Physics · OCR H556 · Topic 4.3

Electrical circuits

Series and parallel circuits, internal resistance, potential dividers.

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

Every sub-topic OCR 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. 4.3.1Series and parallel circuitsY12
  2. 4.3.2Internal resistanceY12
  3. 4.3.3Potential dividersY12

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 65 questions in this topic, free, no account needed.

Electric circuits

Easy

Resistors in parallel · Year 12 · 1 mark

  1. Resistors of 6 Ω and 3 Ω are connected in parallel. What is their combined resistance?
    (1)

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  1. 1R\dfrac{1}{R}=1R1{}= \dfrac{1}{R_1}+1R2{}+ \dfrac{1}{R_2}. The answer is always less than the smaller resistor.B: 2.00 Ω2.00\ \mathrm{\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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