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GCSE Physics Double Award notes

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- 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)
- 5. 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
- 6. 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)
- 7. 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
- 8. 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
- 9. 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
- 10. 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
- 11. The Electric Equation Eels 14

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