Topic 5: Electric currents 5.1 Electric potential difference, current and resistance

Electric potential difference

5.1.1 Define electric potential difference.

5.1.2 Determine the change in potential energy when a charge moves between two points at different potentials.

5.1.3 Define the electron-volt.

5.1.4 Solve problems involving electric potential difference.

Topic 5: Electric currents5.1 Electric potential difference, current and resistance

Define electric potential difference.

Recall from previous classes that there are two types of charge: (+) and (-).

You may also recall the old charge rule that says like charges repel and unlike charges attract.

The basic charge that we will work with is the electron. The electron is a negative charge.

The algebraic symbol for charge is q and it is measured in coulombs (C).

An electron and a proton have the following charges:

Thus qe = -e and qp = e.


qe = -1.610-19 C electron, proton and elementary

chargeqp = +1.610-19 C

electronproton

elementary e = +1.610-19 C



EXAMPLE:

How many electrons are in a charge of q = -15 C?SOLUTION:

q = (-1510-6 C)(1 electron/-1.610-19 C) q = 9.41013 electrons

EXAMPLE:

What is the net charge of a mole of sulfur S+2 ions?

SOLUTION:

(1 mol S+2)(2 e/S+2)(6.021023/mol)(1.610-19 C/e)

q = 1.9105 C.

FYI

The word electron/s is often abbreviated e-.


We begin our study of electric potential difference by looking at a chemical cell.

To make a chemical cell, or a battery, you can begin with a container of weak acid, and two electrodes made of different metals.

Different metals dissolve in acids at different rates.

When a metal dissolves, it enters the acid as a positive ion, leaving behind an electron.

We call the weak acid the electrolyte. We call the least negative metal the anode. We call the most negative metal the cathode.


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Think of a chemical cell as a device that converts chemical energy into electrical energy.

If we connect conductors and a light bulb to the anode and the cathode we see that electrons begin to flow in what’s called an electric current.


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Why do the electrons run from cathode to anode in the external circuit instead of the reverse?

They try to escape the more highly negative cathode for the less negative anode.


Each time an electron leaves the cathode, the acid creates another in the cathode.

Each time an electron enters the anode, the acid neutralizes an electron in the anode.

This process is maintained until one of the metals or the electrolyte is used up completely.

Thus, the battery will continue to supply electrons to the external circuit until its chemical energy runs out.


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PRACTICE: A current isn’t really just one electron moving through a circuit. It is in reality more like a chain of them, each one shoving the next through the circuit. Recalling the rule of charges which says “like charges repel, and unlike charges attract,” explain why the current flows from cathode to anode in terms of the force of repulsion.Consider electrons at A and B.Both electrons feel a repulsive force at their respective electrodes.



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PRACTICE: A current isn’t really just one electron moving through a circuit. It is in reality more like a chain of them, each one shoving the next through the circuit. Recalling the rule of charges which says “like charges repel, and unlike charges attract,” explain why the current flows from cathode to anode in terms of force of repulsion.But A’s is bigger than B’s because there are more negatives repelling at A than B.A wins the “push of war.”



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PRACTICE: Consider an electron at the point A and an electron at the point B. Which one has more potential energy?The electron at B has already traveled through the circuit and is returning to the battery so it can do no more work on the circuit (like lighting up the bulb).The electron at A still has the full circuit to travel and can thus still do work (like lighting up the bulb).Thus the electron at A has more potential energy.



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Back in the days of Ben Franklin scientists understood that there were two types of electricity: positive and negative.

What they didn’t know was which one was actually free to travel through a circuit.

The influential Ben Franklin guessed wrongly that it was the positive charge.

Thus, his vision of our chemical cell looked like the one on the next slide, not the one on this slide.


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Positive charge flow is called conventional current and is still used in circuit analysis.


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PRACTICE: Consider positive charges at points A and B. Which one has more potential energy?The charge at B has already traveled through the circuit and is returning to the battery so it can do no more work on the circuit (like lighting up the bulb).The charge at A still has the full circuit to travel and can thus still do work (like lighting up the bulb).Thus the positive charge at A has more potential energy.



