Students investigate the efficiency of an electric motor - Edexcel - GCSE Physics - Question 4 - 2018 - Paper 1

To sketch the graph of an alternating current of frequency 0.5 Hz, draw a sine wave that completes one full cycle (from 0 to a maximum positive value, back to 0, to a maximum negative value, and back to 0) over a period of 2 seconds. The graph should oscillate between positive and negative values, showing the alternating nature of the current.

An efficiency of 40% means that 40% of the energy supplied to the electric motor is converted into useful work output, while the remaining 60% is lost as wasted energy, typically as heat due to friction and other factors. Thus, the motor is only capable of utilizing 40% of the energy it consumes effectively.

To find the energy supplied to the motor, we use the formula:

$ext{Energy} = ext{Power} imes ext{Time}$

Converting time from minutes to hours: 3 ext{ minutes} = rac{3}{60} = 0.05 ext{ hours}

Thus, substituting the values: $ext{Energy} = 15 ext{ W} imes (0.05 ext{ hours})$

Calculating this gives us: Energy = 0.75 ext{ Wh}$$ Therefore, the unit is watt-hours (Wh).

To find the cost of running the motor, we first calculate the energy consumed in kilowatt-hours (kWh):

Converting watt-hours to kilowatt-hours: 0.75 ext{ Wh} = rac{0.75}{1000} = 0.00075 ext{ kWh}

The cost can then be calculated using the rate of electricity: Cost = Energy (kWh) × Cost per kWh Cost = 0.00075 ext{ kWh} imes 20 ext{ pence/kWh} Cost = 0.015 ext{ pence}.