Electricity and Magnetism

Problems Set 4 Current and resistance. Ohms law. Electric Power.

Ill put a summary of the main ideas and formulas that you need here at the top, simply to have it all in one place. You dont need all of this at once to do the first problems.

Current and resistance

In an isolated conductor there is random motion of the free charges inside the conductor because of thermal energy. This means, roughly, that there is as much charge moving in one

direction as the opposite. If you imagine following the motion of just one charge it looks like a random walk. The charge changes direction randomly as it collides with the stationary atoms of the lattice that make up the solid conductor. One analogue to this would be the motion of air

molecules in a room. These keep colliding with each other, the walls and the floor. On a large scale the air doesnt move anywhere but is still. On a microscopic scale theres a lot of random motion of molecules. To get the air moving as a whole in a given direction i.e. to get wind you need a pressure difference between one region and another. To get the free charges in a conductor to move, en masse, from one region towards another, i.e. to get an electric current,

you need a potential difference between these regions. So the quantity that in this analogue corresponds to pressure difference is potential difference.

If you have a potential difference between two points, say the ends of a conductor wire, theres then an electric field (or component of electric field) along the direction between the points.

There is then an electric force that pushes (positive) free charges in that direction, so we get a flow of charges i.e. an electric current. To maintain the current we need to maintain the potential

difference. In practice this is done by a voltage source that maintains a potential difference between its terminals. So to get a current flowing in the wire we connect one end to one terminal of the source and the other end to the other terminal. Note that we then dont have a static situation anymore and the electric field inside the conductor is no longer zero.

Lets look at this also from the energy perspective from the point of view of the moving charge. If a (positive) charge in the conductor wire starts at point 1 at potential V1 and potential energy qV1 and ends up at point 2 at a lower potential V2 and lower potential energy qV2 it has lost

energy qV1 - qV2. Where did this energy go?

If this motion happened in vacuum the potential energy lost in going from 1 to 2 would appear as kinetic energy gained, as the electric force kept accelerating the charge, along its motion from 1 to 2. However, youll find later that there is no gain in kinetic energy. The charge moves at the same speeds, called the drift velocity, at points 1 and 2 (assuming a wire of constant cross-section.) Instead, as the charge moves along the wire it keeps colliding with the atoms of

the lattice of the wire material. In these collisions some of its kinetic energy is transferred to the atoms of the lattice. This appears as heat (random kinetic energy or vibration) of the lattice atoms. After a collision the charge again gains kinetic energy as it is accelerated by the electric

field and then loses part of it in the next collision. The upshot of this is that on the average the charge can be taken to move with a (constant) average speed, the drift velocity, and an

(constant) average kinetic energy.

This reminds of the case of terminal velocity of a falling body in air from the Mechanics course: As the body dropped downward losing gravitational potential energy it was accelerated until it reached a terminal velocity. At his velocity the force of air resistance upward (due to collisions

with the air molecules) was equal to the gravitational force downward, due to the gravitational field, and the body then kept dropping with this constant terminal velocity.

For the electrical case, the property of the wire material that resists the flow of charges is expressed by the resistance of the wire.

So the answer to the question, where did the lost potential energy qV1 - qV2 go, is that it turned

to heat of the conductor material through collision with its atoms. How much the conductor material resists the flow of charges from a higher potential to a lower is summed up by Ohms law, which tells you how large the current I will be for a given potential difference V

V = IR

so here V = V1 - V2 is the potential difference between points 1 and 2 and R is the resistance between the points.

Current I is defined as

dtdqI .

It gives the amount of charge flowing through a cross-section of the wire. The unit of current is C/s = A (Ampere). From above we see that the unit of resistance is V/A = (Ohm).

For the resistance of a piece of wire of length l and cross-section area A we find AlR /

where is the resistivity of the wire material.

Usually the resistance of a wire of a good conductor material like copper is very small and in circuits we can often ignore the resistance of the wire. Actual resistors are still pieces of some conductor material but with much larger resistance than a small length of copper wire.

Resistors in series and in parallel

As for capacitors resistors can be connected in parallel and in series. The combination can be

thought of as a single resistor of equivalent resistance eqR . For two resistors 1R and 2R in series

we have

21 RRReq

and for two resistors in parallel

21

111

RRReq .

