Year 12 Current Electricity Lesson 3 Potential difference and power Miss O’Donnell.

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Year 12 Current Electricity Lesson 3 Potential difference and power Miss O’Donnell

Transcript of Year 12 Current Electricity Lesson 3 Potential difference and power Miss O’Donnell.

Page 1: Year 12 Current Electricity Lesson 3 Potential difference and power Miss O’Donnell.

Year 12 Current ElectricityLesson 3

Potential difference and power

Miss O’Donnell

Page 2: Year 12 Current Electricity Lesson 3 Potential difference and power Miss O’Donnell.

What questions are we looking at today?

• What do we mean by potential difference?

• How can we calculate electrical power?

• How do energy transfers take place in electrical devices?

Page 3: Year 12 Current Electricity Lesson 3 Potential difference and power Miss O’Donnell.

What do we mean by potential difference?

• Defined as … the work done (or energy transferred) per unit charge

• Unit of pd is the VOLT• Volt is equal to 1 Joule per Coulomb• V= W/Q• Work is done when charge Q flows through a

component• W = QV• V = I x R

Page 4: Year 12 Current Electricity Lesson 3 Potential difference and power Miss O’Donnell.

Quick Questions

• 30J of work is done when 5C of charge pass through a component – what is the pd?

• If the pd is 12V across a component and 3C of charge is passed through it – what is the energy transfer from the battery to the component?

Page 5: Year 12 Current Electricity Lesson 3 Potential difference and power Miss O’Donnell.

Electrical power & current

• A component has a pd V across its terminals and a current I passing through it.

• In time ∆t– the charge through the component Q=I ∆t– the work done by the charge carriers W=QV– By substitution W= (I ∆t)V = IV∆t

• Because Power = Energy/Time = IV∆t/ ∆t• Power = IV

Page 6: Year 12 Current Electricity Lesson 3 Potential difference and power Miss O’Donnell.

Questions• Calculate the energy transfer in 1200s in a component

when the pd across it is 12V and the current is:– 2A– 0.05A

• A 6V, 12W light bulb is connected to a 6V battery. Calculate:– The current through the bulb– The energy transfer to the light bulb in 1800s

• ANSWERS: – 29kJ– 720J– 2A– 22kJ

Page 7: Year 12 Current Electricity Lesson 3 Potential difference and power Miss O’Donnell.

Questions 2

• A 230V electrical appliance has a power rating of 800W. Calculate:– The energy transfer in the appliance in 1 minute– The current taken by the appliance– Which of the following fuse values would be suitable

for this appliance? 3A, 5A & 13A• ANSWERS:– 48kJ– 3.5A– 5A

Page 8: Year 12 Current Electricity Lesson 3 Potential difference and power Miss O’Donnell.

Energy Transfer in Electrical Devices

• The emf of a source of electricity is defined as the electrical energy produced per unit charge passing through the source.

• Unit is the VOLT – same as pd• For a source of emf ε in a circuit, the electrical

energy produced when a charge Q pass through the source is Qε

• This energy is transferred to other parts of the circuit and some may be dissipated by the source itself due to internal resistance – we will explore this further next lesson

Page 9: Year 12 Current Electricity Lesson 3 Potential difference and power Miss O’Donnell.

Electric current effects

• What effects do electric currents have when they pass through components and appliances?– Heating – when they pass through a component

with resistance. WHY?– Magnetic – in electric motors and loudspeakers– Kinetic – force the charge carriers though wires

Page 10: Year 12 Current Electricity Lesson 3 Potential difference and power Miss O’Donnell.

Questions 3• A battery has an emf of 9V and negligible internal

resistance. It is capable of delivering a total charge of 1350C. Calculate:– The maximum energy the battery could deliver– The power it would deliver to the components of the

circuit if the current through it was 0.5A– How long the battery would last for, if it were to supply

power at the rate calculated in the last part of the question• ANSWERS:– 12kJ– 4.5W– 2700s