V = IR V = voltage in volts (aka potential difference) I = Current in...

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Ohms Law V = IR V = voltage in volts (aka potential difference) I = Current in amps R = resistance in ohms ( Ω )

Transcript of V = IR V = voltage in volts (aka potential difference) I = Current in...

Ohms Law

• V = IR

• V = voltage in volts (aka potential

difference)

• I = Current in amps

• R = resistance in ohms (Ω)

Current

• How would you define it?

• Current—the movement of electric

charge through a medium

Current, More Precisely• Current—the rate at which electric charge

flows through a given area. Described by

the letter I

• SI Unit: Amps (amperes) “A”

• So picking a point, the amount of charge that

flows past that point in a given amount of

time.

+

++

++ ++ +++

Current

I = DQ / Dt• Current = charge passing through area

time interval

• SI Unit: amperes (almost always called

amps), abbreviated A.

• 1 A = 1 C/s

Ex 1 Homework

• If the current in a wire of a CD player is

5.00mA, how long would it take for 2.0C

of charge to pass through a point?

If the current in a wire of a CD player is 5.00mA,

how long would it take for 2.0C of charge to pass

through a point?

• t = Q/I

• t = 2 C/.005 A

• 400 sec

Where do the charges come from?

• Charge carriers are the electrons in the

conducting wire. They transmit the energy

Conventional Current

• Current is the flow of any charge, so it

can be positive or negative flow.

• In conductors, what tends to be flowing?

Possible Charge Flows

• In conductive metals—electrons

• In body fluids/solutions—ions

• In particle accelerators—protons

• All cause current! But direction

described in relation to a positive charge.

Why does current flow?

What is potential energy?

• stored energy due to an object’s position

or condition in a field of force.

Think: Where is the electrical potential energy of a

positive test charge (q+) higher, at the point A or B?

Why?

Point A. Because of it’s

location, it is not where

it “wants” to be. It took

work to get it there!

Describe:

• High energy location for a positive test charge

is

– furthest from (-)source charge; near a (+)source

charge

• low energy location for a positive test charge is

– near a (-)source charge; far from (+)source charge

Circuits

• This explains why conventional current

flows opposite to the flow of electrons

• From positive terminal to negative terminal

• Describes flow in relation to + test charge

• If a charge moves in an electric field it’s

potential energy changes

• Charge flows from Hi potential to Low

potential, if there is a conducting path.

Internal vs External Circuit

Potential Difference

What describes internal circuit?

What describes external circuit?

• Sort of like Sisyphus.

0 PE

PE

Heat, Sound

Squashed bugs

Potential Difference

0 PE

PE

Voltage

RiseVoltage

Drop

• All charges moving everywhere transmit the

energy to the light bulb.

0 PE

PE

Voltage

RiseVoltage

Drop

example• 6 V battery.

• Internal Circuit (the electrochemical cell)– Provides chemical energy to move test charge from low

energy negative terminal internally to high energy positive terminal

• External Circuit (Wires attached to terminal)– Test charge: natural movement

• Away from positive, towards negative

– As + test charge moves from + terminal to – terminal, it loses 6V of potential energy for every coulomb of charge

– Since energy can’t be destroyed, it is transformed (light turns on)

VOLTAGE• Battery is the Voltage source

• Referred to as – Voltage

– Potential Difference

– Electrical Potential

– Electromotive Force

– Electrical pressure

• battery voltage is the amount of work (energy) done per Coulomb of charge (J/C =1 V)

• Voltage sources only maintain the difference in potential in the circuit.

• SI Unit: Volts (V) or J/C

FYI: Electrical Potential Energy vs

Electrical Potential• Electrical Potential Energy: Joules

– Dependent on magnitude of charge and location of

charge

– Describes work to move a charge

• Electrical Potential: Voltage or Joules/Coulomb

– is the Joules of potential energy per charge

– Describes affect of field at a particular location

– 1J/C = 1 Volt

• Most alkaline dry cells contain a zinc casing

which serves as the negative terminal.

– The zinc is oxidized to Zn+2 ions during the chemical

reactions.

– This oxidation process produces two electrons per

zinc atom that accumulate at the negative terminal.

– There is a carbon rod that is inserted in the middle of

the cell that serves as the positive terminal of the cell.

– The carbon rod is not consumed in the chemical

reactions.

Resistance

Viva la resistance.

• When a light bulb is connected to a

battery, what effects how much current

flows through the circuit?

• Voltage (provides energy “work” to move

the charge) Direct

• Resistance—the opposition of motion of

charge through a conductor.

• determines how much current will flow in a

circuit with known voltage source:

• Quantitatively:

R =

• The SI unit for resistance is the ohm (W).

DV

I

• Materials that have a constant value for

resistance over a large range of potential

differences or voltages is said to be

ohmic.

• In addition to the load the wire itself offers

resistance

• “internal friction” due to collisions

• Lets examine the wire

1) Length of conductor

2) Cross sectional area of conductor

3) Temperature

4) Nature of the material

4 Factors effecting

Resistance of the

conductor

If the following are increased,

resistance will:– Length

– increase

– Cross-sectional area

– decrease

– Temperature

– increase

– Nature of the material

1) Length of conductor

2) Cross sectional area of conductor

3) Conducting material

4) Temperature

The calculations;

Ohms Law

R = V/I or more commonly V = IR

Where V = Voltage in Volts “V”

I = Current in Amps “A”

R = Resistance in Ohms “W ”

• Uses the natural resistance of the body.