Power dissipated by a resistor Lets put the idea that the potential energy lost by a charge as it moves in the wire is converted to heat into mathematical form. Say an amount dq of charge moves from point 1 to point 2 and

therefore loses potential energy dqVVVdq )( 21 and this happens in a time dt . The amount of

energy turned into heat per time is then PVIdtdqV so the amount of heat per time or

heating power generated by the resistor is then P. Using Ohms law we can write this in three alternative ways

RVRIIVP /22

Kirchoffs second law Kirchoffs law is just a way of writing the conservation of energy for a circuit loop in a way that is useful for calculations. The idea is very simple but is often expressed as a complicated

sentence in textbooks. The idea is this: Say a charge moves in a circuit and completes a closed loop so that it ends up where it started from. Its potential energy at the end is then the same as in the beginning as it is at the same point it started from. So, theres no change in potential energy in going around the loop. Theres also no change in kinetic energy because the current at that point is fixed and hence also the speed of the charge at that point. The change in energy of the charge going around the loop is therefore zero, and in particular the change in potential

will be zero.

(We will idealize things a little here and assume that the conductor wire has zero resistance. This is usually perfectly ok. For two points 1 and 2 along the wire with zero resistance between them Ohms law gives 021 IRVVV or 21 VV . So the potential has the same value

everywhere in a piece of conductor wire. ) Lets do this in excruciating detail. Going through this once will probably be enough for you to get the idea and learn the method.

Look at the simple circuit in the figure. The potential difference between the terminals of the

voltage source is , so ab VV = .

In the second figure there are points a-f indicated. Also indicated is the direction in which we go around the loop, and I have here chosen clockwise. A guess for the direction of the current I is also indicated.

The method is to choose a starting point, say point a where the potential is aV , and then go

stepwise from one point to the next and for each step add the increase or decrease in potential until you come full circle back to the starting point. The final potential will then be the initial

potential aV plus all the changes for each step from one point to the next. Finally the difference

in the end point potential and initial potential will have to be zero if these are the same point.

So start at point a where the potential is aV .

Move to point b. The potential at b is then ab VV

Move to point c. Theres no change in potential as bc VV so ac VV

Move to point d through the resistor

1R . You will then go in the direction of the indicated current

I . Current in a resistor flows from higher potential to lower potential so dV is smaller than cV .

Ohms law says 1IRVV dc or 1IRVV cd .

Substituting ac VV we have

at point d 1IRVV ad .

Move to point e. Theres no change in potential so 1IRVV ae

Move to point f. We are again going from a higher potential to a lower potential so the potential

again decreases, this time by 2IR so 2IRVV ef which gives

21 IRIRVV af .

As af VV we get

021 IRIR

1. To get the scale of things calculate how many electrons flow through a cross-section of a

wire per second when the wire carries a small current of 1.0 mA.

2.

3.

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HenryTypewriterI1 = I2

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HenryTypewriterleft

HenryTypewriterdown

HenryTypewriterDecreases 3V

HenryTypewriterzero

4.

5.

6.

HenryTypewriterI3 = I4 > I1 = I2

HenryTypewriterCounterClockwise

HenryTypewriterDecrease

HenryTypewriterKinetic Energy

7.

8.

9.

10.

11.

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HenryTypewriterR1 > R2 => P1 > P2

HenryTypewriterI4 = I7 >

12.

13.

HenryTypewriterVab = Vcd

14. Suppose you have a 12 V battery and a supply of identical resistors. What you actually want is a

4 V voltage (perhaps as the input voltage for some gadget or circuit stage). How can this be accomplished?

15.

16.

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18.

19.

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HenryTypewriterP2 > P1

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20.

21.

22.

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HenryTypewriterI3 > I1 = I2 = I4 = I5

HenryTypewriterB and C go out. A brighter

23.

24.

HenryTypewriterb: more +

HenryTypewriterLeft, from high potential to low

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HenryTypewriterB out, no close path.

HenryTypewriterA dimmer. Rtotal >> => I

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26.

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HenryTypewriterI decreases

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27. A source in an open circuit

28. A source in a complete circuit

29.

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HenryTypewriterVa'b' = E - Ir = 8V

HenryTypewriterI = 0 A, V = 12V

HenryTypewriterI = E / (R+r) = 2 A

30.

31. A heater has three settings for heating power and it will dissipate 500 W, 1000 W or 1500 W, depending on the setting of the switch K. The switch K can be set so that it connects with points A, B or C. Which setting corresponds to which power? What are the resistance values for R1, R2 and R3? The voltage source provides 230 V as indicated.

HenryTypewriterVab = E - Ir = 0

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HenryTypewriterI = 6A

HenryTypewriterVa'b' = Vab = 8V

HenryTypewriterI = 2A

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HenryTypewriterA: 1500W, B: 1000W, C: 500W

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HenryTypewriterI = 0A

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HenryTypewriterVa'b' = 0V

HenryTypewriterVab = 12V