ch 25 notes - fcps 25 note… · title: microsoft powerpoint - ch 25 notes.ppt author: daniel.burr...

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley

PowerPoint® Lectures forUniversity Physics, Twelfth Edition

– Hugh D. Young and Roger A. Freedman

Lectures by James Pazun

Chapter 25

Current, Resistance, and Electromotive Force


Goals for Chapter 25

• To consider current and current density

• To study the intrinsic property of resistivity

• To use Ohm’s Law and study resistance and resistors

• To connect circuits (mentally, virtually, or with actual parts) and find emf

• To examine circuits and determine the energy and power in them

• To describe the conduction of metals microscopically, on an atomic scale


How to achieve these goals . . .Study this PowerPoint

Read the Chapter

Do the Homework

33, 35, 37, 41, 45, 49, 53, 61


Introduction• Electrons leave one terminal

of a battery, pass through wire of low resistance, reach a light bulb with a special calibrated resistor sealed in a bulb of inert gas, and then return to the opposite terminal of the battery.


The direction of current flow

In the absence of an external field, electrons move randomly in a conductor. If a field exists near the conductor, its force on the electron imposes a drift.

The electrons move randomly at speeds at about 106 m/s, but the drift velocity, vd, is only at about 10-4 m/s.

Then, why does the light turn on almost instantly?


Current flowingPositive charges would move with the electric field, electrons move in opposition.The motion of electrons in a wire is analogous to water coursing through a river. This fits the metaphor used earlier.We define the current, I, as the flow of positive charge in a conductor. This is called conventional current.


Current flowingWe define the current through the cross-sectional area A to be the net charge flowing through the area per unit time.Thus, if a net charge dQflows through an area in a time dt, the current Ithrough the area is I = dQ/dtThe SI unit for current is the ampere (A), which is one coulomb per second (C/s).


Current, Drift Velocity, and Current DensityWe can express current in terms of the drift velocity of the moving charges. Lets assume the moving charges are positive, that way the velocity is with the electric field.Suppose a concentration, n, of charged particles per unit volume move with the same drift velocity in a time interval dtthrough a wire with cross sectional area A, like in Fig 25.3.During that time interval the particles move a distance of vddt, which is the length of the shaded volume in Fig. 25.3. The charged particles that exit the right side of the shaded region during dt were the particles within the shaded region at the beginning of the interval.The area of the shaded region is Avddt and the number of charges inside the region is nAvddt.


Current, Drift Velocity, and Current DensityIf each particle has a charge q, the charge dQ that flows out of the right end of the shaded region is:

dQ = q(nAvddt) = nqvdAdtThe current will be:

I = dQ/dt = nqvdAThe current per unit cross-secctional area J (called the current density):

J = I/A = nqvd

The sign of the moving charges will not change I or J: q = |q|Study equation 25.5 on page 946 for the vector form of J.


Current flow requires conductors throughoutIn Figure 25.4, a negative terminal of a battery extends through wire to a bare post inside the open tube. Another open tube next to the first one also contains a bare post with wire running back to entry of a light bulb resistor. The exit of the light bulb resistor continues through wire back to the positive terminal on the battery.

If the tubes are immersed in a conducting fluid, the bulb will light. If the fluid is nonconducting, the light will remain off.

Consider Example 25.1.


Resistivity is intrinsic to a metal sample (like density is)

The resistivity of a material is the ratio of the magnitudes of the electric field to the current density.

A perfect conductor has a resistivity of zero while a perfect insulator has a resistivity of infinity.

EJ

ρ =


Words and DefinitionsConductivity—the reciprocal of resistivity

Semiconductors—materials that have resistivities that are intermediate between those of metals and those of insulators

Linear (Ohmic) Conductor—obeys Ohm’s “Law” and ρ is constant regardless of temperature

Nonlinear (Nonohmic) Conductor—material whose resistivity depends greatly on the temperature of the material

Metallic Conductor—resistivity increases with temperature


Resistivity and temperatureAs temperature increases, so does the likelihood of electrons colliding in a conductor.Over a small temperature range (0-100oC) the resistivity can be calculated by:

ρ(T) = ρo[1 + α(T – To)]Where ρo is the resistivity at a reference temperature To and ρ is the resistivity at temperature T. Alpha is the temperature coefficient of resistivity, different for each material.


25.1&25.2 SummaryCurrent is the amount of charge flowing through a specified area, per unit time. The SI unit of current is the ampere, equal to one coulomb per second. The current through an area depends on the concentration n and charge q of the charge carriers, as well as on the magnitude of their drift velocity. The current density is current per unit cross-sectional area. Current is conventionally described in terms of a flow of positive charge, even when the actual charge carriers are negative or of both signs. The resistivity of a material is the ratio of the magnitudes of electric field and current density. Good conductors have small resistivity; good insulators have large resistivity. Ohm’s law, obeyed approximately by many materials, states that ρ is a constant independent of the value of E. Resistivity usually increases with temperature; for small temperature changes this variation is represented approximately by Eq. (25.6), where α is the temperature coefficient of resistivity.


