Physics · Pearson Edexcel 9PH0 · Topic 3

Electric circuits

Current, potential difference and resistance, resistivity, circuits, potential dividers and internal resistance.

  • 18 sub-topics
  • Taught in Year 12
  • 209 practice questions
  • 18 animated lessons
Practice questions (209) Lessons Worksheets All Edexcel Physics topics

What the specification says

Every sub-topic Pearson Edexcel can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 31Electric current as rate of flow of chargeY12
  2. 32Potential difference as work done per unit chargeY12
  3. 33Resistance and Ohm's lawY12
  4. 34Current and conservation of chargeY12
  5. 35Potential difference and conservation of energyY12
  6. 36Resistors in series and parallelY12
  7. 37Electrical powerY12
  8. 38I–V characteristicsY12
  9. 39Resistivity, R = ρl/AY12
  10. 40CP2: Determine the electrical resistivity of a materialY12
  11. 41I = nqvA and the range of resistivitiesY12
  12. 42Potential along a uniform current-carrying wireY12
  13. 43Potential dividersY12
  14. 44Potential dividers with thermistors and LDRsY12
  15. 45E.m.f. and internal resistanceY12
  16. 46CP3: Measure the e.m.f. and internal resistance of a cellY12
  17. 47Resistance and temperatureY12
  18. 48Resistance and illuminationY12

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 209 questions in this topic, free, no account needed.

Electric circuits

Easy

Resistance of a wire · Year 12 · 2 marks

A wire is 0.80 m long with cross-sectional area 2.0×10−7 m22.0 \times 10^{-7}\ \mathrm{m^2}. Its material has resistivity 4.9×10−7 Ω m4.9 \times 10^{-7}\ \Omega\,\mathrm{m}.
  1. Calculate the resistance of the wire.
    (2)
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  1. Resistivity links resistance to the length and area of a wire.RR=ρLA{}= \dfrac{\rho L}{A}=4.9×10−7×0.802.0×10−7{}= \dfrac{4.9 \times 10^{-7} \times 0.80}{2.0 \times 10^{-7}}
  2. Give the unit, ohms.R=2.0 ΩR = 2.0\ \Omega

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Electric circuits

Medium

Potential difference and conservation of energy · Year 12 · 4 marks

Figure 1 shows a 9 V supply of negligible internal resistance connected to resistors of 10 Ω and 40 Ω in series.
9 V10 Ω40 Ω
Figure 1
  1. (a)
    Calculate the current in the circuit.
    (2)
  2. (b)
    Calculate the p.d. across each resistor, and show that they add up to the supply p.d.
    (2)

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Part (a)2 marks

  1. I=910+40I = \frac{9}{10 + 40}
  2. I=0.180 AI = 0.180\ \mathrm{A}

Part (b)2 marks

  1. 1.80 V and 7.20 V1.80\text{ V and } 7.20\text{ V}
  2. They add to 9 V: Kirchhoff's second law (conservation of energy).

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Electric circuits

Hard

Drift velocity · Year 12 · 6 marks

A copper wire of diameter 0.5 mm carries a current of 1.5 A. Copper has 8.5 × 10²⁸ free electrons per m³.
  1. (a)
    Calculate the mean drift velocity of the electrons.
    (3)
  2. (b)
    The wire is joined to a thinner wire of the same metal. Explain what happens to the drift velocity.
    (2)
  3. (c)
    Explain why semiconductors have much higher resistivity than metals.
    (1)

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Part (a)3 marks

  1. AA=1.96×10−7 m2{}= 1.96 \times 10^{-7}\ \mathrm{m^2}
  2. v=InAqv = \frac{I}{nAq}
  3. vv=5.62×10−4 m s−1{}= 5.62 \times 10^{-4}\ \mathrm{m\,s^{-1}}

Part (b)2 marks

  1. The current is the same in both (series).
  2. A is smaller, so v is larger (v ∝ 1/A).

Part (c)1 mark

  1. They have far fewer charge carriers per unit volume (n is much smaller).

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