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We define the electric potential difference ∆V as the amount of work done in moving a positive charge q from a point of lower potential energy to a point of higher potential energy.

The units for electric potential difference are volts (V) or, as can be seen from the formula, (J C-1).


FYI

Electric potential difference is often abbreviated p.d.

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∆V = ∆EP/q potential difference


The instrument used to measure electric potential difference is called the voltmeter.


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Anode (+) Cathode (-)01.600.0 This cell has

a potential difference of 1.6 V.

RED = high

potential

BLACK = low

potential

EXAMPLE:

A battery’s voltage is measured as shown.

(a) What is the uncertainty in it’s measurement?

SOLUTION:

For measuring devices always use the place value of the least significant digit as your raw uncertainty.

For this voltmeter the voltage is measured to the tenths place so we give the raw uncertainty a value of ∆V = 0.1 V.

Solve problems involving electric potential difference.


tenths place

09.400.0

Solve problems involving electric potential difference.

EXAMPLE:

A battery’s voltage is measured as shown.

(b) What is the fractional error in this measurement?

SOLUTION:

Fractional error is just ∆V/V. For this particular measurement we then have∆V/V = 0.1/9.4 = 0.011 (or 1.1%).

FYI

When using a voltmeter the red lead is placed at the point of highest potential.


09.4


Think of a battery as an engine that uses chemical energy to take positive charges and move them from low to high potential within the battery so that they can do work outside the battery in the external circuit.

To simplify the drawing of circuits schematic symbols have been developed. For this circuit we have:


Electrolyte


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As an analogy to gravitational potential energy, think of the chemical cell as an elevator, powered by a chemical engine.

Inside the cell, positive charges at low potential are raised through chemical energy to high potential.

Outside the cell, positive charges at high potential are released into the external circuit to do their electrical work.FYI

Remember that there is a whole chain of electrons moving at once. One pushes the next…

external circuit

chemical cell


(-)

(+)

Determine the change in potential energy when a charge moves between two points at different potentials.


EXAMPLE:

200 C of charge is brought from an electric potential of 2.0 V to an electric potential of 14 V through use of a car battery. What is the change in potential energy of the charge?

SOLUTION:

From ∆V = ∆EP/q we see that ∆EP = q∆V. Thus

∆EP = q(V – V0)

∆EP = (20010-6)(14 – 2)

∆EP = 0.0024 J.FYI

Note the increase rather than the decrease.



PRACTICE:

An elementary charge is carried from the cathode to the anode within a chemical cell whose potential difference is 1.64 V. What is the change in potential energy of the charge?

SOLUTION:

We see that q = +1.610-19 C and ∆V = 1.64 V.

From ∆V = ∆EP/q we see that ∆EP = q∆V so that

∆EP = (1.610-19)(1.64)

∆EP = 2.610-19 J.

Determine the change in potential energy when a charge moves between two points at different potentials.EXAMPLE: An evacuated glass tube is shown with a potential difference of 1000 V between the two plates. A thin-wire filament that emits electrons when heated up is located behind a hole in the negative plate.

(a) Find the change in potential energy of the electron as it passes through the hole and travels between the plates.

SOLUTION:

We see that q = -1.610-19 C and ∆V = +1000 V.

From ∆EP = q∆V we see that

∆EP = (-1.610-19)(+1000)

∆EP = -1.610-16 J.


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(b) Find the change in kinetic energy of the electron as it passes through the hole and travels between the plates.

SOLUTION:

From ∆EK + ∆EP = 0 we get

∆EK = -∆EP = +1.610-16 J.

Note that the loss in potential energy is equal to the gain in kinetic energy.


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(c) Given that the mass of an electron is 9.1110-31 kg, find the speed of the electron if its speed at the negative plate is close to zero.

SOLUTION:

From the previous problem ∆EK = +1.610-16 J.

Thus 1.610-16 = EK,f – EK,0 = (1/2)mv2 - (1/2)mu2

1.610-16 = (1/2)(9.1110-31)v2

v = 1.9107 m s-1.


