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How are voltage, current, resistance and power related in an analogue circuit?

Analogue electronics: voltage, current and resistance, Ohm's law, electrical power, and combining resistors in series and in parallel.

An SQA National 5 Engineering Science answer on analogue electronics, covering voltage, current and resistance, Ohm's law V equals IR, electrical power P equals IV, and how to combine resistors in series and in parallel in a circuit.

Generated by Claude Opus 4.810 min answer

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  1. What this key area is asking
  2. Analogue signals
  3. Ohm's law
  4. Electrical power
  5. Resistors in series and parallel
  6. Why analogue electronics matters
  7. Try this

What this key area is asking

The SQA wants you to work with analogue quantities - voltage, current and resistance - using Ohm's law, calculate electrical power, and combine resistors in series and in parallel.

Analogue signals

Ohm's law

Rearranging gives I=VRI = \dfrac{V}{R} and R=VIR = \dfrac{V}{I}. Resistance is the opposition to current: a higher resistance gives a smaller current for the same voltage. For a resistor at constant temperature, a graph of voltage against current is a straight line through the origin.

Electrical power

Power is the rate at which a component transforms electrical energy. If you know the current and voltage, use P=IVP = IV. If you know the current and resistance, use P=I2RP = I^2R. If you know the voltage and resistance, use P=V2/RP = V^2/R. All three are in the data booklet.

Resistors in series and parallel

In series the same current flows through every resistor, so their oppositions add. In parallel the current splits between branches, giving it more paths, so the total resistance falls. A quick sanity check: a series total must be bigger than each resistor; a parallel total must be smaller than each resistor. A useful special case is two equal resistors in parallel - the total is always exactly half of one of them, because the two equal branches share the current equally.

You should also be able to read the voltage and current rules that go with these combinations. In a series circuit the current is the same at every point and the supply voltage is shared between the components. In a parallel circuit the voltage across each branch is the same (the full supply) and the current splits between the branches. These rules are the mirror image of each other, so it pays to learn them as a pair alongside the resistance rules.

Why analogue electronics matters

Analogue electronics is the foundation for the rest of the area. The voltage divider, input transducers, transistor switching and the operational amplifier all rely on Ohm's law and on the series and parallel rules. Getting these calculations automatic frees you to focus on the circuit design.

Try this

Q1. A 6.0 V6.0 \text{ V} supply drives 0.20 A0.20 \text{ A} through a resistor. Calculate the resistance. [2 marks]

  • Cue. R=V/I=6.0/0.20=30 ΩR = V/I = 6.0/0.20 = 30 \text{ }\Omega.

Q2. Resistors of 100 Ω100 \text{ }\Omega and 150 Ω150 \text{ }\Omega are in series. Find the total resistance. [1 mark]

  • Cue. RT=100+150=250 ΩR_T = 100 + 150 = 250 \text{ }\Omega.

Q3. A lamp transfers 24 W24 \text{ W} when 2.0 A2.0 \text{ A} flows through it. Calculate the voltage across it. [2 marks]

  • Cue. V=P/I=24/2.0=12 VV = P/I = 24/2.0 = 12 \text{ V}.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA N5 style3 marksA resistor has a voltage of 9.0 V across it and a current of 0.030 A through it. Calculate its resistance.
Show worked answer →

Use Ohm's law to link voltage, current and resistance.

Relationship: V=IRV = IR, so R=VIR = \dfrac{V}{I}.

Substitution: R=9.00.030=300 ΩR = \dfrac{9.0}{0.030} = 300 \text{ }\Omega.

Markers reward selecting Ohm's law, rearranging for resistance, correct substitution, and a final answer in ohms (Ω\Omega). Watch the decimal: 0.0300.030 A, not 3030 A.

SQA N5 style3 marksA 12 V motor draws a current of 2.5 A. Calculate the power it transfers.
Show worked answer →

Power links voltage and current.

Relationship: P=IVP = IV.

Substitution: P=2.5×12=30 WP = 2.5 \times 12 = 30 \text{ W}.

Markers reward selecting the power relationship, correct substitution and a final answer in watts (W). If only resistance were given you could use P=I2RP = I^2R or P=V2/RP = V^2/R instead.

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