Efficiency (Cambridge (CIE) A Level Physics): Revision Note

An electric motor has an input power, P_in, useful output power, P_out, power lost P_lost, and an efficiency η.

What is the output power of the motor? 

A. $\frac{\eta }{P_{in}}$      

B. $\frac{-\eta P_{lost}}{\eta -1}$      

C. $\eta P_{lost}$      

D. $-\eta P_{lost}\left(\eta -1\right)$

Answer: B

Step 1: State the efficiency equation

$\mathrm{efficiency}=\frac{\mathrm{useful}\mathrm{power}\mathrm{output}}{\mathrm{total}\mathrm{power}\mathrm{input}}$

Step 2: Substitute the terms given in the question

$\eta =\frac{P_{out}}{P_{in}}=\frac{P_{out}}{P_{out}+P_{lost}}$

Step 3: Manipulate the equation

• Multiply by P_out + P_lost on both sides

$\eta \left(p_{out}+p_{lost}\right)=P_{out}$

• Expand the brackets

$\eta P_{out}+\eta P_{lost}=P_{out}$

• Minus P_out from both sides

  • It is helpful to first minus ηP_lost

$\eta P_{out}=P_{out}-\eta P_{lost}$

$\eta P_{out}-P_{out}=-\eta P_{lost}$

• Factor out P_out

$P_{out}\left(\eta -1\right)=-\eta P_{lost}$

• Divide by η − 1

$P_{out}=\frac{-\eta P_{lost}}{\left(\eta -1\right)}$

Efficiency can be in a ratio or percentage format. If the question asks for an efficiency as a ratio, give your answer as a fraction or decimal. If the answer is required as a percentage, remember to multiply the ratio by 100 to convert it, e.g. Ratio = 0.25, Percentage = 0.25 × 100 = 25 %

The diagram shows a pump called a hydraulic ram.

In one such pump, the long approach pipe holds 700 kg of water. A valve shuts when the speed of this water reaches 3.5 m s^-1. The kinetic energy of this water is used to lift a small quantity of water by a height of 12m. The efficiency of the pump is 20%.

Which mass of water could be lifted 12 m?

A. 6.2 kg               

B. 4.6 kg               

C. 7.3 kg               

D. 0.24 kg

Answer: C

Step 1: List the known quantities

• Mass of water in approach pipe = 700 kg

• Speed of water in approach pipe, v = 3.5 m s^-1

• Height of lifted water, h = 12 m

Step 2: Consider the energy transfer taking place

• Energy is transferred from the kinetic store of the water to its gravitational potential store

Step 3: Consider the efficiency of the energy transfer

• The transfer is 20% efficient

• Therefore, 20% of the kinetic input energy is output as gravitational potential energy

$0.2E_k=E_p$

$0.2\times \frac{1}{2}m{v}^2=mgh$

Step 4: Calculate the mass of water lifted water

$0.2\times 0.5\times 700\times {\left(3.5\right)}^2=m\times 9.81\times 12$

$857.5=m\times 117.72$

$m=\frac{857.5}{117.72}$

$m=7.3\mathrm{kg}$

Equations for kinetic and potential energies are important for these types of questions. Also, familiarise yourself with the different equations for power depending on what quantities are given.