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# Ele ctric Curr ent PowerPoint PPT Presentation

Ele ctric Curr ent. Chapter 34. Flow of Charge. When the ends of an electric conductor are at different potentials, charge flows from one end to another - just like heat due to temperature difference - just like water flows downhill (difference in gravitational potential energy).

Ele ctric Curr ent

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## Electric Current

Chapter 34

### Flow of Charge

When the ends of an electric conductor are at different potentials, charge flows from one end to another

- just like heat due to temperature difference

- just like water flows downhill (difference in gravitational potential energy)

### Flow of Charge

• This difference in potential is often referred to as potential difference or voltage

• Voltage (V) is measured in volts (V).

• You can think of the source of voltage like a pump.

### Voltage Sources

• Supplies the voltage for any circuit.

• Acts like an “electric pump”.

• Examples:

• Batteries

• Generators

• Wall plug

### Types of Current

There are two types of current:

AC – alternating current

- electrons flow first in one direction and then in the opposite direction

- this happens by alternating the polarity of the generator or voltage source

- the voltage of AC in America is normally 120 volts

DC – direct current

– electrons flow in a single direction

– a battery in a circuit produces DC because the terminals of the battery always have the same signs.

### Electric Current

• Potential difference causes a flow of charge (electrons) in a conductor

• We call this electric current (I)

• Current is measured in amperes (A)

Charge

Current = time

I = q/t

A car’s starter motor draws 50 A. How much charge flows if the motor runs for 0.75 s?

Given:I = 50At = 0.75 s

Unknown:q (charge)

Equation:I = q/t

Substitute:50 A = q/(0.75 s)

Solve:q = 37.5 C

How long does it take for 52 C to pass through a wire carrying a current of 8.0 A?

Given:q = 52 CI = 8.0 A

Unknown:t (time)

Equation:I = q/t

Substitute:8.0 A = 52 C/t

Solve:t = 52 C/8.0 A

t = 6.5 s

What will happen to bulbs 1 and 2 when you disconnect the wires at various points?

### Activity 1

Consensus:

Current requires a closed loop

What type of object, when inserted into the loop, will allow the two test bulbs to light?

### Activity 2

Consensus:

Current requires a closed loop made entirely of conductors.

New Term:

Continuous Conducting Path

Test Circuit

What parts of a socket and bulb are conductors and which are insulators? What is the conducting path through the bulb?

Clips

Base

Fig 2. Clip – side view

Plates

### Activity 3

Consensus:

Trace the Continuous Conducting Path (CCP) through the bulb.

What parts of a socket and bulb are conductors and which are insulators? What is the conducting path through the bulb?

Clips

Base

Fig 2. Clip – side view

Plates

### Activity 3

Consensus:

Trace the Continuous Conducting Path (CCP) through the bulb.

### Electric Power

Where have we seen “power” before?

It was the rate work was done (energy was transferred).

P = W/t

All power is measured in Watts (W).

### Electric Power

Electric power is similar. It’s the rate electrical energy is transferred into another form.

P = IV

Electric power is often measured in the kilowatt, because the watt is so small.

### Electric Power

What do you know of that is electrical and has a watt measurement on it?

### Electric Energy

Electric energy can be calculated the same as any other form.

E = Pt

Electric energy, however, is measured in kilowatt-hours (kWh)

Your parents’ energy bill is measured in kilowatt-hours. Usually, electric companies charge 7-12 cents per kWh.

How much power is used by a calculator that operates on 8 volts and 0.1 ampere? If it is used for two hours, how much energy does it use?

Given:V = 8 VI = 0.1 A

Unknown:P (power)

Equation:P = IV

Substitute:P = (0.1 A)(8 V)

Solve:P = 0.8 W

How much power is used by a calculator that operates on 8 volts and 0.1 ampere? If it is used for two hours, how much energy does it use?

Given: V = 8 V I = 0.1 A P = 0.8 W t = 2 hr

Unknown:E (energy)

Equation:E = Pt

Substitute:E = (0.8 W)(2 hr)

Solve:E = 1.6 Wh

E = .0016 kWh

Will a 1200-watt hair dryer operate on a 120-volt line if the current is limited to 15 amperes by a safety fuse? Can two hair dryers operate on this line?

Given:P = 1200 WV = 120 V

Unknown:I (current)

Equation:P = IV

Substitute:1200 W = I (120 V)

Solve:I = 1200 W/120 V

I = 10 A

Can you operate the hair dryer? What about two hair dryers?

How much energy is expended in lighting a 100-watt bulb for 30 minutes?

Given:P = 100 Wt = 30 min

Unknown:E (energy)

Equation:E = Pt

Substitute:E = (100 W)(0.5 hr)

Solve:E = 50 Wh

E = 0.05 kWh

How much does it cost to operate a 100-watt lamp continuously for one week if the power utility rate is 10 cents per kilowatt-hour?

Given:P = 100 Wt = 1 week

Unknown:E (energy)

Equation:E = Pt

Substitute:E = (100 W)(168 hr)

Solve:E = 16800 Wh

E = 16.8 kWh

How much does it cost to operate a 100-watt lamp continuously for one week if the power utility rate is 10 cents per kilowatt-hour?

Given:E = 16.8 kWhcost = 10c/kWh

Unknown:T(total cost)

Equation:T = E(cost)

Substitute:T = (16.8 kWh)(10c/kWh)

Solve:T = 168 cents

T = \$1.68

### Electric Resistance

• Current depends not only on the amount of voltage impressed upon it but on the amount of resistance in the conductor.

• Different conductors offer different amounts of resistance

• Resistance is measured in ohms (W)

### Ohm’s Law

• Georg Simon Ohm, a German physicist, tested different wire circuits to see what effect resistance had on the current.

• He discovered that:

Voltage

Current = ---------------

Resistance

We call this Ohm’s Law!

### Ohm’s Law

Ohm’s Law can also be written using symbols:

I = V/R

A triangle can help when solving problems with Ohm’s Law.

What is the resistance of an electric frying pan that draws 12 amperes of current when connected to a 120-volt circuit?

Given:I = 12 AV = 120 V

Unknown:R (resistance)

Equation:R = V/I

Substitute:R = 120 V/12 A

Solve:R = 10 W

How much current is drawn by a lamp that has a resistance of 100 ohms when a voltage of 50 volts is impressed across it?

Given:R = 100 WV = 50 V

Unknown:I (current)

Equation:I = V/R

Substitute:I = 50 V/100 W

Solve:I = 0.5 A