(a) Describe, with a labelled diagram, the structure of a metal-wire strain gauge. (b) In a strain gauge, the increase in

Q#8 (Past Exam Paper – March 2017 Paper 42 Q7)

(a) Describe, with a labelled diagram, the structure of a metal-wire strain gauge. [3]

(b) In a strain gauge, the increase in resistance ΔR depends on the increase in length ΔL.

The variation of ΔR with ΔL is shown in Fig. 7.1.

Fig. 7.1

The strain gauge is connected into a circuit incorporating an ideal operational amplifier

(op-amp), as shown in Fig. 7.2.

Fig. 7.2

(i) The strain gauge is initially unstrained with resistance 300.0 Ω.

Use data from Fig. 7.1 to calculate the increase in length ΔL of the strain gauge that gives rise to a potential of +2.00 V at point A in Fig. 7.2. [3]

(ii) The strain gauge undergoes a further increase in length beyond the value in (b)(i).

State and explain which one of the light-emitting diodes, X or Y, will be emitting light. [4]

[Total: 10]

Solution:

(a)

The metal-wire strain gauge consists of a folded fine metal wire mounted on a flexible insulating (plastic) envelope.

(b)

(i)

{A potential divider circuit is formed. So, we can use the potential divider formula. Bringing the p.d.s together:

Ratio of p.d. = Ratio of resistances

The p.d. across the 153.0 Ω resistor is 2.00 V and the p.d. across the both the strain gauge and the resistor is 6.00 V.

Let the resistance of the strain gauge be R.

p.d. across 153.0 Ω resistor / total p.d. = 153.0 Ω / total resistance}

2.00 / 6.00 = 153.0 / (R + 153.0)

OR

{p.d. across strain gauge = 6.00 – 2.00 = 4.00 V

p.d. across strain gauge / total p.d. = R / total resistance}}

4.00 / 6.00 = R / (R + 153.0)   (so R = 306.0)

{Simplifying:

2 (R + 153) = 3R

2R + 306 = 3R

3R – 2R = 306

R = 306.0 Ω}

{Initially, the resistance of the strain gauge was 300 Ω.

Change in resistance:}

ΔR = 306.0 – 300.0 = 6.0 (Ω)

{From the graph, this change in resistance corresponds to a change in length of}

ΔL = 8(.0) × 10^-5 m

(ii)

R or ΔR increases    

                                     

V⁺ < V^- or VA < 2.00 or V⁺ / VA decreases      

output is negative / –5 V      

                          

diode X emits light / is ‘on’

{This is a comparator circuit.

A further increase in length would result in an increase in the resistance of the strain gauge.

From Ohm’s law (V = IR), the p.d. across the strain gauge would be greater (than 4.00 V) and thus, the p.d. across the resistor would be less – point A would be at a lower (than 2.00 V) potential.

This results in the input potential (= 2.00 V) to V^- to be greater than the input potential (less than 2.00 V) to V⁺. Thus, the output potential of the op-amp would be negative.

Current flows from a relatively high(er) potential (here, earth) to a relatively low(er) potential (here, the output of the op-amp since it is negative). So, currents flows upward – diode X would light up.}

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