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A-LevelPhysicsD.C. circuits9702/41 · May/June 2019 · Q78 Marks

The circuit for an inverting amplifier incorporating an ideal operational amplifier (op-amp) is shown in Fig. 7.1. [Figure 7.1]

  1. (a)(i)For the circuit of Fig. 7.1: explain why point P is known as a virtual earth[3]
  2. (a)(ii)calculate the gain of the amplifier.[2]
  3. (b)When the op-amp is saturated, the potential difference across the LED is 2.3 V. Calculate the minimum resistance of resistor R so that the current in the LED is limited to 30 mA.[3]

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Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

  1. [1] (amplifier) gain is very large/infinite
  2. [1] for amplifier not to saturate, V+ = V- or feedback (loop) ensures V+ = V-
  3. [1] V+ is at earth/0 V so V-is (almost) at earth/0 V
  4. [1] gain = (-)5200/800 or (−)5.2/0.80
  5. [1] = -6.5
  6. [1] (at saturation,) VOUT = 5 V
  7. [1] p.d. across R = 5 – 2.3 = 2.7 (V)
  8. [1] resistance = 2.7/(30 × 10⁻³) = 90 Ω
  9. or
  10. [1] Rdiode = 2.3/0.030 = 77 Ω
  11. [1] Rtotal = 5.0/0.030 = 167 Ω
  12. [1] Rresistor (= 167 – 77) = 90 Ω
  13. or
  14. [1] Rdiode = 2.3/0.030 = 77 Ω
  15. [1] 77/(Rresistor + 77) × 5 = 2.3
  16. [1] Rresistor = 90 Ω
  17. or
  18. [1] Rdiode = 2.3/0.030 = 77 Ω
  19. [1] Rresistor = 77 × (2.7/2.3)
  20. [1] Rresistor = 90 Ω

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About this question

Topic
D.C. circuits, A-Level Physics (syllabus 9702). Structured question, 8 marks.
Source
A-Level Physics, May/June 2019 Paper 4 Variant 1, question 7.

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