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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).

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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.


Worksheet 1


Worksheet 1


Worksheet 1


Worksheet 1


Worksheet 1


Worksheet 1


Worksheet 1


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


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