Chapter 17Current and Resistance

Bright Storm on Electric Current

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Current

• Practical applications were based on static electricity.

• A steady source of electric current allowed scientists to learn how to control the flow of electric charges in circuits.

Introduction

Electric Current

• The current is the rate at which the charge flows through a surface.– Look at the charges flowing perpendicularly through a

surface of area A.

• The SI unit of current is Ampere (A)– 1 A = 1 C/s

Section 17.1

Instantaneous Current

• The instantaneous current is the limit of the average current as the time interval goes to zero:

• If there is a steady current, the average and instantaneous currents will be the same.

• SI unit: A

Section 17.1

Electric Current, Cont.• The direction of the current is the

direction positive charge would flow.– This is known as conventional

current direction.• In a common conductor, such as

copper, the current is due to the motion of the negatively charged electrons.

• It is common to refer to a moving charge as a mobile charge carrier.– A charge carrier can be positive

or negative.

Section 17.1

EXAMPLE 17.1 Turn On the Light

Goal Apply the concept of current. Problem The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C. Find (a) the average current in the lightbulb and (b) the number of electrons that pass through the filament in 5.00 s. Strategy Substitute into Equation 17.1a for part (a), then multiply the answer by the time given in part (b) to get the total charge that passes in that time. The total charge equals the number N of electrons going through the circuit times the charge per electron.

SOLUTION

(a) Compute the average current in the lightbulb. Substitute the charge and time to find the current:

Iav = ΔQ

= 1.67 C

= 0.835 A Δt 2.00 s

(b) Find the number of electrons passing through the filament in 5.00 s.

The total number N of electrons times the charge per electron equals the total charge, IavΔt.

(1) Nq = IavΔt

Substitute and solve for N.

N(1.60 10-19 C/electron) = (0.835 A)(5.00 s)

N = 2.61 1019 electrons

LEARN MORE

Remarks In developing the solution, it was important to use units to ensure the correctness of equations such as Equation (1). Notice the enormous number of electrons passing through a given point in a typical circuit. Question In comparing wires of identical shape and carrying the same current but made of different metals, is the drift velocity in each wire inversely proportional to the number density of charge carriers?

Yes. To have the same current from k times as many carriers, it takes k

times the drift speed. Yes. To have the same current from k times as

many carriers, it takes 1/k times the drift speed. No. The drift velocity

is directly proportional to the number density of carriers. No. The drift velocity is the same for all conductors.

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Current and Drift Speed

• Charged particles move through a conductor of cross-sectional area A.

• n is the number of charge carriers per unit volume.

• n A Δx is the total number of charge carriers.

Section 17.2

Current and Drift Speed, Cont.

• The total charge is the number of carriers times the charge per carrier, q– ΔQ = (n A Δx) q

• The drift speed, vd, is the speed at which the carriers move.– vd = Δx/ Δt

• Rewritten: ΔQ = (n A vd Δt) q

• Finally, current, I = ΔQ/Δt = nqvdA

Section 17.2

Current and Drift Speed, Final

• If the conductor is isolated, the electrons undergo random motion.

• When an electric field is set up in the conductor, it creates an electric force on the electrons and hence a current.

Section 17.2

Charge Carrier Motion in a Conductor

• The zig-zag black line represents the motion of a charge carrier in a conductor.– The net drift speed is small.

• The sharp changes in direction are due to collisions.

• The net motion of electrons is opposite the direction of the electric field.

Section 17.2

Electrons in a Circuit

• Assume you close a switch to turn on a light.• The electrons do not travel from the switch to the

bulb.• The electrons already in the bulb move in response

to the electric field set up in the completed circuit.• A battery in a circuit supplies energy (not charges) to

the circuit.

Electrons in a Circuit, Cont.

• The drift speed is much smaller than the average speed between collisions.

• When a circuit is completed, the electric field travels with a speed close to the speed of light.

• Although the drift speed is on the order of 10-4 m/s, the effect of the electric field is felt on the order of 108 m/s.

Section 17.2

Circuits

• A circuit is a closed path of some sort around which current circulates.

• A circuit diagram can be used to represent the circuit.

• Quantities of interest are generally current and potential difference.

Section 17.3

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Meters in a Circuit – Ammeter

• An ammeter is used to measure current.– In line with the bulb, all the charge passing through the

bulb also must pass through the meter.

Section 17.3

Meters in a Circuit – Voltmeter

• A voltmeter is used to measure voltage (potential difference).– Connects to the two contacts of the bulb

Section 17.3

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Georg Simon Ohm

• 1787 – 1854• Formulated the concept

of resistance• Discovered the

proportionality between current and voltages

Section 17.4

Bright Storm on Ohm’s Law

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Resistance

• In a conductor, the voltage applied across the ends of the conductor is proportional to the current through the conductor.

