P2.4.2 Current, Charge and Power

P2 Physics

KS4 ADDITIONAL SCIENCEMr D Powell

Mr Powell 2012Index

Connection

• Connect your learning to the content of the lesson

• Share the process by which the learning will actually take place

• Explore the outcomes of the learning, emphasising why this will be beneficial for the learner

Demonstration

• Use formative feedback – Assessment for Learning

• Vary the groupings within the classroom for the purpose of learning – individual; pair; group/team; friendship; teacher selected; single sex; mixed sex

• Offer different ways for the students to demonstrate their understanding

• Allow the students to “show off” their learning

Activation

• Construct problem-solving challenges for the students

• Use a multi-sensory approach – VAK• Promote a language of learning to

enable the students to talk about their progress or obstacles to it

• Learning as an active process, so the students aren’t passive receptors

Consolidation

• Structure active reflection on the lesson content and the process of learning

• Seek transfer between “subjects”• Review the learning from this lesson and

preview the learning for the next• Promote ways in which the students will

remember• A “news broadcast” approach to learning

Mr Powell 2012Index

P2.4.2 Current, Charge and Power

a) When an electrical charge flows through a resistor, the resistor gets hot.

b) The rate at which energy is transferred by an appliance is called the power: P = E/t

c) Power, potential difference and current are related by the equation: P = VI

d) Energy transferred, potential difference and charge are related by the equation: E = VQ (HT Only)

P is power in watts, WE is energy in joules, Jt is time in seconds, sI is current in amperes (amps), AV is potential difference in volts, VQ is charge in coulombs, C

Mr Powell 2012Index

P5.4 – Electrical Power & PD

a) When an electrical charge flows through a resistor, the resistor gets hot.

b) The rate at which energy is transferred by an appliance is called the power:

P = E/t

c) Power, potential difference and current are related by the equation:

P = VI

P is power in watts, WE is energy in joules, Jt is time in seconds, sI is current in amperes (amps), AV is potential difference in volts, VQ is charge in coulombs, C

If the value for energy is less that 1 J it is displayed in millijoules (mJ), and when it reaches 1000 J it is displayed in kilojoules (kJ).

If the value for power is less that 1 W it is displayed in milliwatts (mW).

If the value for current is less that 1 A it is displayed in milliamps (mA).

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Power, what do we mean?

Mr Powell 2012Index

Mr Powell 2012Index

Calculating Power

Millions of millions of electrons pass through the circuit of an artificial heart every second. Each electron transfers a small amount of energy to it from the battery. So the total energy transferred to it each second is large enough toenable the device to work.

For any electrical appliance:

the current through it is a measure of the number of electrons passing through it each second (i.e. the charge flow per second),

the potential difference across it is a measure of how much energy each electron passing through it transfers to it (i.e. the electrical energy transferred per unit charge),

the power supplied to it is the energy transferred to it each second. This is the electrical energy it transforms every second.

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Coulombs

Electrons are charged particles and each of them have a charge of 1.9 x 10-19 C.

It is a simple property which cannot be removed or changed. It is a useful to us as charges make particles move i.e. opposites attract.

If we add a load of them together and think of them as a single “sphere of charge” or ball we get a whole coulomb of charge and can think about defining the ampere or amp 1A = 1C/s

C

1.9 x 10-19 C x 6.25 x 1018 electrons = 1C

e- e-

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Power Delivery

When we talk about Power what we mean is “the amount of energy delivered per second”

1 Joule / 1 Second = 1 Watt

It then makes sense that the Power used by a component can be found from the product of current throughand voltage across the component;

Power = Voltage x CurrentP = V x I

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Example

In this circuit the Voltmeter = 4V and the Current = 1A

Power = Voltage x Current

P = V x I

P = 4V x 1A

P = 4J/C x 1C/s

P = 4J/s

P = 4W

4V

1A

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Analogy

Another way of thinking about it is saying that the current carries the energy;

4V

1A

C CC C

C C

C CC

1J

1J

1J

1J

C1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

As the Coulombs of Charge move they release their energy as heat and light (through the bulb)

C

1J

= 1 Coulomb of charge

= 1 Joule of energy

= 1 Second of time

Mr Powell 2012Index

Analogy 2

If the voltage increases, more energy is delivered so the power increases;

5V

1A

C CC C

C C

C CC

1J

1J

1J

1J

C1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

Power = 5V x 1A = 5J/s = 5W

1J 1J 1J 1J

C

1J

= 1 Coulomb of charge

= 1 Joule of energy

= 1 Second of time

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Analogy 3 – “Power Modelling”

If the current increases, more energy is delivered so the power increases;

4V

2A

C CC C

C C

C CC

C1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

Power = 4V x 2A = 8J/s = 8W

C CC C

CC

C CC

C

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J

1J 1J1J

1J

C

1J

= 1 Coulomb of charge

= 1 Joule of energy

= 1 Second of time

Mr Powell 2012Index

Homework - Current Flowing?

Power = Voltage x Current

P = VI

I = P/V

I = 154W / 230V

I = 154J/s / 230J/C

I = 0.664 C/s

I = 0.7 C/s

I = 0.7 A

We can work out the current flow from the total power use;

So if this is the current flow we can now pick a fuse to match i.e. 3A

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Fuses in use...

Domestic appliances are often fitted with a 3A, or a 5A or a 13 A fuse.

