Describe, with a labelled diagram, the structure of a metal-wire strain gauge.

Question 8

(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]

Reference: Past Exam Paper – March 2017 Paper 42 Q7

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.}

Two large horizontal metal plates are separated by 4 mm. The lower plate is at a potential of –80 V.

Question 13

Two large horizontal metal plates are separated by 4 mm. The lower plate is at a potential of –80 V.

Which potential should be applied to the upper plate to create an electric field of strength

60 000 V m^-1 upwards in the space between the plates?

A –320 V                     B –160 V                     C +160 V                     D +320 V

Reference: Past Exam Paper – November 2016 Paper 11 & 13 Q32

Solution:

Answer: A.

Electric field strength E = V / d

Where V is the potential difference between the plates

E = V / d

V = E × d = 60 000 × 4×10^-3

Potential difference V = 240 V

The potential difference between the plates is 240 V. The bottom plate is at a potential of –80 V. The difference between the lower plate and the upper plate should be 240 V.

The upper plate be       EITHER greater than –80 V by 240 V

OR it could be smaller than –80 V by –240 V

But since the electric field is upwards, the top plate must be at a lower potential than the bottom plate (since electric field is drawn from high(er) potential to low(er) potential).

Therefore a potential of (–80 – 240 =) –320 V is required on the top plate [A is correct].

We need to consider the direction of the field. Merely calculating correct potential difference would make C a possible answer. But when the direction of the field is considered, only A can be correct.

It may be helpful to remember that the field gives the direction of the force on a positive charge. For a positive charge to experience a force upwards, the top plate must be at a more negative potential than the bottom plate.