single-Loop Circuit, Ideal Battery:
In the figure, let us start at point a, whose potential is Va, and mentally go clockwise around the circuit until we are back at a, keeping track of potential changes as we move.
Our starting point is at the low-potential terminal of the battery. Since the battery is ideal, the potential difference between its terminals is equal to E.
As we go along the top wire to the top end of the resistor, there is no potential change because the wire has negligible resistance.
When we pass through the resistor, however, the potential decreases by iR.
We return to point a by moving along the bottom wire. At point a, the potential is again Va. The initial potential, as modified for potential changes along the way, must be equal to our final potential; that is,

44. Calculating the Current in a Single-Loop Circuit, Ideal Battery:
For circuits that are more complex than that of the previous figure, two basic rules are usually followed for finding potential differences as we move around a loop:

45. Real Batteries: can be represented by a ideal battery of emf ξ, and a small resistance r = to the internal resistance of the battery.

47. Other Single-Loop Circuits, Internal Resistance:
The figure above shows a real battery, with internal resistance r, wired to an external resistor of resistance R. According to the potential rule,

48. When the charge ΔQ flows through the source of emf ξ, its potential energy is increased by the amount ΔQξ .
The charge than flows through the resistor were this potential energy is dissipated as thermal energy.
The rate at which energy is supplied by the source is the power output of the source:

49. Terminal Voltage of A Real Battery.
V = ξ – Ir.
Batteries are often rated in ampere-hours (Ah), which is the total charge batteries can deliver.
1Ah = (1C/s) x (3600s) = 3600 C
The total energy stored in a battery is the product of the emf and the total charge it can deliver:
W = Qξ

50. For a battery of emf ξ, and internal resistance r, what external resistance would produce maximum power delivered to resistor?
P = I2R
Set dP/dR = 0
r = R

51. ConcepTest 17.5a Lightbulbs
Two lightbulbs operate at 120 V, but one has a power rating of 25 W while the other has a power rating of 100 W. Which one has the greater resistance?
1) the 25 W bulb
2) the 100 W bulb
3) both have the same
4) this has nothing to do with resistance

52. ConcepTest 17.5a Lightbulbs
Two lightbulbs operate at 120 V, but one has a power rating of 25 W while the other has a power rating of 100 W. Which one has the greater resistance?
1) the 25 W bulb
2) the 100 W bulb
3) both have the same
4) this has nothing to do with resistance
Since P = V2 / R , the bulb with the lower power rating has to have the higher resistance.
Follow-up: Which one carries the greater current?

53. ConcepTest 17.5b Space Heaters I
Two space heaters in your living room are operated at 120 V. Heater 1 has twice the resistance of heater 2. Which one will give off more heat?
1) heater 1
2) heater 2
3) both equally

54. ConcepTest 17.5b Space Heaters I
Two space heaters in your living room are operated at 120 V. Heater 1 has twice the resistance of heater 2. Which one will give off more heat?
1) heater 1
2) heater 2
3) both equally
Using P = V2 / R, the heater with the smaller resistance will have the larger power output. Thus, heater 2 will give off more heat.
Follow-up: Which one carries the greater current?