If you don’t know which one to use for an appliance, you can work it out from thepower rating of the appliance and its potential difference (voltage).

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Simple Energy Transfers

When you use an electrical appliance, it transforms electrical energy into other forms of energy. The power of the appliance, in watts, is the energy it transforms, in joules, per second. The more per second the higher the power.

Take a short note of this key point...

Mr Powell 2012Index

Summary Questions

power

currentp.d.

current

a)i) 12V x 3A = 36Wii) 230V x 2A = 460W

b)i) 24V / 12A = 2A -> 3A fuseii) 800W/230V= 3.47A -> 5A fuse

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DeviceTime

taken (s)

Energy supplied

(J)

Useful energy transferred

(J)useful

power (W) Efficiency

Lamp 1000 100 000 20 000 20

Microwave oven 120 96 000 48 000 0.5

Motor 300 6000 20 0.33

Computer 3000 900 000 33 0.11

Power & Efficiency...

The table below shows the time taken by different electrical devices to transfer a given amount of energy supplied. The useful energy transferred in this time is also stated. Copy and Complete the table....

0.2

400

18 182

99 000

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P5.5 – Electrical Energy & Charge p172

FT: Revisit of Q = It

d) Energy transferred, potential difference and charge are related by the equation:

E = VQ (HT Only)

P is power in watts, WE is energy in joules, Jt is time in seconds, sI is current in amperes (amps), AV is potential difference in volts, VQ is charge in coulombs, C

Mr Powell 2012Index

Flow of Charge (FT)

When an electrical appliance is on, electrons are pushed through the appliance by the potential difference

The potential difference causes a flow of charge particles (free electrons). The rate of flow of charge is the electric current through the appliance. 1A = 1C/s

The unit of charge, the coulomb (C).

The charge passing along a wire or through a component in a certain time depends on: the current, and the time.

We can calculate the charge using the equation:

Q=It

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Worked Examples..... FT

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Energy and Potential Difference (HT)

When a resistor is connected to a battery, electrons are made to pass through the resistor by the battery.

Each electron repeatedly collides with the vibrating atoms of the resistor, transferring energy to them.

The atoms of the resistor therefore gain kinetic energy and vibrate even more. The resistor becomes hotter.

When charge flows through a resistor, electrical energy is transformed into heat energy.

The energy transformed in a certain time in a resistor depends on: the amount of charge that passes through it, and the potential difference across the resistor.

E = VQQ=ItE = VIt

Mr Powell 2012Index

A*/B Questions – copy & complete

1. A cordless drill operates using a 14.4 V battery pack. The battery is rated at 2 amp hours which means that it can deliver a current of two amps for a period of 1 hour. How much energy flows from the battery?

2. A 12 volt heater takes a current of 3.6 A. It is left to heat up an aluminium block for a period of 45 minutes. How much heat energy is transferred to the aluminium block?

3. What current is consumed by a 60 W light bulb operating on the 230 V mains?

1 hour = 3600 s Charge = current x time = 2 A x 3600 s = 7200 CEnergy = charge x voltage = 7200 C x 14.4 V = 103680 J

Time = 45 x 60 = 2700 sEnergy = VIt = 2700 x 12 x 3.6 = 116640J

I = P/V = 60 W / 230 V = 0.26 A

Mr Powell 2012Index

Summary Questions

P2.4.2 Current, Charge and Power

a) When an electrical charge flows through a resistor, the resistor gets hot.

b) The rate at which energy is transferred by an appliance is called the power:

P = E/t

c) Power, potential difference and current are related by the equation:

P = VI

d) Energy transferred, potential difference and charge are related by the equation:

E = VQ (HT Only)

P is power in watts, WE is energy in joules, Jt is time in seconds, sI is current in amperes (amps), AV is potential difference in volts, VQ is charge in coulombs, C

P2.4.2 Current, Charge and Power

a) When an electrical charge flows through a resistor, the resistor gets hot.

b) The rate at which energy is transferred by an appliance is called the power:

P = E/t

c) Power, potential difference and current are related by the equation:

P = VI

d) Energy transferred, potential difference and charge are related by the equation:

E = VQ (HT Only)

P is power in watts, WE is energy in joules, Jt is time in seconds, sI is current in amperes (amps), AV is potential difference in volts, VQ is charge in coulombs, C

P2.4.2 Current, Charge and Power

a) When an electrical charge flows through a resistor, the resistor gets hot.

b) The rate at which energy is transferred by an appliance is called the power:

P = E/t

c) Power, potential difference and current are related by the equation:

P = VI

d) Energy transferred, potential difference and charge are related by the equation:

E = VQ (HT Only)

P is power in watts, WE is energy in joules, Jt is time in seconds, sI is current in amperes (amps), AV is potential difference in volts, VQ is charge in coulombs, C

Suggested ideas for practical work to develop skills and understanding include the following:

measuring oscilloscope traces

demonstrating the action of fuse wires

using fluctuations in light intensity measurements from filament bulbs to determine the frequency of a.c.

measuring the power of 12 V appliances by measuring energy transferred (using a joulemeter or ammeter and voltmeter) in a set time.

Additional Guidance...

A lot of energy is wasted in filament bulbs as heat.

Less energy is wasted in power-saving lamps such as Compact Fluorescent Lamps (CFLs).

There is a choice when buying new appliances in how efficiently they transfer energy.

Should be able to calculate the current through an appliance from its power and the potential difference of the supply, and from this determine the size of fuse needed.