Download electric current

Electric Current
chapters 22 23
Electric Current


Whenever electric charges of like signs
move, an electric current is said to exist
The current is the rate at which the
charge flows through this surface


Look at the charges flowing perpendicularly
to a surface of area A
Q
I 
t
The SI unit of current is Ampere (A)

1 A = 1 coulomb/sec
What makes the charge
flow?


….a potential difference in energy
levels
….similar to


the way heat flows from hot surface
to cold surface
The way objects fall from location of
high PE to location of low PE

Potential difference
charge
flows when there is a difference in
voltage across the ends of a
conductor
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
Technically, negative charges leave the (-) end of the
battery and return to the (+) end BUT
the current is said to flow the opposite way : (+) to ()
Blame Ben Franklin for this definition for the direction
More about current


Ex: 5 amps = movement of 5
coulombs of charge in one second
or about 31.25 billion billion
electrons each second.
The net charge of any wire is zero
because just as many electrons
leave the wire that enter it.
Charges form a conga
line..



Everything already contains
electrons so…
A circuit doesn’t provide electrons
but..
It does provide all the
requirements for charges to get
energized and start moving and
doing work
Sources of potential
difference




A voltage source provides a sustained
potential difference to allow charges to
flow.
Examples: dry cells, wet cells, i.e.
batteries, and generators/powerplants
Dry cells and wet cells are the result of
a chemical reaction.
Generators/powerplants convert
mechanical energy to electrical energy.
Voltage



This is like the electric pressure pushing the
electrons.
Example: The potential difference
between two slots in a household outlet
is ~120 volts or 120 joules/coulomb of
charge.
Voltage is the potential energy /unit of
charge that pushes the electrons.
Analogy for electricity







Charges….water molecules
Electric current…flow of water current
Voltage source…pump
Voltage…pressure
Wires…pipes
Do work of lighting a bulb…do work of
turning a water wheel to grind grain
Resistance to electricity…rocks, gunk in
pipes
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
V
R
I
Resistance, cont

Units of resistance are ohms (Ω)



1Ω=1V/A
Resistance in a circuit arises due to
collisions between the electrons
carrying the current with the fixed
atoms inside the conductor
Mostly depends on material,
thickness and length of wire

Electric resistance can slow
down the flow of electrons.

Thick wires have less resistance.

Short wires have less resistance.
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
Units for Ohm’s law


V=IR
Voltage = current x resistance
equatio
n
V=
I
R
variable
voltage
current
resistance
unit
volt
ampere
ohm
Unit
symbol
v
A
Ω
Basic
units
Joules/
coulomb
Coulomb Not
/sec
applicable
Ohm’s Law and shock

Example of 1.5 volt battery

Touching battery and exposed Christmas
bulb wires in class produces no shock or
sensation at all


Resistance of dry hands was very high,
resulting current very low
If you touched battery and wires with
wet hands…


Might feel something sensation
Resistance of wet hands is lower, resulting
current is higher
Birds on exposed wire,
no shock??


IF the bird has one foot on our original wire, and the
other foot on, for example, the ground or on a different
wire with less voltage, THEN the bird would be
electrocuted. The electricity would pass through the bird
on its way from the high-voltage line to the lowervoltage line or the ground.
BUT as long as both of the bird’s feet are on the same
wire (or wires of the same voltage), the bird is safe. The
current doesn’t have anywhere else to go, so the
electricity won’t pass through the bird–it stays on the
path of least resistance, the wire.
Person dangling from high
tension wire?