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Determine the change in potential energy when a charge moves between two points at different potentials.PRACTICE: Show that for a general free charge q having mass m and an initial speed of zero, qV = (1/2)mv2, where V is the potential difference the charge accelerates through as it loses potential energy.

From the previous problems recall ∆EK = -∆EP and ∆EP = -qV.

Thus ∆EK = -∆EP = -(-qV)

qV = EK,f – EK,0

qV = (1/2)mv2 - (1/2)mu2

qV = (1/2)mv2.


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FYIA loss of EP is always a gain in EK.


We can tailor the previous formula to fit the case where q = e, the elementary charge. This formula is in the physics data booklet.


Ve = (1/2)mv2 kinetic energy of an accelerated elementary charge

PRACTICE: Find the final velocity of a proton that has been accelerated from rest through a potential difference of 125 V. mp = 1.6710-27 kg.

Ve = (1/2)mv2

125(1.610-19) = (1/2)(1.6710-27)v2

v = 1.5105 m s-1.

Define the electron-volt.

We define a new unit of energy called the electron volt (eV) as the energy gained by a single electron in moving through a potential difference of exactly 1 V.

From ∆EP = q∆V we see that ∆EP = (1.610-19)(1) or


1 eV = 1.610-19 J electron volt

PRACTICE:

From the last example we discovered that ∆EP = 2.610-19 J. Convert this energy into electron- volts.

SOLUTION:

Since 1 eV = 1.610-19 J we have ∆EP = (2.610-19 J)(1 eV/1.610-19 J)

∆EP = 1.6 eV.

Electric current and resistance

5.1.5 Define electric current.

5.1.6 Define resistance.

5.1.7 Apply the resistance equation.

5.1.8 State Ohm’s law.

5.1.9 Compare ohmic and non-ohmic behavior.

5.1.10 Derive and apply expressions for electrical power dissipation in resistors.

5.1.11 Solve problems involving potential difference, current and resistance.


Define electric current.

Electric current I is the time rate ∆t at which charge ∆q moves past a particular point in a circuit.

From the formula it should be clear that current is measured in coulombs per second (C s-1) which is called an ampere (A).


I = ∆q/∆t electric currentor I = q/t

PRACTICE:

Many houses have 20-amp(ere) service. How many electrons per second is this?

SOLUTION:

20 A is 20 C per s so we only need to know how many electrons are in 1 C. (20 C)(1 e- / 1.610-19 C) = 1.31020 e-.



A simple model may help clarify how electrons move through a circuit.

Think of conductors as pipes that hold electrons.The chemical cell pushes an electron out of the (-) side. This electron in turn pushes the next, and so on, because like charges repel.

This electromotive force is transfered simultaneously to every charge in the circuit.


EXAMPLE: Suppose each (-) represents an electron. Find the current in electrons per second and then convert it to amperes.

SOLUTION:

Choose a reference point and start a timer.The rate is about 7 electrons each 12 s or I = 7/12 = 0.58 e-

s-1.

I = (0.58 e- s-1)(1.610-19 C/1 e-) = 9.310-20 A.




Define electric current.EXAMPLE: Explain why when a wire is cut current stops everywhere and does not leak into the air.

SOLUTION:

In order to free an e- from a conductor takes quite a bit of energy. This is why when a metal is conducting electricity and you stand near it you don’t get electrocuted by e- jumping off of the metal. This is also why when you cut the wire e- do not leak out into the surrounding environment.

Finally, if the chain is broken the push stops, so the current stops everywhere.


Define electric current.EXAMPLE: Sketch in the electron current and the conventional current in each of the circuits.

SOLUTION:

Electron current originates at the cathode (-).

Conventional current originates at the anode (+).

Finally, as we just discussed, the current is not just a single charge. Rather it is a whole chain of charge that is moving as one stream.


lamp switch

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lamp switch

(+) (-)

electron current

conventional currentFYI

All currents start and stop as one in a circuit.


Consider the simple circuit of battery, lamp, and wire. As we have just learned, in order to make the bulb light the circuit has to be complete.

To measure the voltage of the circuit we mere- ly connect the voltmeter while the circuit is in operation.


