GCSE Physics Double Award notes

1. Electricity
Energy and Power

2. Aims of the lesson
• State the definition of electrical energy and
power
• Understand how to derive the ‘energy’ and
‘power’ equations from previous electrical
equations
• Use these equations to solve simple
problems
2

3. Notes in the text book
• Page 131 - 134

4. What is the purpose of electricity?
An electric current is a convenient way of
transmitting energy from one place to another
The energy is provided by the battery (the
source)
The charged electrons carry the energy and are
‘pushed’ by the potential difference of the
battery
The energy the electrons carry is transferred by a
component (light in a bulb)

6. Electrical Energy
One joule of energy is required to transfer one
coulomb of charge between two points having a
potential difference of one volt
Energy = Voltage x Charge
E=VxQ
joules = volts x coulombs

8. Electrical Energy
One coulomb is the amount of charge required to pass a
current of one ampere past a point in one second
Q = I x t
Substitute the value of Q into the energy equation-
Electrical Energy = Voltage x Current x Time
E = Q
VxIxt
joule = volts x amperes x seconds
Or E = ItV (the TV channel)

9. Electrical Energy
Example Question
A current of 4 A flows through the element of a
radiator working on a 250 V supply for half an
hour. How much heat energy is produced?
Energy transferred = I x t x V
= 4 x (30 x 60) x 250
= 1 800 000 J
= 1800 kJ or 1.8 MJ

10. Electrical Power
Power is the rate of energy transfer for an
appliance
Power = Energy / Time
=IxtxV/t
=IxV
Power = Current x Voltage
P = I x V
Watts = Ampere x Volt
Pg 132 Q 28 - 31

12. More EQUATIONS V
Remember Ohm’s Law? V = I x R R I
The Power equation can be re-written using
resistance
Substitute in- V = IxR
So P = I xV (I
I2 X R X R)
And re-arrange Ohm’s Law for Current
I = V/R
So = I
P = (V / R) 2 / V
VX R
Robinho, the
substitute of
substitutes

13. Example
A large battery is connected to a resistor of
resistance 1000 ohms. The potential difference
across the resistor is 50 V. What is the power
dissipated by the resistor?
P = V2 / R
= 502 / 1000
= 2500 / 1000
= 2.5 W
Page 134 Q 32 - 35

14. The Electric Equation Eels
14