Wheatstone bridges and Potentiometers

ChemBoss

Can anyone help me with 6 and 7 please?

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Reply 1

uberteknik

Original post by ChemBoss

Can anyone help me with 6 and 7 please?

This is just a different way of looking at potential dividers.

Q6

The potentiometer output voltage is picked-off across the lower part (13mm) of the resistance at that wiper position measure w.r.t. the supply 0V terminal.

The potentiometer is linear (same resistance per unit length along the whole resistance track).

You can do this in one of two ways:

a) use the ratio of wiper length (13mm) to whole track length (25mm) and multiply by the full potentiometer resistance (470 ohms).

b) divide the pot resistance (470 ohms) by the full track length (25mm) to get an ohm/mm resistance and then multiply by the needed (13mm) length to get the lower part resistance.

In both cases, the resistance answer should be the same.

So then use that lower resistance together with the total series resistance (470+330) in the potential divider formula used in the previous question you asked about to find the required voltage.

Q7

The output voltage is measured between two potential dividers (left and right) half's of the Wheatstone Bridge. The output voltage will be balanced (zero potential) when the two halves are also balanced. i.e. the resistance values are mirrored so that the potential voltage at each of the output terminals when measured with respect to the supply common, are both the same. (i.e. no voltage is measured between the o/p terminals)

So the thermistor resistance needs to be 270 ohms to balance the bridge. You are given the resistance @ 0^oC (200 ohms) so use that and the resistance coefficient (0.004^oC^-1) to find the temperature needed to make up the remaining 70 ohms.

Spoiler

(edited 9 years ago)

Reply 2

ChemBoss

Thank you!