It’s like being part of a huge
VandeGraf machine!
Don’t make contact with ‘ground’!
Turn power off, then rescue…
http://www.dailymail.co.uk/news/article-2244478/Paraglider-tangled-high-powered-powerlines-man-dices-death-spending-hours-dangling-100m-ground.html
Squirrels on the other hand…


When one end of a squirrel
touches ground or a transformer
and the other end is still on an
electrified wire…….
The circuit is complete!
Surprising reason for
some power outages!
Charge Carrier Motion in a
Conductor

The zig-zag black
line represents the
motion of 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
Electrons in a Circuit



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
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
Meters in a Circuit –
Voltmeter

A voltmeter is used to measure voltage
(potential difference)

Connects to the two ends of the bulb
Fun fact: 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
Superconductor, final


Good conductors
do not necessarily
exhibit
superconductivity
One application is
superconducting
magnets
Electrical Energy and
Power

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
Energy Transfer in the
Circuit


Consider the
circuit shown
Imagine a
quantity of
positive charge,
Q, moving
around the circuit
from point A back
to point A
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
QV

The chemical energy of the battery
decreases by the same amount
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
Electrical Energy and
Power, cont



The rate at which the energy is used is
the power
Q

V  I V
t
Electrical appliances are often described
in terms of power required or watts
Power determines the brightness of a
lamp

More voltage, more current or both
What does the electric
company sell?


Electrons? No, all matter has that
Electricity? Power?


Be more specific
Energy!


Our units - Joules
Their units – kWhr

Kilowatt x hr
Electrical Energy and
Power, final

The SI unit of power is Watt (W)


Watt = joules/sec
What’s a kilowatt-hour ?


amount of energy used in one hour at
a rate of 1 kilojoule/sec (1kW)
1 kWh = 3.60 x 106 J
Using electrical energy




Produced at power plant
To your house
converted to any variety of
energies by appliances
Ex: transfer to heat energy


Toaster, iron, curling iron, heater
100 % efficient or not?
Calculate cost of operation

Ex: 1500 watt microwave oven

Estimate hours of operation/ day


Determine rate charged by electric co
per kwhr


~ 1.5 hours (must be in hours!)
From electric bill- $0.18/kWhr or 18cents
Cost = kW x hrs x rate


= (1500/1000) x (1.5) x ($0.18)
= $0.405 or about 41cents per day
Chapter 35

Electric Circuits
What makes a good
circuit?



Must use all conductors
Must include voltage source
Complete path




From (+) to (-) end of voltage source
No short circuits
No breaks, gaps
All electrical devices working
What is a circuit?

Path along which the charges flow
Electrical devices, voltage sources,
wiring can be connected in a
variety of ways
• Series – single loop or path
• Parallel – circuit has branches
Observations about
Series circuits


Same current throughout the single loop
Voltage at battery = sum of voltage drops
across the bulbs = V1 +V2,etc




Add more bulbs, bulbs get dimmer, less power
One bulb is unscrewed, all bulbs go out –
disadvantage
total resistance increases as more bulbs are
added to the circuit
 R total = R1 + R2 + etc.
Advantage – uses less power, batteries last
longer
Observations about
parallel circuits

Same voltage throughout the circuit


Current at the battery=sum of currents in all
branches




Voltage at battery = v1 = v2 = etc.
I total = I1 + I2 + etc.
One bulb unscrewed, rest stay on – advantage
Add more bulbs, total resistance decreases
Add more bulbs, total current increases

Disadvantage – can overload the circuit causing
overheating of wires

Can protect the circuit by including circuit breaker or fuse in
the main branch
Resistors/resistances in
circuits

In series



Total or equivalent R = R1 +R2 + etc
total increases with more resistors
In parallel

Total or equivalent R


1 / R total = 1/R1 + 1/R2 + etc.
Total decreases with more resistors
Sample problem

