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Electric Current and Circuits (1).notebook

Electric Current

In the late 1700's Luigi Galvani and Alessandro Volta carried out experiements dealing with the contraction of frogs' leg muscles.

Volta's work led to the invention of the electric battery (voltaic cell) which produced the first steady flow of charged particles.

In honor of his work in the field of electricity, the electrical unit known as the volt, V, was named in his honor.

VoltaGalvani

http://www.energyquest.ca.gov/scientists/volta.html


The rate of flow of electric charge is called electric current.

Electric current in a wire can be defined as the amount of charge that passes through it per unit time at any point.

Electric current is measured in coulombs per second. One coulomb per second is called an ampere (amp).

Current is often measured in mA (milliamps) and μA (microamps).

I is electric current q is quantitiy of charge t is time interval

Electric Current

1 mA = 1 x 103 A 1μA = 1 x 10-6 A


Conventional Current vs. Electron Flow

+

Conventional Current vs. Electron Flow

http://www.youtube.com/watch?v=IpaEGhjpZgc&feature=related

http://www.youtube.com/watch?v=nZjMARe6APs&feature=relmfu

http://www.youtube.com/watch?v=IpaEGhjpZgc&feature=related

http://www.youtube.com/watch?v=nZjMARe6APs&feature=relmfu


Sample Problems

1. A steady current of 3.5 A flowed in a wire for 2.0 minutes. How much charge passed through the circuit?


2. A battery was charged using a current of 5.7 A. How long did it take to charge the battery if 1.2 x 105 C of charge passed through it?


3. What is the current if 2000 Na+ ions were to flow across a cell membrane in 9.8 μs? The charge on a sodium ion is the same as a proton.


Electric Circuits

An electric circuit is a closed loop or continuous path that consists of a device that will increase the potential energy of electric charges, such as

generators

batteries

photovoltaic cells

that reduces the potential energy of the charges while converting the electrical energy into a form of "useful" energy (sound, light, heat).

In order to keep the charge flowing, a potential difference must be maintained.

connected to a device (radio, lamp, toaster)

(devices that change sunlight directly into electricity)


D

q

high potential

low potential

D

q

D

Electric Potential in Circuits

A battery powered electric circuit has locations of high and low potential. Within the battery, there is an electric field established between the two terminals, directed from the positive towards the negative terminal.

Work is required to move a positive charge through the battery from the negative terminal to the positive terminal, thus increasing the potential energy of the charge. (The charge is moving against the electric field.)

It is for this reason that the positive terminal is described as the high potential terminal.

The movement of the positive charge through the wires from the positive terminal to the negative terminal would occur naturally. (No work is required to move the chargein the direction of the electric field.)

The charge loses potential energy as it moves through the wires. It is for this reason that the negative terminal is described as the low potential terminal.

This assignment of high and low potential to the terminals of the battery presumes that we are using

conventional current.

D

q

high potential


An electric circuit is nothing more than an energy conversion system.

The reaction of the chemicals inside a battery produces chemical energy that is used to do work on positive charge to move it from the low potential terminal to the high potential terminal.

Chemical energy is transformed into electric potential energy within the internal circuit (the battery).

Once at the high potential terminal, the positive charge will then move through the external circuit and do work on a light bulb, a motor, heater coils, etc., transforming its electric potential energy into useful forms for which the circuit was designed.


Charge flows all the way around the circuit. The battery pushes the charge around the circuit. As the charge passesthrough the lamp, it makes it light up.

Below is a simple electric circuit. It has a switch, a bulb and a battery. These components are connected togetherwith metal connecting wires.

A Simple Circuit

A simple switch is made of a metal lever that can join up with a metal contact. When you press the switch, the two pieces of metal touch and the current can flow through it. When you open it, this breaks the circuit.


Resistance and Ohm's Law

The relationhsip between the current,I, in a metal wire and the potential difference, V, applied to its ends, was experimentally determined by Georg Ohm.