• The constant of proportionality is the resistance of the conductor.

Section 17.4

Resistance, Cont.

• Units of resistance are ohms (Ω)– 1 Ω = 1 V / A

• Resistance in a circuit arises due to collisions between the electrons carrying the current with the fixed atoms inside the conductor.

Section 17.4

Ohm’s Law

• Experiments show that for many materials, including most metals, the resistance remains constant over a wide range of applied voltages or currents.

• This statement has become known as Ohm’s Law.– ΔV = I R

• Ohm’s Law is an empirical relationship that is valid only for certain materials.– Materials that obey Ohm’s Law are said to be ohmic.

Section 17.4

Ohm’s Law, Cont.• An ohmic device• The resistance is constant

over a wide range of voltages.

• The relationship between current and voltage is linear.

• The slope is related to the resistance.

Section 17.4

Ohm’s Law, Final

• Non-ohmic materials are those whose resistance changes with voltage or current.

• The current-voltage relationship is nonlinear.

• A diode is a common example of a non-ohmic device.

Section 17.4

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Resistivity

• The resistance of an ohmic conductor is proportional to its length, L, and inversely proportional to its cross-sectional area, A.

– ρ is the constant of proportionality and is called the resistivity of the material.

– See table 17.1

Section 17.4

EXAMPLE 17.3 The Resistance of Nichrome Wire

Goal Combine the concept of resistivity with Ohm's law. Problem (a) Calculate the resistance per unit length of a 22-gauge Nichrome wire of radius 0.321 mm. (b) If a potential difference of 10.0 V is maintained across a 1.00-m length of the Nichrome wire, what is the current in the wire? (c) The wire is melted down and recast with twice its original length. Find the new resistance RN as a multiple of the old resistance RO. Strategy Part (a) requires substitution into Equation 17.5 after calculating the cross-sectional area, whereas part (b) is a matter of substitution into Ohm's law. Part (c) requires some algebra. The idea is to take the expression for the new resistance and substitute expressions for N and AN, the new length and cross-sectional area, in terms of the old length and cross-section. For the area substitution, remember that the volumes of the old and new wires are the same.

SOLUTION

(a) Calculate the resistance per unit length. Find the cross-sectional area of the wire: A = πr2 = π(3.210 10-4 m)2 = 3.24 10-7 m2

Obtain the resistivity of Nichrome, solve for R/ , and substitute:

R = ρ

= 1.5 10-6 Ω · m

= 4.6 Ω/m A 3.24 10-7 m2

(b) Find the current in a 1.00-m segment of the wire if the potential difference across it is 10.0 V:

Substitute given values into Ohm's law:

I = ΔV

= 10.0 V

= 2.2 A R 4.6 Ω

(c) If the wire is melted down and recast with twice its original length, find the new resistance as a multiple of the old.

Find the new area AN in terms of the old area AO, using the fact the volume doesn't change and N = 2 O.

VN = VO → AN N = AO O → AN = AO( O/ N)

AN = AO( O/2 O) = AO/2

Substitute to find the new resistance:

RN = ρ N

= ρ(2 O)

= 4 ρ O

= 4RO AN (AO/2) AO

LEARN MORE

Remarks The resistivity of Nichrome is about 100 times that of copper, a typical good conductor. Therefore, a copper wire of the same radius would have a resistance per unit length of only 0.052 Ω/m, and a 1.00-m length of copper wire of the same radius would carry the same current (2.2 A) with an applied voltage of only 0.115 V. Because of its resistance to oxidation, Nichrome is often used for heating elements in toasters, irons, and electric heaters. Question Would replacing the Nichrome with copper wire of the same size result in a higher current or lower current?

the same current lower current higher current

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Temperature Variation of Resistivity

• For most metals, resistivity increases with increasing temperature.– With a higher temperature, the metal’s

constituent atoms vibrate with increasing amplitude.

– The electrons find it more difficult to pass through the atoms.

Section 17.5

Temperature Variation of Resistivity, Cont.

• For most metals, resistivity increases approximately linearly with temperature over a limited temperature range.

– ρ is the resistivity at some temperature T– ρo is the resistivity at some reference temperature To

• To is usually taken to be 20° C– is the temperature coefficient of resistivity

Section 17.5

Temperature Variation of Resistance

• Since the resistance of a conductor with uniform cross sectional area is proportional to the resistivity, you can find the effect of temperature on resistance.