Three resistors are connected in series. If placed in a circuit
with a 12-volt power supply. Determine the equivalent
resistance, the total circuit current, and the voltage drop
across and current at each resistor.
V 1 = I1 •
R1 V1 =
(0.31579
A) • (11 )
V1 =
3.47 V
V 2 = I2 •
R2 V2 =
(0.31579
A) • (7 )
V2 =
2.21 V
V 3 = I3 •
R3 V3 =
(0.31579
A) • (20 )
V3 =
6.32 V
The analysis begins by using the resistance values for the individual resistors in order to determine
the equivalent resistance of the circuit.
Req = R1 + R2 + R3 = 11 + 7 + 20 = 38
Now that the equivalent resistance is known, the current through the battery can be determined
using Ohm's law equation. In using the Ohm's law equation ( V = I • R) to determine the current
in the circuit, it is important to use the battery voltage for V and the equivalent resistance for R.
The calculation is shown here:
Itot = Vbattery / Req = (12 V) / (38 ) = 0.31579 Amp
The 1.5 Amp value for current is the current at the battery location. For a series circuit with no
branching locations, the current is everywhere the same. The current at the battery location is the
same as the current at each resistor location. Subsequently, the 0.316 Amp (rounded) is the value
of I1, I2, and I3.
Ibattery = I1 = I2 = I3 = 0.316 Amp (rounded)
There are three values left to be determined - the voltage drops across each of the individual
resistors. Ohm's law is used once more to determine the voltage drops for each resistor - it is
simply the product of the current at each resistor (calculated above as 0.31579 Amp) and the
resistance of each resistor (given in the problem statement). The calculations are shown
below.
As a check of the accuracy of the mathematics performed, it is wise to see if the calculated values
satisfy the principle that the sum of the voltage drops for each individual resistor is equal to the
voltage rating of the battery. In other words, is Vbattery = V1 + V2 + V3 ?
Is Vbattery = V1 + V2 + V3 ?
Is 12 V = 3.47 V + 2.21 V + 6.32 V ?
Is 12 V = 12 V?
Yes!!
Sample problem parallel

Three resistors are connected in parallel. If placed in a
circuit with a 12-volt power supply. Determine the
equivalent resistance, the total circuit current, and the
voltage drop across and current in each resistor.
I1 = V1 / R1
I1 = (12 V)
/ (11 )
I1 = 1.091
Amp
V2 = V 2 /
R2 I2 = (12
V) / (7 )
I2 = 1.714
Amp
V3 = V 3 /
R3 I3 = (12
V) / (20 )
I3 = 0.600
Amp
The analysis begins by using the resistance values for the individual resistors
in order to determine the equivalent resistance of the circuit.
1 / Req = 1 / R1 + 1 / R2 + 1 / R3 = (1 / 11 ) + (1 / 7 ) + (1 / 20 )
1 / Req = 0.283766 -1
Req = 1 / (0.283766 -1)
Req = 3.52
(rounded from 3.524027 )
Now that the equivalent resistance is known, the current in the battery can be determined using the Ohm's law equation.
In using the Ohm's law equation ( V = I • R) to determine the current in the circuit,
it is important to use the battery voltage for V and the equivalent resistance for R. The calculation is shown here:
Itot = Vbattery / Req = (12 V) / (3.524027 )
Itot = 3.41 Amp
(rounded from 3.4051948 Amp)
The 12 V battery voltage represents the gain in electric potential by a charge as it passes through the battery.
The charge loses this same amount of electric potential for any given pass through the external circuit.
That is, the voltage drop across each one of the three resistors is the same as the voltage gained in the battery:
V battery = V 1 = V 2 = V 3 = 12 V
There are three values left to be determined - the current in each of the individual resistors.
Ohm's law is used once more to determine the current values for each resistor
- it is simply the voltage drop across each resistor (12 Volts) divided by the resistance of each resistor
(given in the problem statement). The calculations are shown below.
As a check of the accuracy of the mathematics performed, it is wise to see if the calculated values satisfy the principle that
the sum of the current values for each individual resistor is equal to the total current in the circuit (or in the battery).
In other words, is Itot = I1 + I2 + I3 ?
Is Itot = I1 + I2 + I3 ?
Is 3.405 Amp = 1.091 Amp + 1.714 Amp + 0.600 Amp ?
Is 3.405 Amp = 3.405 Amp?
Yes!!
Parallel circuits and
overloading



More and more appliances operated in parallel
in one circuit…
 Washer, dryer, fridge, etc. at same time?
 Too many Christmas lights?
 Blowdryer, hair straightener, hot curlers all
at once?
 Causes current in main branch to increase,
wires can overheat
Circuit breaker ‘pops’, opens a switch when
specified maximum current is reached in main
branch
Ex: circuit breakers for 15A, 20A, etc.