Question

The power usage (P = (E^2)/R) of a strip heater is to be determined by measuring heater resistance and heater voltage drop simultaneously. The resistance is to be measured using an ohmmeter having a resolution of 1 Ohm and an error stated to be within 1% of its reading, and voltage is to be measured using a voltmeter having a resolution of 1 V and a stated error of 1% of reading.

It is expected that the heater will have a resistance of 100 Ohm and use 100 W of power. Determine the uncertainty in power determination to be expected with this equipment at the zeroth-order level (i.e. ignoring the individual instrument errors). Also, recompute the uncertainty at the design stage.

Answer 1

Answer:

Here, P = (E^2)/R and error of voltmeter reading = 1% & error of ohmmeter reading = 1% as given in the question. Also, least count value (resolution) of the voltmeter = 1 V & least count value (resolution) of the ohmmeter = 1 ohm are given in the question.

The uncertainty in power determination to be expected with these equipments at the zeroth-order level is as follows:

= 2 x (least count value of the voltmeter) + (least count value of the ohmmeter) = ( 2 x 1 + 1 ) W = 3 W (Answer).

This means that irrespective of the individual instrument errors, there will always be an uncertainty of 3 W in power determination using these instruments.

The uncertainty in power determination to be expected with these equipments at the design stage ( i.e., as given in the question, R = 100 ohm, P = 100 W, so, E = 100 V ) is as follows:

= 2 x (error of voltmeter reading) + (error of ohmmeter reading) = 2 x 1% + 1% = 3% of Power consumption.

So, at the design stage in this question, = 3% of 100 W = 3% x 100 W = 3 W (Answer).

Note : According to the data in this question, coincidentally these two different uncertainties turned out to be equal. This need not be the case in general. If we had error of voltmeter reading = 2% & error of ohmmeter reading = 1% then at the design stage would be 2 x 2% + 1% = 5% i.e., = 5% of 100 W = 5% x 100 W = 5 W, whereas, the zeroth-order level uncertainty in power determination will still be at 3 W with these two individual instruments of the same resolution as the one given in the question.

Cheers!