Physics · AQA 7408 · Topic 3.5

Electricity

Current, potential difference and resistance, I–V graphs, resistivity, circuits, potential dividers and internal resistance.

  • 6 sub-topics
  • Taught in Year 12
  • 101 practice questions
  • 6 animated lessons
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What the specification says

Every sub-topic AQA 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. 3.5.1 Current electricity
  2. 3.5.1.1Basics of electricityY12
  3. 3.5.1.2Current–voltage characteristicsY12
  4. 3.5.1.3ResistivityY12
  5. 3.5.1.4CircuitsY12
  6. 3.5.1.5Potential dividerY12
  7. 3.5.1.6Electromotive force and internal resistanceY12

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 101 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

Resistors in series and parallel · Year 12 · 5 marks

Figure 1 shows a 12 V supply, of negligible internal resistance, connected to a 4 Ω resistor in series with a parallel pair of 20 Ω and 60 Ω.
12 V4 Ω20 Ω60 Ω
Figure 1
  1. (a)
    Calculate the total resistance of the circuit.
    (2)
  2. (b)
    Calculate the current in the 60 Ω resistor.
    (3)

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Part (a)2 marks

  1. 1R\frac{1}{R}=120+160{}= \frac{1}{20} + \frac{1}{60}⇒R{}\Rightarrow R=15.0 Ω{}= 15.0\ \Omega
  2. RT=19.0 ΩR_T = 19.0\ \mathrm{\Omega}

Part (b)3 marks

  1. I=0.632 AI = 0.632\ \mathrm{A}
  2. V∥V_{\parallel}=0.632×15.0{}= 0.632 \times 15.0=9.47 V{}= 9.47\text{ V}
  3. I60=0.158 AI_{60} = 0.158\ \mathrm{A}

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Electric circuits

Hard

Potential divider · Year 12 · 9 marks

Figure 1 shows a potential divider. The supply has e.m.f. 9.0 V and negligible internal resistance. R₁ = 6.8 kΩ and R₂ is a thermistor of resistance 2.2 kΩ at room temperature.
R₁R₂9.0 VV out
Figure 1
  1. (a)
    Calculate the current in the circuit.
    (2)
  2. (b)
    Calculate the output p.d. across the thermistor.
    (2)
  3. (c)
    Calculate the power dissipated in the thermistor.
    (2)
  4. (d)
    The thermistor is warmed and its resistance halves. Calculate the new output p.d. and explain the change.
    (3)

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Part (a)2 marks

  1. The resistors are in series, so add them. Convert kΩ to Ω.II=VR1+R2{}= \frac{V}{R_1 + R_2}=9.0(6.8+2.2)×103{}= \frac{9.0}{(6.8 + 2.2) \times 10^3}
  2. I=0.00100 AI = 0.00100\ \mathrm{A}

Part (b)2 marks

  1. The p.d. is shared in the ratio of the resistances.VoutV_{out}=R2R1+R2V{}= \frac{R_2}{R_1 + R_2}V=2.26.8+2.2×9.0{}= \frac{2.2}{6.8 + 2.2} \times 9.0
  2. Vout=2.20 VV_{out} = 2.20\ \mathrm{V}

Part (c)2 marks

  1. Use the current found in the first part.P=I2R2P = I^2R_2
  2. P=0.00220 WP = 0.00220\ \mathrm{W}

Part (d)3 marks

  1. Use the new resistance.VoutV_{out}=1.106.8+1.10×9.0{}= \frac{1.10}{6.8 + 1.10} \times 9.0
  2. Vout=1.25 VV_{out} = 1.25\ \mathrm{V}
  3. Explain in terms of the ratio.The thermistor now has a smaller share of the total resistance, so it takes a smaller share of the supply p.d.

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