Section 17.5

EXAMPLE 17.4 A Platinum Resistance Thermometer

Goal Apply the temperature dependence of resistance. Problem A resistance thermometer, which measures temperature by measuring the change in resistance of a conductor, is made of platinum and has a resistance of 50.0 Ω at 20.0°C. (a) When the device is immersed in a vessel containing melting indium, its resistance increases to 76.8 Ω. From this information, find the melting point of indium. (b) The indium is heated further until it reaches a temperature of 235°C. What is the ratio of the new current in the platinum to the current Imp at the melting point?

Strategy For part (a), solve for T - T0 and get α for platinum, substituting known quantities. For part (b), use Ohm's law.

SOLUTION

(a) Find the melting point of indium: Solve for T - T0.

T - T0 = R - R0

= 76.8 Ω - 50.0 Ω

= 137°C αR0 [3.92 10-3 (°C)-1][50.0 Ω]

Substitute T0 = 20.0°C and obtain the melting point of indium: T = 157°C

(b) Find the ratio of the new current to the old when the temperature rises from 157°C to 235°C.

Write the equation with R0 and T0 replaced by Rmp and Tmp, the resistance and temperature at the melting point.

R = Rmp[1 + α(T-Tmp)]

According to Ohm's law, R = ΔV/I and Rmp = ΔV/Imp. Substitute these expressions into the equation.

ΔV = ΔV

[1 +α(T - Tmp)] I Imp Cancel the voltage differences, invert the two expressions, and then divide both sides by Imp:

I =

1

Imp 1 +α(T - Tmp)

Substitute T = 235°C, Tmp = 157°C, and the value for α, obtaining the desired ratio:

I = 0.766

Imp

LEARN MORE

Remarks As the temperature rises, both the rms speed of the electrons in the metal and the resistance increase. Question What happens to the drift speed of the electrons as the temperature rises? (Select all that apply.)

It becomes smaller, so that the resistance increases and current

decreases. It becomes larger, so that the resistance decreases and current

increases. It remains the same. It becomes smaller because the atoms of the metal vibrate more strongly and interfere more with electron

drift motion. It becomes larger because the electrons become more energetic and move faster.

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Electrical Energy in a Circuit

• In a circuit, as a charge moves through the battery, the electrical potential energy of the system is increased by ΔQΔV.– The chemical potential energy of the battery decreases by

the same amount.• As the charge moves through a resistor, it loses this

potential energy during collisions with atoms in the resistor.– The temperature of the resistor will increase.

Section 17.6

Energy Transfer in the Circuit

• Consider the circuit shown.

• Imagine a quantity of positive charge, DQ, moving around the circuit from point A back to point A.

Section 17.6

Energy Transfer in the Circuit, Cont.

• Point A is the reference point.– It is grounded and its potential is taken to be zero.

• As the charge moves through the battery from A to B, the potential energy of the system increases by DQDV.– The chemical energy of the battery decreases by

the same amount.

Section 17.6

Energy Transfer in the Circuit, Final

• As the charge moves through the resistor, from C to D, it loses energy in collisions with the atoms of the resistor.

• The energy is transferred to internal energy.• When the charge returns to A, the net result is that

some chemical energy of the battery has been delivered to the resistor and caused its temperature to rise.

Section 17.6

Bright Storm on Power

Power

• In a conductor carrying a current, the electric potential of the charges is continually decreasing.

• Positive charges move from regions of high potential to regions of low potential.

• ΔUcharges = q ΔV is negative– Often only the magnitude is desired

• The power delivered to the circuit element is the energy divided by the elapsed time.

Section 17.1

Electrical Energy and Power, Cont.

• The rate at which the energy is lost is the power.

• From Ohm’s Law, alternate forms of power are

Section 17.6

Electrical Energy and Power, Final

• The SI unit of power is Watt (W).– I must be in Amperes, R in ohms and DV in Volts

• The unit of energy used by electric companies is the kilowatt-hour.– This is defined in terms of the unit of power and

the amount of time it is supplied.– 1 kWh = 3.60 x 106 J

Section 17.6

EXAMPLE 17.5 The Cost of Lighting Up Your Life

Goal Apply the electric power concept and calculate the cost of power usage using kilowatt-hours. Problem A circuit provides a maximum current of 20.0 A at an operating voltage of 1.20 102 V. (a) How many 75 W bulbs can operate with this voltage source? (b) At $0.120 per kilowatt-hour, how much does it cost to operate these bulbs for 8.00 h? Strategy Find the necessary power with = IΔV then divide by 75.0 W per bulb to get the total number of bulbs. To find the cost, convert power to kilowatts and multiply by the number of hours, then multiply by the cost per kilowatt-hour.