• People have a general resistance of

around 500 000 W.

• Sweat causes dramatic change in

resistance (as low as 100W.)

• Machine measures the GSR.

• We said that temperature affects

resistance.

• Some materials have zero resistance

below a certain temperature (called their

critical temperature)

• Materials that display such properties are

called superconductors.

Ex 2: What is the potential

difference required for 20 amps of

current to flow through a 5 ohm

resistor?

What are you solving for?

V = IR

V = (20 A) (5Ω)

V = 100 V

Ex 3 The resistance of a steam iron is

19.0W. What is the current in the iron when

it is connected across a potential difference

of 220.V?

Ex 3 The resistance of a steam iron is

19.5W. What is the current in the iron when

it is connected across a potential difference

of 220.V?

V = IR

I = V/R

I = 220V/19W

I = 11.28 A

• Something to think about……

• What is the function of a battery?

• How is current flow described?

• What is the unit for voltage, current,

resistance, and power?

• What factors affect resistance in a

conductor?

• What is the relationship between voltage,

current, and resistance?

What is the difference between

a 40W light bulb and a 75W

light bulb?

What is power? In general……

• The amount of work done per time period.

• How have we discussed the idea of work in

relation to electricity?

• Work is done by the electric field on

electrons (or other charged particles) as

they move from one area to another due to

potential difference.

Electric Power

• Electric power is the rate at which charge

carriers do work.

Electric Power

• P = W/t or P = E/t = J/s

– Where W is Work and E is Energy

• Power is measured in Watts (W).

• Given this relationship: P = IV, how is this

still watts?

• P= (C/s)(J/C) = J/s

Ex. 4

• A 1000.-watt heater operates at 115 V.

Calculate the current, resistance, and

energy generated in 1.00 hour.

Ex. 4 A 1000.-watt heater operates at 115 V.

Calculate the current, resistance, and energy

generated in 1.00 hour.

I = P/V

I = 1000W/115V = 8.70A

R = V/I

R = 115V /8.70A =13.2Ω

E = Pt

E = (1000W)(3600s) = 3.6 x 106 J

Ex. 4 A 1000.-watt heater operates at 115 V.

Calculate the current, resistance, and energy

generated in 1.00 hour.

I = P/V

I = 1000W/115V = 8.70A

R = V/I

R = 115V /8.70A =13.2Ω

E = Pt

E = (1000W)(3600s) = 3.6 x 106 J

Ex 5

• The Nutri Ninja Junior offers a resistance of 16 ohms when a current

of 7.5A is flowing through the circuit. What is the power rating of the

Nutri Ninja delicious smoothies? ASK YOURSELF WHAT YOU

KNOW

• R = 16Ω I = 7.5 A P = ?

• Solve for V using V = IR

• 7.5A x 16Ω = 120V

• Solve for P using P = IV

• 7.5A x 120V = 900W

Ex 6

Compare the resistance of a 40 Watt light

bulb to a 75 Watt light bulb when

connected to a 120 V circuit. Which bulb

has a higher resistance? How would we

calculate this?

Ex 6Compare the resistance of a 40 Watt light bulb to a 75 Watt

light bulb when connected to a 120 V circuit. Let’s

calculate this.

Knowns: P = 40W P = 75W V = 120V R = ?

I have volts and power, how do I find resistance?

I know V = IR and P = VI

Rearrange V = IR to solve for I, substitute this into P = VI

Thus: I = V/R therefore P = V x V/R or P = V2/R

Rearrange P = V2/R to solve for R

R = V2/P

R = (120V)2/40W = 360Ω

R = (120V)2/75W = 192Ω

The higher the Wattage the lower the resistance!!!!!!

Electric Power

• So what do we pay for on our electric bill?

• Rate of energy usage in Watts

• More specifically KW/hr instead Watts

(which is J/s)

Ex. 7

• How much does using a 1500W hair dryer

cost each month (30 days) if you use it for

5 minutes each day, and the price for 1

kilowatt hour is $0.10?

• SOLVE

Ex 7. How much does using a 1500W hair dryer

cost each month (30 days) if you use it for 5

minutes each day, and the price for 1 kilowatt hour

is $0.10?

Figure out total time (in hours

since final answer is based on kW

per hour)

Convert Watt to kiloWatts

Determine Energy Usage per hour

Multiply Energy Usage times cost

Ex 7. How much does using a 1500W hair dryer

cost each month (30 days) if you use it for 5

minutes each day, and the price for 1 kilowatt hour

is $0.10?

Time in Hours:

(30 days x 5min/day x 1hr/60min) = 2.5hr

kiloWatts

1500W x 1kW/1000W = 1.5 kW

Energy per hour

E = Pt (1.5 kW) (2.5hr) = 3.75 kWHr

Cost

Cost = Price x kWHr

Cost = ($0.10/kWHr) (3.75 kWHr) = $0.38