Ohm

He found that current is directly proportional to potential difference.

I α V

This means that if you connect a wire to a 9 V battery, the current will be three times what it would be if the wire were connected to a 3 V battery.

The amount of current that flows in a circuit is also dependent on the resistance offered by the circuit.

Imagine you are walking down the hall during first period. You might meet a few people along the way, but you won't encounter much opposition.

Compare that situation to walking down the hall a few seconds after the lunch bell rings you will encounter a lot of resistance!

Current, I is inversely proportional to resistance, R.

I α 1 R

Example

http://www.youtube.com/watch?v=d8BB89y4m4g&feature=related


If we combine the two relationships, we get:

The unit of resistance is the ohm. The symbol for ohm is the Greek letter omega, Ω.

The equation is usually written as V = IR and it is referred to as "Ohm's Law."

1 Ω = 1 V A

Not all materials obey the law, but we will assume the materials invovled in

our problems do.

I = V R

I is current V is potential difference R is resistance


Connecting wires generally have very low resistance compared to the coils or filaments in some electrical devices like heaters and light bulbs.

Resistors are devices designed to have a specific resistance and are often used in electronic devices to control the amount of current that flows.

Some resistors have their resistance values written ontheir exteriors. Others have a color code that allows us to calculate their resistance.

Resistors


Fourband identification is the most commonly used for color coding resistors. It consists of four colored bands that are painted aorund the body of the resistor.

The first two numbers are the first two significant digitsof the resitance value, the third is a mulitplier and the fourth is the tolerance of the value.

Each color corresponds to a certain number, shown inthe chart below. The tolerance for a 4band resistor will be 2%, 5% or 10%.

A useful mnemonic for remembering the first ten color codes matches the first letter of the color code, by order of increasing magnitude. There are many variations:

• Bright Boys Rave Over Young Girls But Veto Getting Wed • B. B. R O Y of Great Britain has a Very Good Wife • Better Be Right Or Your Great Big Venture Goes West


Sample Problems

1. What is the resistance of a toaster if 1.10 x 102 V produces a current of 3.1 A?


2. A 4.5 V battery is connected to a bulb whose resistance is 2.5 Ω. What is the current?


3. A hair dryer draws 11.0 A when plugged into a 1.20 x 102 V line. How much charge passes through in 10.0 minutes?


Experimentally it has been found that the resistance, R, of a metal is directly proportional to its length, L, and inversely proportional to its crosssectional area, A.

R = ρL A

R resistanceρ resistivityL lengthA crosssectional area

Material Resistivity (Ωm)

silver 1.59 x 108 copper 1.68 x 108 aluminum 2.65 x 108tungsten 5.6 x 108 iron 9.71 x 108 platinum 10.6 x 108

The resistivities of some metals are listed below.

ρ is the Greek letter rho and it is a proportionality constant called resistivity that depends on

the material from whichthe wire is made.


To find the crosssectional area of the the wire,use the following formula:

where d represents the diameter of the wire


Sample Problems

1. What is the resistance of a 3.5 m length of aluminum wire 1.5 mm in diameter?


2. What is the length of a copper wire that has a cross sectional area of 3.4 x 106 m2 and a resistance of 7.1 x 102 Ω?


3. What is the radius of a 1.00 m length of tungsten wire whose resistance is 0.25 Ω?


Electric PowerElectric power measures the rate at which electric energyis transformed into another form of energy such as light or heat.

The energy transformed when a charge, Q, moves through a potential difference, V, is QV. Powercan be calculated as follows:

P = QV t

Remember that I = Q/t, so the formula becomes:

P = IVP powerI electric currentV potential difference

The unit of electric power is the watt, W.

1W = 1 J s


By substituting V/R for I, or IR for V, we can get two more equations for electric power.