SOLUTION

(a) Find the number of bulbs that can be lighted. Substitute to get the total power.

total = IΔV = (20.0 A)(1.20 102 V) = 2.40 103 W

Divide the total power by the power per bulb to get the number of bulbs.

Number of bulbs = total

= 2.40 103 W

= 32.0 bulb 75.0 W

(b) Calculate the cost of this electricity for an 8.00-h day.

Find the energy in kilowatt-hours.

Energy = t = (2.40 103 W)( 1.00 kW )(8.00 h) 1.00 103 W

= 19.2 kWh

Multiply the energy by the cost per kilowatt-hour.

Cost = (19.2 kWh)($0.12/kwH) = $2.30

LEARN MORE

Remarks This amount of energy might correspond to what a small office uses in a working day, taking into account all power requirements (not just lighting). In general, resistive devices can have variable power output, depending on how the circuit is wired. Here, power outputs were specified, so such considerations were unnecessary. Question Considering how hot the parts of an incandescent light bulb get during operation, guess what fraction of the energy emitted by an incandescent light bulb is in the form of visible light.

80% 50% 10%

EXAMPLE 17.6 The Power Converted by an Electric Heater

Goal Calculate an electrical power output and link to its effect on the environment through the first law of thermodynamics. Problem An electric heater is operated by applying a potential difference of 50.0 V to a Nichrome wire of total resistance 8.00 Ω. (a) Find the current carried by the wire and the power rating of the heater. (b) Using this heater, how long would it take to heat 2.50 103 moles of diatomic gas (e.g., a mixture of oxygen and nitrogen, or air) from a chilly 10.0°C to 25.0°C? Take the molar specific heat at constant volume of air to be (5/2)R. Strategy For part (a), find the current with Ohms law and substitute into the expression for power. Part (b) is an isovolumetric process, so the thermal energy provided by the heater all goes into the change in internal energy, ΔU. Calculate this quantity using the first law of thermodynamics and divide by the power to get the time.

SOLUTION

(a) Compute the current and power output. Apply Ohms law to get the current.

I = ΔV

= 50.0 V

= 6.25 A R 8.00 Ω

Substitute to find the power. = I2R = (6.25 A)2(8.00 Ω) = 313 W

(b) How long does it take to heat the gas?

Calculate the thermal energy transfer from the first law. Note that W = 0 because the volume doesn't change

Q = ΔU = nCvΔT = (2.50 103 mol)((5/2) · 8.31 J/mol · K)(298 K − 283 K)

= 7.79 105 J

Divide the thermal energy by the power to get the time.

t = Q

= 7.79 105 J

= 2.49 105 s

313 W

LEARN MORE

Remarks The number of moles of gas given here is approximately what would be found in a bedroom. Warming the air with this space heater requires only about 40 minutes. The calculation, however, doesn't take into account conduction losses. Recall that a 20-cm-thick concrete wall permitted the loss of more than 2 megajoules an hour by conduction! Question If the heater wire is replaced by a wire with lower resistance, the time required to heat the gas is:

increased. unchanged. decreased.

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Superconductors• A class of materials and

compounds whose resistances fall to virtually zero below a certain temperature, TC

– TC is called the critical temperature

• The graph is the same as a normal metal above TC, but suddenly drops to zero at TC

Section 17.7

Superconductors, Cont.

• The value of TC is sensitive to – Chemical composition– Pressure– Crystalline structure

• Once a current is set up in a superconductor, it persists without any applied voltage.– Since R = 0

Section 17.7

Superconductor Timeline• 1911

– Superconductivity discovered by H. Kamerlingh Onnes• 1986

– High temperature superconductivity discovered by Bednorz and Müller

– Superconductivity near 30 K• 1987

– Superconductivity at 96 K and 105 K• Current

– Superconductivity at 150 K– More materials and more applications

Section 17.7

Superconductor, Final

• Good conductors do not necessarily exhibit superconductivity.

• One application is the construction of superconducting magnets.

Section 17.7

Electrical Activity in the Heart

• Every action involving the body’s muscles is initiated by electrical activity.

• Voltage pulses cause the heart to beat.• These voltage pulses are large enough to be

detected by equipment attached to the skin.

Section 17.8

Operation of the Heart• The sinoatrial (SA) node

initiates the heartbeat.• The electrical impulses

cause the right and left artial muscles to contract.

• When the impulse reaches the atrioventricular (AV) node, the muscles of the atria begin to relax.

• The ventricles relax and the cycle repeats.

Section 17.8

Electrocardiogram (EKG)• A normal EKG• P occurs just before the

atria begin to contract.• The QRS pulse occurs in the

ventricles just before they contract.

• The T pulse occurs when the cells in the ventricles begin to recover.

Section 17.8

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