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celllampvoltmeterin parallel


To measure the current of the circuit we must break the circuit and insert the ammeter so that it intercepts all of the electrons that normally travel through the circuit.


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ammeterin series

Define resistance.

If you have ever looked inside an electronic device you have no doubt seen what a resistor looks like.

A resistor’s working part is usually made of carbon, which is a semiconductor.

The less carbon there is the harder it is for current to flow through the resistor.

As the animation shows, carbon is spiraled away to cut down the cross-sectional area, thereby increasing the resistance to whatever value is desired.


Define resistance.

Some very precise resistors are made of wire and are called wire-wound resistors.

And some resistors can be made to vary their resistance by tapping them at various places. These are called variable resistors and potentiometers.

Thermistors are temperature- dependent resistors, changing their resistance in response to their temperature.

Light-dependent resistors (LDRs) change their resistance in response to light intensity.


Define resistance.

Electrical resistance R is a measure of how hard it is for current to flow through a material. Resistance is measured in ohms () using an ohm-meter.

FYIA reading of 0.L on an ohmeter means “overload”. The resistance is too high to record with the meter.


330.4 0.L 0.0

This resistor has a resistance of 330.4 .

Define resistance.

The different types of resistors have different symbols.


fixed-value resistor

variable resistor

potentiometer2 leads

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3 leads

Define resistance.

The different types of resistors have different symbols.


thermisterLight-

dependent resistor

(LDR)2 leads

2 leads

As temperature increases resistance decreases As brightness

increases resistance decreases

Define resistance.

The precise definition of resistance is as follows:

The resistance R of a material is the ratio of the potential difference V across the material to the current I flowing through the material.

The units from the formula are (V A-1) which are called ohms ().


R = V/I electric resistance

PRACTICE:

A fixed resistor has a current of 2.5 mA when it has a 6.0 V potential difference across it. What is its resistance?

R = V/I = 6/2.510-3 = 2400 (or 2.4 k).

Apply the resistance equation.

In order to better understand electrical resistance, consider two identical milk shakes.

In the first experiment the straws have the same diameter, but different lengths.

In the second experiment the straws have the same length, but different diameters.

Thus we see that R L/A.


R L R 1/A

Resistance is a measure of how hard it is to pass something

through a material.


Of course conductors and resistors are not hollow like straws. And instead of milk shake current we have electrical current.

Even through solids R L/A.But R also depends on the material through which the electricity is flowing.

For example the exact same size of copper will have much less resistance than the carbon.

It turns out that the complete resistance equation looks like this:

The Greek rho () is the resistivity of the particular material the resistor is made from.


COPPER

L

A

CARBON A

L

R = L/A resistance equation


Here is a table of resistivities and thermal coefficients. Note that from the resistance equation resistivity is measured in ( m).


Resistivities and Temperature Coefficients for Various Materials at 20C(m) (C -1)Material (m) (C -1)Material

Conductors Semiconductors

2.8210-8 4.2910-3Aluminum Carbon 360010-8 -5.010-4

1.7010-8 6.8010-3Copper Germanium 4.610-1 -5.010-2

1010-8 6.5110-3Iron Silicon 2.5102 -7.010-2

98.410-8 0.8910-3Mercury

10010-8 0.4010-3Nichrome Nonconductors

7.810-8 6.010-3Nickel Glass 1012

1010-8 3.9310-3Platinum Rubber 1015

1.5910-8 6.110-3Silver Wood 1010

5.610-8 4.510-3Tungsten


From the previous table it should be clear that resistance depends on temperature. The IBO does not require us to explore this facet of resistivity.


PRACTICE: What is the resistance of a 0.00200 meter long carbon core resistor having a core diameter of 0.000100 m? Assume the temperature is 20 C.r = d/2 = 0.0001/2 = 0.00005 m.A = r2 = (0.00005)2 = 7.85410-9 m2.

From the table = 360010-8 m.

R = L/A = (360010-8)(0.002)/7.85410-9 = 9.17 .

A

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State Ohm’s law.

