The Electric Battery

Electric Currents

See howstuffworks. Another example of conservation of energy.

Electric Current

Connecting wires (and/or lamps, etc.) to a battery permits electric charge to flow. The current that passes any point in the wire in a time t is defined by

OSE: I = Q/t,

where Q is the amount of charge passing the point. One ampere of current is one coulomb per second.

We use this symbol for a battery (the short line is negative):

+ -

Here’s a really simple circuit.

Don’t try it at home! (Why not?)

The current is in the direction of flow of positive charge. Opposite to the flow of electrons, which are usually the charge carriers.

Because electric charge is conserved, the current at any point in a circuit is the same as the current at any other point in the same circuit at that instant in time.

+ -

current

+ -

current electrons

An electron flowing from – to + gives rise to the same “conventional current” as a proton flowing from + to -. “Conventional” means our convention is always to consider the effect of + charges.

Example: A steady current of 2.5 A flows in a wire for 4.0 min.(a) How much charged passed through any point in the circuit?

Q

It

Q I t

ΔQ = 2.5 4.0 60 Minutes are not SI units!

(b) How electrons would this be?

total chargenumber of electrons =

charge of an electron

ΔQ = 600 C

21

-19

600 Cnumber of electrons = = 3.8×10 electrons

1.6×10 Celectron

“This is a piece of cake so far!”

Don’t worry, it gets “better” later.

Ohm’s Law

It is experimentally observed that the current flowing through a wire depends on the potential difference (voltage) causing the flow, and the resistance of the wire to the flow of electricity. The observed relationship can be written

OSE: V = I R,

and this is often called Ohm’s law.

Ohm’s law is not “fundamental.” It is not really a “law” in the sense of Newton’s Laws. It only works for conductors, and some things that conduct electricity do not obey Ohm’s law.The unit of resistance is the ohm, and is equal to 1 Volt / 1 Ampere.

Example

A small flashlight bulb draws 300 mA from its 1.5 V battery.(a) What is the resistance of the bulb?

V = I R

R = 1.5 / 300x10-3

R = V / I

R = 5.0

(b) If the voltage dropped to 1.2 V, how would the current change? V = I R

I = V / R

I = 1.2 / 5.0

I = 0.24 A

(“If it’s this easy now, does that mean I’ll pay later?”)

Every circuit component has resistance. This is the symbol we use for a “resistor:”

All wires have resistance. Obviously, for efficiently carrying a current, we want a wire having a low resistance. In idealized problems, we will consider wire resistance to be zero.Lamps, batteries, and other devices in circuits have resistance.

Resistors are often intentionally used in circuits. The picture shows a strip of five resistors (you tear off the paper and solder the resistors into circuits).

The little bands of color on the resistors have meaning. Here are a couple of handy web links:http://www.dannyg.com/examples/res2/resistor.htmhttp://xtronics.com/kits/rcode.htm

You light me up.http://jersey.uoregon.edu/vlab/Voltage/http://jersey.uoregon.edu/vlab/Voltage/volt1.html

Resistivity

where is a “constant” called the resistivity of the wire material, L is the wire length, and A its cross-sectional area. This makes sense: a longer wire or higher-resistivity wire should have a greater resistance. A larger area means more “space” for electrons to get through, hence lower resistance.

It is also experimentally observed that the resistance of a metal wire is well-described by

OSE: R = L / A,

Suppose you want to connect your stereo to remote speakers.(a) If each wire must be 20 m long, what diameter copper wire should you use to make the resistance 0.10 per wire.

Resistivities range from roughly 10-8 ·m for copper wire to 1015 ·m for hard rubber. That’s an incredible range of 23 orders of magnitude, and doesn’t even include superconductors (we might talk about them some time). Example

R = L / A

A = L / R

A = (d/2)2 geometry!

(d/2)2 = L / R

(d/2)2 = L / R

d/2= ( L / R )½ don’t skip steps!

d = 2 ( L / R )½

d = 2 [ (1.68x10-8) (20) / (0.1) ]½

In the spirit of not skipping steps, you are welcome to show all units!

d = 0.0021 m = 2.1 mm

(b) If the current to each speaker is 4.0 A, what is the voltage drop across each wire?

V = I R

V = (4.0) (0.10)

V = 0.4 V

Example

A wire of resistance R is stretched uniformly until it is twice its original length. What happens to its resistance? Hint: the volume of wire material stays the same.

Hint: R = L / A.

Resistivity depends on temperature—providing an important research tool—but we won’t explore that.

Electric Power

Last week we defined power as the work done by a force divided by the time it took to do the work.

PF = WF / t

We put a “bar” above the PF to indicate it is really an average power. We had probably better do that again, hadn’t we.

OSE: FF

WP = .

t

We’d better use the same definition this semester! So we will.

We focus here on the interpretation that power is energy transformed per time, instead of work by a force per time.

However, we begin with the work aspect. We know the work done in moving a charge through a potential difference: Wif = PEif= q Vif

The power, which is the work per time transformed by an electric device when a charge Q moves through a potential difference in a time t, is

P = PEif / t = q Vif / t

OSE: P = q Vif / t

I’m a bit bothered by the way ’s seem to fly in and out.

P = q V / t = (q / t) V.

Let’s get lazy and drop the in front of the V, but keep in the back of our heads the understanding that we are talking about potential difference. Then

P = I V .

But wait! A while ago, we wrote I = Q/t. The in front of the Q and t refer to the net flow of charge in some amount of time. That’s just what we have here in q / t, so

But “everybody else” uses P = IV so it is probably good to follow the crowd.

why isn’t this an OSE…

Also, using Ohm’s “law” V=IR, we can write P = I2R = V2/R. It is not clear if these should be OSE’s, because they are derived from OSE’s. To simplify matters, I’ll make the following an OSE:

OSE: P = IV = I2R = V2/R.

