(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. 8 ΔR/Ω 6+ 4 2- 0 0 2 4 6 8 10 ΔL/10⁻⁵ m Fig. 7.1 The strain gauge is connected into a circuit incorporating an ideal operational amplifier (op-amp), as shown in Fig. 7.2. +6.00 VO strain gauge +5 V +2.00 V OVO 153.0 Ω A + -5 V X Y Fig. 7.2
- (a)Describe, with a labelled diagram, the structure of a metal-wire strain gauge.[3]
- (b)(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]
- (b)(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]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2017). The mark or mark type is shown in brackets.
- [B1] correct grid shape (of wire)
- [B1] fine wire/foil strip
- [B1] plastic/insulating envelope containing the wire
- [C1] 2.00/6.00 = 153.0 / (R + 153.0) or 4.00 / 6.00 = R / (R + 153.0) (so R = 306.0)
- [C1] ΔR = 306.0 – 300.0 = 6.0 (Ω)
- [A1] so ΔL = 8(.0) × 10⁻⁵ m
- [B1] R or ΔR increases
- [M1] V⁺ < V⁻ or V_A < 2.00 or V⁺ / V_A decreases
- [A1] output is negative / –5 V
- [A1] diode X emits light/is 'on'
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