The German Ohm studied resistance of materials in the 1800s and in 1826 stated:

“Provided the temperature is kept constant, the resistance of very many materials is constant over a wide range of applied potential differences, and therefore the potential difference is proportional to the current.”

In formula form Ohm’s law looks like this:


V I Ohm’s lawor V/I = CONST or V = IR

FYIDon’t write Ohm’s as R = V/I because this just the definition of resistance.

Compare ohmic and non-ohmic behavior.

A material is considered ohmic if it behaves according to Ohm’s law. In other words the resistance stays constant as the voltage changes.


EXAMPLE: Three graphs for three electrical components are shown. Label appropriate graphs with the following labels: ohmic, non-ohmic, R increasing, R decreasing, and R constant.

SOLUTION:

First label the resistance dependence.

R = V/I so R is just the slope of the V vs. I graph.

Ohm’s law states the R is constant. Thus only one graph is ohmic.

V

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EXAMPLE: The graph shows the applied voltage V vs. resulting current I through a tungsten filament lamp.

Find R when I = 0.5 mA and 1.5 mA. Is this filament ohmic or non-ohmic?

SOLUTION:

At 0.5 mA: V = 0.08 V R = V/I = 0.08/0.510-3 = 160 . At 1.5 mA: V = 0.6 V R = V/I = 0.6/1.510-3 = 400 .Since R is not constant the filament is non-ohmic. (Or observe that V vs. I is non-linear.)



EXAMPLE: The graph shows the applied voltage V vs. resulting current I through a tungsten filament lamp.

Give a plausible explanation as to why the lamp filament is non-ohmic.

SOLUTION:

The temperature coefficient for tungsten is positive, as it is for all of the conductors in the table.

Therefore, the hotter the filament the higher R.But the more current, the hotter and brighter a lamp filament burns.

Thus, the bigger the I the bigger the R.

Solve problems involving potential difference, current and resistance.


EXAMPLE: The graph shows the I-V characteristic for a non- ohmic component. In the graph sketch in the I-V characteristic for a 40 ohmic component in the range of 0.0 V to 6.0 V.

SOLUTION:

”Ohmic” means V = IR and R is constant (and graph is linear).

Thus V = 40I or I = V/40.

If V = 0, I = 0/40 = 0.0.If V = 6, I = 6/40 = 0.15 A.But 0.15 A = 150 mA.

Derive and apply expressions for electrical power dissipation in resistors.

Way back when, we defined the electrical potential difference as V = ∆EP/q.

We rewrote this as ∆EP = Vq.

Recall that power P is energy production (or usage or delivery) per unit time. Thus for electrical power we have:

P = ∆EP/t

= Vq/t

= V(q/t)

P = VI.


FYIThe IBO expects you to be able to derive P = VI just like this.


Thus we can put it all together.


PRACTICE: Recall that the definition of resistance is R = V/I. Using this relationship together with the one we just derived (P = VI) derive the following two formulas:

(a) P = I2R and (b) P = V2/R.

SOLUTION:

From R = V/I we get V = IR.For (a): P = IV = I(IR) = I2R.

From R = V/I we get I = V/R.For (b): P = IV = (V/R)(V) = V2/R.

P = VI = I2R = V2/R electrical power



PRACTICE:

The graph shows the V-I characteristics of a tungsten filament lamp.

What is its power consump- tion at I = 0.5 mA and at I = 1.5 mA?

SOLUTION:

At 0.5 mA, V = 0.08 V.P = IV = (0.510-3)(0.08) = 4.010-5 W.

At 1.5 mA, V = 0.6 V.P = IV = (1.510-3)(0.6) = 9.010-4 W.



PRACTICE: A battery is connected to a 25-W lamp as shown.

What is the lamp’s resistance?

SOLUTION:

Suppose we connect a voltmeter to the circuit.

We know P = 25 W. We know V = 1.4 V.From P = V2/R we get

R = V2/P = 1.42/25 = 0.078 .

01.400.0

Topic 5: Electric currents 5.1 Electric potential difference, current and resistance

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Transcript of Topic 5: Electric currents 5.1 Electric potential difference, current and resistance