OCR · A-Level · Physics

    Course version: H556 · Version from Sept 2015

    (i) potential divider equations e.g. Vout = [R₂/(R₁ + R₂)]Vin and V₁/V₂ =…

    Full specification wording

    Potential dividers: (i) potential divider equations e.g. Vout = [R₂/(R₁ + R₂)]Vin and V₁/V₂ = R₁/R₂

    A potential divider splits a supply voltage between components in series.

    Checked against the OCR H556 specification (opens in a new tab) on .

    • Point 3 of 4
    • 4 key terms

    Study notes

    From the checked teaching notes for this specification point.

    The idea

    A potential divider splits a supply voltage between components in series. For two resistors R₁ and R₂ in series across a supply Vin, the same current flows through both, so the voltage across each is proportional to its resistance. The output taken across R₂ is Vout = [R₂/(R₁ + R₂)]Vin, and the ratio of the voltages across the two resistors is V₁/V₂ = R₁/R₂. For example, with Vin = 12 V, R₁ = 200 Ω and R₂ = 400 Ω, Vout = [400/(200 + 400)] × 12 V = 8 V, and V₁ = 4 V so V₁/V₂ = 4/8 = 0.5 = 200/400. The equations assume ideal components and no significant load current drawn from the output.

    What earns the marks

    • State that in a series divider the current is the same through each resistor, so the potential difference across each resistor is proportional to its resistance.
    • Apply Vout = [R₂/(R₁ + R₂)]Vin to find the output voltage across the lower resistor R₂.
    • Apply V₁/V₂ = R₁/R₂ to compare the potential differences across two series resistors.
    • Substitute values with consistent units and calculate the output voltage or voltage ratio correctly.
    • Recognise that the simple divider equations assume negligible current is drawn from the output, so a connected load changes the output voltage.

    Common mistakes and how to put them right

    • Swapping the numerator so the output is taken across the wrong resistor.

      Put it right: Vout across R₂ uses R₂ in the numerator, Vout = [R₂/(R₁ + R₂)]Vin.

    • Adding resistances in parallel when the divider resistors are in series.

      Put it right: The denominator is the series sum R₁ + R₂.

    • Assuming the output voltage is unchanged when a load is connected.

      Put it right: A load draws current and reduces the output voltage unless its resistance is much larger than the divider resistances.

    Exam tips

    • Write down the equation before substituting so the numerator and denominator are clearly assigned to the correct resistors.
    • Check the ratio answer by seeing whether the larger resistor has the larger share of the supply voltage.
    • Keep units consistent, converting kΩ to Ω only if the question requires it, and state the unit with the final answer.

    Try the lesson

    The start of the Potential dividers: (i) potential divider equations e.g. Vout = [R₂/(R₁ + R₂)]Vin and V₁/V₂ = R₁/R₂ lesson, made from checked teaching notes for this specification point.

    Listen

    A potential divider splits a supply voltage between components in series. For two resistors R₁ and R₂ in series across a supply Vin, the same current flows through both, so the voltage across each is proportional to its resistance. The output taken across R₂ is Vout = [R₂/(R₁ + R₂)]Vin, and the ratio of the voltages across the two resistors is V₁/V₂ = R₁/R₂. For example, with Vin = 12 V, R₁ = 200 Ω and R₂ = 400 Ω, Vout = [400/(200 + 400)] × 12 V = 8 V, and V₁ = 4 V so V₁/V₂ = 4/8 = 0.5 = 200/400. The equations assume ideal components and no significant load current drawn from the output. A strong answer earns marks for these points. State that in a series divider the current is the same through each resistor, so the potential difference across each resistor is proportional to its resistance. Apply Vout = [R₂/(R₁ + R₂)]Vin to find the output voltage across the lower resistor R₂. Apply V₁/V₂ = R₁/R₂ to compare the potential differences across two series resistors. Substitute values with consistent units and calculate the output voltage or voltage ratio correctly. A common slip: Swapping the numerator so the output is taken across the wrong resistor. Instead, Vout across R₂ uses R₂ in the numerator, Vout = [R₂/(R₁ + R₂)]Vin. Exam tip: Write down the equation before substituting so the numerator and denominator are clearly assigned to the correct resistors.

    See it

    A top-down flowchart. The first box states that the supply Vin is across R₁ and R₂ in series. An arrow leads to a box saying the same current flows through both resistors. An arrow leads to a box saying the voltage across each resistor is proportional to its resistance. From this box two arrows branch out: one to a box giving the output across R₂ as Vout = [R₂/(R₁ + R₂)]Vin, and one to a box giving the voltage ratio V₁/V₂ = R₁/R₂. Both of these boxes lead to a box stating the equations assume negligible current drawn from the output. A final arrow leads to a box saying a connected load draws current and reduces the output voltage.
    Shows how a series potential divider leads to the two divider equations and the assumption that no significant output current is drawn.

    Key words

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    Potential divider

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    Voltage division proportional to resistance

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    Using and rearranging divider equations

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    Effect of loading on a divider output

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    See every definition
    Potential divider
    A circuit that splits a supply voltage between components in series.
    Voltage division proportional to resistance
    In a series divider the current is the same through each resistor, so the potential difference across each resistor is proportional to its resistance.
    Using and rearranging divider equations
    Applying Vout = [R₂/(R₁ + R₂)]Vin to find the output voltage across the lower resistor R₂, and V₁/V₂ = R₁/R₂ to compare the potential differences across two series resistors.
    Effect of loading on a divider output
    The simple divider equations assume negligible current is drawn from the output, so a connected load changes the output voltage.

    Label the diagram

    Some boxes have lost their names. Pick the right name for each number.

    The diagram again, with 3 boxes numbered instead of named. A top-down flowchart. The first box states that the supply Vin is across R₁ and R₂ in series. An arrow leads to a box saying the same current flows through both resistors. An arrow leads to a box saying the voltage across each resistor is proportional to its resistance. From this box two arrows branch out: one to a box giving the output across R₂ as Vout = [R₂/(R₁ + R₂)]Vin, and one to a box giving the voltage ratio V₁/V₂ = R₁/R₂. Both of these boxes lead to a box stating the equations assume negligible current drawn from the output. A final arrow leads to a box saying a connected load draws current and reduces the output voltage.

    • 1Box 1

    • 2Box 2

    • 3Box 3

    0 of 3 labelled

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