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Refer back to the equation derived from the definition of potential difference: This can be rearranged as: Thus: The ratio of charge to time, , is the current I in the component. Therefore: P = VI By substituting from the resistance equation V = IR, we get the alternative equations for power: KEY EQUATIONS Equations for power: WORKED EXAMPLE 6 Calculate the rate at which energy is transferred by a 230 V mains supply that provides a current of 8.0 A to an electric heater. Step 1 Use the equation for power: P = VI with V = 230 V and I = 8.0 A. Step 2 Substitute values: P = 8 × 230 = 1840 W (1.84 kW) 7 a A power station produces 20 MW of power at a voltage of 200 kV. Calculate the current supplied to the grid cables. Step 1 Here we have P and V and we have to find I, so we can use P = VI. Step 2 Rearranging the equation and substituting the values we know gives: Hint: Remember to convert megawatts into watts and kilovolts into volts. So, the power station supplies a current of 100 A. b The grid cables are 15 km long, with a resistance per unit length of 0.20 Ω km−1. How much power is wasted as heat in these cables? Step 1 First, we must calculate the resistance of the cables: resistance R = 15 km × 0.20 Ω km−1 = 3.0 Ω Step 2 Now we know I and R and we want to find P. We can use P = I2R: power wasted as heat, Hence, of the 20 MW of power produced by the power station, 30 kW is wasted – just 0.15%. 8 A bathroom heater, when connected to a 230 V supply has an output power of 1.0 kW. Calculate the resistance of the heater. Step 1 We have P and V and have to find R, so we can use Step 2 Rearrange the equation and substitute in the known values: Note: The kilowatts were converted to watts in a similar way to the previous example. Questions 17 Calculate the current in a 60 W light bulb when it is connected to a 230 V power supply. 18 A power station supplies electrical energy to the grid at a voltage of 25 kV. Calculate the output power of the station when the current it supplies is 40 kA. Power and resistance A current I in a resistor of resistance R transfers energy to it. The resistor dissipates energy heating the resistor and the surroundings.. The p.d. V across the resistor is given by V = IR. Combining this with the equation for power, P = VI, gives us two further forms of the equation for power dissipated in the resistor: Which form of the equation we use in any particular situation depends on the information we have available to us. 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