Truth in advertising?* Your power company doesn’t sell you power. It sells energy. Energy is power times time, so a kilowatt-hour (what you buy from your energy company) is an amount of energy.

*Just getting your attention. Not really complaining.

Example

An electric heater draws 18.0 A on a 120 V line. How much power does it use and how much does it cost per 30 day month if it operates 3.0 h per day and the electric company charges 10.5 cents per kWh. For simplicity assume the current flows steadily in one direction.What the heck is a kWh?

What’s the meaning of this assumption about the current?

We’ll get to the second question in a minute. The current in your household wiring doesn’t flow in one direction, but because we haven’t talked about current other than a steady flow of charge, we’ll make the assumption (which doesn’t wreck the calculation.)

Remember your steps to solving problems? The first step is to think. Maybe the step should really be titled “figure out what kind of problem it is.”

The problem asks for power. Maybe that identifies the problem type! Next maybe we had better go to our OSE’s and see what they say about power.

.2 2FF i f

WP = P = q ΔV / Δt P = IV = I R = V /R

t

The first two equations are general and tell us power has something to do with energy transformations per time. The last set of three is specific to current flowing in a circuit.We’re given current and voltage. It should be clear how to calculate power.

P = I V

P = (15.0 A) (120 V)

P = 1800 W = 1.8 kW

How much does it cost. We are given cost per kWh and we calculated k above. What is this kWh (and why the odd capitalization?

(1 kW) (1 h) = (1000 W) (3600 s)

= (1000 J/s) (3600 s)

= 3.6 x 106 J

So a kWh is a “funny” unit of energy. K (kilo) and h (hours) are lowercase, and W (James Watt) is uppercase.

To get the cost, find the total energy and multiply by the cost per energy “unit.”

Cost = (1.8 kW) (30 days) (used 3 h/day) ($0.105/kWh)

Cost = (1.8 kW) (30 days) (used 3 h/day) ($0.105/kWh)

I wonder how much energy we actually used?

Pavg = WF /t

Energy Transformed = WF = Pavg t

Pavg and the P with the bar above it both mean average power.

Cost = $17

Energy Transformed = (1800 J/s) (30 d) (3 h used/d) (3600 s/h)

Energy Transformed = 583,200,000 Joules used

OK, “used” is not an SI unit, but I stuck it in there to help me understand. And joules don’t come by the ton.

That’s a ton of joules! Good bargain for $17. That’s about 34,000,000 joules per dollar.

One last quibble. You know from energy conservation that you don’t “use up” energy. You just transform it from one form to another. Unfortunately, in the first semester we skipped the (important) section where you should have studied this.

Pavg time t

Example

A typical lightning bolt can transfer 109 J of energy across a potential difference of perhaps 5x107 V during a time interval of 0.2 s. Estimate the total amount of charge transferred, the current and the average power over the 0.2 s. learn about lightning

at howstuffworksWhat kind of a problem is this?

You are given energy, potential difference, and time.

You need to calculate charge transferred, current, and average power. OSE’s for current and power are “obvious:”

FF

WΔQI = P =

Δt t

We could calculate power right now, but let’s do this in the order requested. Besides, we can’t get current without Q, charge transferred.

Can you think of anything where we have energy, potential, and time combined?

PEif= q Vif

We need to think in terms of energy transformations rather than work done by forces. The equation above tells us the potential energy stored in clouds can be transferred to the ground (at a different potential) by moving charge from cloud to ground. We are given energy transferred and potential difference, so we can calculate q.Could I think of the cloud-earth system as a giant capacitor which stores energy?

You could, except our capacitor OSE U=QV/2 assumes the same charge on both plates; untrue here.

Etransferred= Qtransferred Vif

Continuing with our energy transformation idea, our OSE becomes

Qtransferred = Etransferred / Vif

Qtransferred = 109 J / 5x107 V

That’s a lot of charge (remember, our typical charges are 10-6 C.

Qtransferred = 20 C

Once we have the charge transferred, the current is easy.

ΔQI =

Δt

20 CI = = 100 A

0.2 s

Average power is just the total energy transferred divided by the total time.

FF

WP =

t

transferredEP =

tMaybe I should make this into an OSE

910 JP =

0.2 s

9P = 5×10 W

P = 5 GW

Power in Household Circuits

I’ll do a little demo on short circuits and electrical safety…

Never replace a properly-rated fuse with a higher rated one (“higher rated” means handles more current.

Alternating Current

This topic is surprisingly complex. We will not explore it here.

We have implicitly been discussing circuits using batteries as a power source. The current is called direct current (DC), because it flows in one direction and does not vary with time.

If you want to do that at home, use the really fine steel wool. The coarse stuff doesn’t work as well. And remember—adult supervision!

Microscopic View of Electric Current

Here’s the “classical” picture of the mechanism for the resistance of a metal:

+ + +

+

+

+ +

+ + ++

-

E electron “drift” velocity

The voltage accelerates the electron, but only until the electron collides with a + ion. Then the electron’s velocity is randomized and the acceleration process begins again.

The free electrons move with velocities on the order of 105 or 106 m/s (fast!) but the directions are random. The “drift velocity” represents the average velocity, or the net “drift” of the electrons. It is on the order of 0.05 mm/s.

So how come when I flip a light switch, the light comes on “right away?”

Predictions made by this theory are typically off by a factor or 10 or so, but it was the best we could do before quantum mechanics.