Resistance is an extensive property (like mass)For a conductor with resistivity ρ, the current density J at a point where the electric field is E is given by: E = ρJ.We are often more interested in the total current instead of J and potential difference instead of E, because those values are easily measured. For the conductor to the right, the current is I = JA and the potential difference V = EL.

If we rearrange the previous two equations and plug them into the equation above we get equation 25.8: V/L = ρI/A or V = ρL/A * IThe ratio of V to I is called the resistance R of a conductor: R = V/ISo the factor ρL/A = R. From this we can get Ohm’s “Law” V = IRR = ρL/A is true for any conductor, whether it obeys Ohm’s “Law” or not. Only when R is constant can we call V = IR Ohm’s Law


Resistors are color-coded for assembly work


Current–voltage relationships

• Ohm’s Law is linear, but current flow through other devices may not be.

• Follow Example 25.2.


Calculating resistances• Refer to Example 25.3 to see the effects of changing temperature.

• Refer to Example 25.4 to calculate the resistance of a hollow tube (unlike a normal wire). Figure 25.10 (below) illustrates this example.


25.3 Summary and HomeworkFor materials obeying Ohm’s law, the potential difference V across a particular sample of material is proportional to the current I through the material. The ratio R is the resistance of the sample. The SI unit of resistance is the ohm Ω. The resistance of a cylindrical conductor is related to its resistivity ρ length L and cross-sectional area A.

53


Electromotive forceThe figure to the right demonstrates why we need a complete circuit for a steady motion of charge in a conductor.

If a charge goes completely around a circuit it will have the same potential energy it started with. If the charge looses energy as it moves through a conductor with resistance, it must gain energy at some point.

Water will not move through a hose or pipe without some force pushing it. In a conductor, that force is called the electromotive force, EMF. EMF is not a real force, it is an energy-per-unit-charge quantity, like potential.


Ideal diagrams of “open” and “complete” circuitsThe electrostatic force is due to the potential at either end. This opposes the flow of current when the EMF source is connected to a circuit. The non-electrostatic force is the “source” of the current within the EMF source.When the charge is moved from b to a, the work done by the non electrostatic force Wn = qE and the potential energy increases by qVab.When the source is not connected to a circuit, the work done on q is qE so qE = qVab or Vab = E.


Ideal diagrams of “open” and “complete” circuitsWhen we connect a wire, the potentials at the terminals set up an electric field within the wire and current flows around the circuit. The potential difference between the ends of the wire is Vab = IR. We can now say: E = Vab = IRThis means the potential rise through the source is numerically equal to the drop in potential as the charge moves around the circuit.This is for an ideal source.


Internal resistanceReal sources do not produce a potential difference in a circuit equal to the emf as described earlier. The produced voltage is incrementally less than that of an ideal source due to internal resistance of the source, r. If this resistance behaves according to Ohm’s “Law”, r is constant and independent of the current I. As the current moves through r itexperiences a drop in potential, Ir. If the ideal emf is E, then the produced voltage Vab = E – Ir.For a real source of emf, the terminal voltage equals the emf if no current is flowing through the source.We can rewrite Ohm’s “Law” as: Vab = E – Ir = IR or I = E/(R + r)


Symbols for circuit diagrams• Shorthand symbols are in use for all wiring components. See

below.


Source in an open circuit I• Consider Conceptual Example 25.5.

• This example is illustrated in Figure 25.16.


Source in an open circuit II• Follow Example 25.6.


Voltmeters and ammeters• Follow Conceptual Example 25.7.


A source with a short circuit• Follow Example 25.8.

• Figure 25.19 (below) illustrates this example.


Potential changes around a circuitThe net change in potential energy must be zero for the entire circuit. Local differences in potential and emf do occur. See Figure 25.20 below.


25.4 Summary and Homer-workA complete circuit has a continuous current-carrying path. A complete circuit carrying a steady current must contain a source of electromotive force (emf) E. The SI unit of electromotive force is the volt (1 V). An ideal source of emf maintains a constant potential difference, independent of current through the device, but every real source of emf has some internal resistance r. The terminal potential difference Vab then depends on current.

Page 975:

33, 35, 37, 41


Energy conversion and power input to a source


Power and energy in circuits• Consider Problem-Solving Strategy 25.1.

• Refer to Example 25.9, illustrated by Figure 25.25 below.

• Refer to Example 25.10.

• Refer to Example 25.11, illustrated by Figure 25.26 below.


Summary and Homework

45, 49, 61

A circuit element with a potential difference a b abV V V− = and a current I puts energy into a circuit if the current direction is

from lower to higher potential in the device, and it takes energy out of the circuit if the current is opposite. The power

P (rate of energy transfer) is equal to the product of the potential difference and the current. A resistor always takes

electrical energy out of a circuit. (See Examples 25.9–25.11.)

abP V I= (25.17)

(general circuit element)

2

2 abab

VP V I I RR

= = = (25.18)

(power into a resistor)

ch 25 notes - fcps 25 note… · title: microsoft powerpoint - ch 25 notes.ppt author: daniel.burr...

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