P = IV P = IV

P = V (V) R

P = V2

R

P = I (IR)

P = I2R


1. Calculate the resistance of a 60.0 W bulb designed for 12 V?

Sample Problems


2. What is the maximum power consumption of a 6.0 V tape player that draws a maximum of 4.50 x 102 mA of current?


3. An 8.00 x 102 W hair dryer has a resistance of 18 Ω. What is the current throught the hair dryer?


If you look at an electric bill, you will see that you pay for energy not power. We can calculate the electric energy used by a device by multiplying power consumption by thetime the device is on.

E = PtE energy P power t time

We could write the units of energy as Ws. The electric company uses a larger unit, the kilowatthour (kWh).

1 kWh = 1000 W x 3600 s = 3.60 x 106 J

Another equation that is often used to calculateelectric energy is:

E = I2Rt


1. An electric heater draws 15.0 A on a 1.20 x 102 V line. How much does it cost to operate the heater for 30.0 days if the heater is used for 3.00 h per day and the electric company charges 10.5 cents per kWh?

Sample Problem

t = 3.00 hours per day x 30.0 days = 90.0 hours

P = IVP = 15.0 x 1.20 x 102P = 1.80 x 103 WP = 1.80 kW

Determine the total number of hours the heateris used.

Determine the amount of power that is used.

Calculate the amount of energy that was used.

E = PtE = 1.80 kW x 90.0 hE = 1.62 x 102 kWh

Now calculate cost.

Cost = (1.62 x 102 kWh)($0.105) kWh Cost = $17.00


2. A small electric furnace operating on 1.00 x 102 V, expends 2.0 kW of power.

a) What current is in the circuit?b) What is the resistance of the furnace?c) What is the cost of operation for 24 h at $0.05/kWh?


Diagramming Circuits

There are standard symbols that are used to representthe elements of a circuit. The diagram that you end up with is called a circuit schematic or circuit diagram.

We will be using the following symbols:

ReminderAlthough electricity is the flow of electrons, called electron flow, it was originally thought

that positive charge flowedin a circuit. The flow of positive charge is called

conventional urrent.

Arrows are used to show the direction of the current.

+

connector (wire)

battery

resistor

ammeter

voltmeter

(or )


In a circuit, resistors can either be arranged in series with one another or parallel to another.

Resistors in Series All the resistors are connected one after the other.There is only one path for the charge to follow.

IT = I1 = I2 = I3

V

R1

R2

R3

+

Since it is one continuous loop, the current throughout the loop is the same. The total current is equal tothe current in resistor 1, resistor 2 and resistor 3.

http://www.youtube.com/watch?v=oL-V-nOVy0w&feature=related


To find the total resistance of a series circuit, sum the resistances in each resistor.

RT = R1 + R2 + R3

The increase in potential across the battery, is equalto the sum of the potential drops at each resistor.

VT = V1 +V2 + V3

Total resistance is also referred to asthe effective or equivalent resistance.


Sample Problem

A 45.0 V potential difference is place across a 5.0 Ωresistor and a 10.0 Ω resistor connected in series.

a) What is the eqivalent resistance of the circuit?b) What is the current through the circuit?c) What is the voltage drop across each resistor?d) What is the total voltage drop across the circuit?

We can use a VIR chart to keep track of our values.

Fill in the known values.

5.010.0

45.0

a) equivalent resistance = total resistance

RT = R1 + R2

(add the values in the third column!)

Fill this value into the chart.

5.010.0

45.0 15.0


We now have two values in the "Total" row. Using Ohm's law, we can determine the third, the total current.

b) current through the circuit = total current

IT = VT/RT

Fill the values of IT, I1 and I2 into the chart.

IT = I1 = I2 Reminder

5.0

10.045.0 15.03.00

3.003.00

5.010.0

45.0 15.0


We now have two values in the "R1" row and the"R2" row. Using Ohm's law, we can determine the third value, V, for each resistor.

c) voltage drop across each resistor

Fill these values into the chart.

