levo

Resistance

Electrical power and energy transfer

  • 3 things to know
  • Component 2

Overview

Recall questions on P = IV and the derived forms I²R and V²/R, choosing the right form for the data given, worked power and energy calculations, the kilowatt hour, and why the power dissipated quadruples when the current doubles.

What you need to know

Answer each one before you reveal it.

Your recallNot started
Key term

State the equation for the power dissipated in a component in terms of current and potential difference.

Key term

Show how P = IV follows from the definitions of current and potential difference.

Key term

State the two forms of the power equation that use resistance.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    State the equation for the power dissipated in a component in terms of current and potential difference.

    Model answer

    The power is P = IV, where P is in watts (W), I is the current in amperes (A) and V is the potential difference in volts (V).

  2. 2

    Show how P = IV follows from the definitions of current and potential difference.

    Model answer

    Power is energy per unit time, W/t, and W = QV, so P = QV/t. Since Q/t is the current, P = IV.

  3. 3

    State the two forms of the power equation that use resistance.

    Model answer

    The two forms are P = I²R and P = V²/R.

  4. 4

    Derive P = I²R from P = IV.

    Model answer

    Substituting V = IR into P = IV gives P = I × IR, so P = I²R.

  5. 5

    Derive P = V²/R from P = IV.

    Model answer

    Substituting I = V/R into P = IV gives P = (V/R) × V, so P = V²/R.

  6. 6

    A 6.0 Ω resistor is connected across a 12 V supply. Calculate the power dissipated in it.

    Model answer

    The power is P = V²/R = 144 ÷ 6.0 = 24 W. The current is 2.0 A, and I²R = 4.0 × 6.0 gives the same 24 W.

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