Answer to Question #154338 in Electric Circuits for Jane Smythe

Question

The efficiency of a small electric motor is being investigated. It is set up to lift a mass while the current and potential difference are both monitored. The measurements are as follows:

Mass raised = 500 g Height raised = 1.2 m

Time taken = 12 s Potential difference = 3.2 V Current = 0.6 A.

(a) Calculate the work done on the mass

(b) Calculate the useful power output of the motor

(c) Calculate the electrical power supplied

(d) Calculate the efficiency of the motor

(e) What happened to the energy that was ‘lost’?

Answer 1

(a) m = 500 g = 0.5 kg

h = 1.2 m

Therefore work done to rise the mass

$W=mgh \\ W=0.5×9.8×1.2 \\ =5.88\; J$

(b) Useful power output

$P= \frac{W}{t} \\ t=12 \;s \\ W=5.88 \;J \\ P= \frac{5.88}{12} \\ =0.49 \; Watt$

(c) Electrical power supplied p= VI

V = potential difference

V = 3.2 V

I = current

I = 0.6 A

$p=3.2 \times 0.6 \\ =1.92 \;W$

(d) efficiency = output power/input power

The output power P = 0.49 W

The input electrical power p = 1.92 W

Therefore efficiency $= \frac{0.49}{1.92}=0.255$

Percentage efficiency $=0.255 \times 100\; \%=25.5\; \%$

(e) The loss of energy converted to heat due to frictional force, the heat generated by the resistance of the circuit, and sound.