5.0

10.045.0

3.003.003.00 15.0

5.0

10.045.0

3.003.003.00 15.0

15.030.0


d) total voltage drop across the resistor

This value was given! Verify to make sure that:

VT = V1 +V2

5.0

10.045.0

3.003.003.00 15.0

15.030.0


Resistors in Parallel

Parallel circuits are made by connecting resistors in such a way that you create several paths/branches through which current can flow. For the resistors tobe truly in parallel, the current must split, then come back together.

In the example given below, the current has threepossible paths it can take.

VR1 R2 R3

+

current splits here

current comes together here

Path 1 Path 2 Path 3

http://www.youtube.com/watch?v=rm89sH_tyY8&feature=related


The total current in the circuit is equal to the sum of the currents in each path.

IT = I1 + I2 + I3

The battery provides the source of potential differencefor the circuit. Each path acts as if the other paths are not present. All the potential drops are the same.

VT = V1 = V2 = V3


The equivalent resistance of a parallel circuit canbe found using the following formula,

Placing a resistor in parallel with an existing resistor always decreases the resistance of the circuit. The resistance decreases because each new resistor provides an additional path for the current to flow.

The equivalent resistance is always less that the resistance of any resistor in the circuit.

NOTE


Derivation of the formula for the equivalent resistance of a parallel circuit.

The total current in the circuit is the sum of the currents through the branches of the circuit.

I = I1 + I2 + I3

The total current through the equivalent resistance, R, is given by I = V/R, but all the potential drops in the circuit are the same.

V = V + V + VR R1 R2 R3

Dividing both sides of the equation by V gives an equation for the equivalent resistance of the paralle resistors.


Sample ProblemThree resistors of 60.0 Ω, 30.0 Ω and 20.0 Ω are connected in parallel across a 90.0 V difference inpotential.

a) Find the equivalent resistance of the circuit.b) Find the current in the entire circuit.c) FInd the current through each branch of the circuit.

We will use a VIR chart to keep track of our values.


a) equivalent resistance


10.0

Note that the equivalent resistance is less than resistance of any resistor

in the circuit.


10.0

b) current in the entire circuit

We now have two values in the "Total" row. Using Ohm's law, we can determine the third, the total current.

IT = VT/RT


10.09.00


We now have two values in the "R1" row, the"R2" row and the "R3" row. Using Ohm's law,we can determine the third value, V, for eachresistor.

c) voltage drop across each resistor

10.09.00


10.09.00

1.503.004.50

Verify to make sure that:

IT = I1 + I2 + I3



Combination Circuits

Find the total resistance and current, then find the individual voltages and currents for each of the resistors in this circuit:

R1 5 ΩR2 7 Ω

R3 10 Ω+

12 V



Next, simplify the circuit. Calculate the equivalent resistance for the parallel part of the circuit. We will call this resistance, RA.

R1 5 ΩR2 7 Ω

R3 10 Ω+

12 V

R1 5 Ω RA

+

12 V


Now, calculate the equivalent resistance of the entire circuit, RT. Do this by following the rule for resistors in series.


R1 5 Ω RA

+

12 V


We now have two values in the "Total" row. Using Ohm's law, we can determine the third, the total current, IT.

IT = VT / RT



All the current that flows through the circuit will also flow through R1 because it is in series with the battery.


V1 = I1R1

We now have two of the three values in the R1 row. We can find V1 using Ohm's law.


If VT is 12 V and the drop in potential at R1 is 6.5 V, then there is still

12 V 6.5 V = 5.5 V

to be used up.

We know that the voltage drops at R2 and R3 are equal because they are connected in parallel. Therefore,

V2 = 5.5 V and V3 = 5.5 V



Determine I2 and I3 using Ohm's Law.

Complete the chart.

Note that the sum of I2 and I3 is ~1.3 A.


R1 2 Ω

R3

1 Ω

R2

5 Ω

+

9 V

R4

3 Ω

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