recall and use V = W / Q

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Potential Difference and Electrical Power (Cambridge IGCSE/A‑Level 9702)

Learning Objectives

  • Recall and use the definition V = W ⁄ Q for potential difference, including its SI unit (J C⁻¹ = V).
  • State Ohm’s law for a **linear** resistor (V = IR) and relate voltage, current and resistance.
  • Derive and apply the four power formulas required by the syllabus:
    • P = VI
    • P = W ⁄ t
    • P = I²R
    • P = V²⁄R
  • Distinguish between electromotive force (ε), terminal voltage (V) and internal resistance (r) of a real source.

1. Potential Difference (Voltage)

The potential difference between two points is the energy transferred per unit charge moved between those points.

$$V = \frac{W}{Q}\qquad\text{(V in J C}^{-1}\text{ = V)}$$
  • V – potential difference (volts, V)
  • W – work done or energy transferred (joules, J)
  • Q – charge moved (coulombs, C)

2. Ohm’s Law (Linear Resistor)

For a **linear** resistor the voltage, current and resistance are related by

$$V = I R$$
  • I – current (amperes, A)
  • R – resistance (ohms, Ω)

This relationship allows the power equation to be rewritten in three additional useful forms.

3. Power – Definitions and Derivations

3.1 Basic definitions

  • Power is the rate at which energy is transferred:
    $$P = \frac{W}{t}$$
  • Combining the definition of voltage with current gives the familiar form:
    $$P = V I$$

3.2 Using Ohm’s law

Substituting V = IR into P = VI yields two further expressions:

  • Replace V:
    $$P = (I R) I = I^{2} R$$
  • Replace I (from I = V⁄R):
    $$P = V\left(\frac{V}{R}\right)=\frac{V^{2}}{R}$$

3.3 Summary of power formulas

Formula When it is most useful
P = VI Voltage and current are known (e.g., household appliances)
P = I²R Current and resistance are known (e.g., heating of a wire)
P = V²⁄R Voltage and resistance are known (e.g., transmission‑line loss)
P = W⁄t Energy and time are given directly

4. Internal Resistance and EMF

A real source (battery or generator) has an internal resistance r. Its terminal voltage V differs from the electromotive force (emf) ε according to

$$V = \varepsilon - I r$$
  • ε – emf, the maximum potential difference when no current flows (V).
  • r – internal resistance of the source (Ω).
  • Power delivered to an external load of resistance R:
    $$P_{\text{load}} = V I = I^{2} R$$
  • Power dissipated inside the source itself:
    $$P_{\text{internal}} = I^{2} r$$

5. Units and Symbols

Quantity Symbol SI Unit (name) Unit symbol
Potential difference (voltage) V joule per coulomb V
Work / Energy W joule J
Charge Q coulomb C
Current I coulomb per second A
Resistance R, r volt per ampere Ω
Power P joule per second W
Time t second s
Electromotive force ε volt V

6. Worked Examples

Example 1 – Using P = VI

A heater is connected to a 240 V supply and draws a current of 10 A. Find its power output.

$$P = V I = 240\ \text{V} \times 10\ \text{A} = 2400\ \text{W} = 2.4\ \text{kW}$$

Example 2 – Using P = I²R

A copper wire of resistance 0.5 Ω carries a current of 8 A. Determine the rate at which the wire heats up.

$$P = I^{2} R = (8\ \text{A})^{2} \times 0.5\ \Omega = 64 \times 0.5 = 32\ \text{W}$$

Example 3 – Transmission‑line loss (P = V²⁄R)

A power line transmits 10 kV over a resistance of 2 Ω. Calculate the power lost as heat in the line.

$$P_{\text{loss}} = \frac{V^{2}}{R} = \frac{(10\,000\ \text{V})^{2}}{2\ \Omega} = \frac{1.0 \times 10^{8}}{2} = 5.0 \times 10^{7}\ \text{W} = 50\ \text{MW}$$

Example 4 – Effect of internal resistance

A 12 V battery has an internal resistance of 0.2 Ω and supplies a load of 4 Ω. Find the terminal voltage and the power delivered to the load.

  1. Total resistance: \(R_{\text{tot}} = r + R = 0.2\ \Omega + 4\ \Omega = 4.2\ \Omega\).
  2. Current: \(I = \dfrac{\varepsilon}{R_{\text{tot}}}= \dfrac{12\ \text{V}}{4.2\ \Omega}=2.86\ \text{A}\).
  3. Terminal voltage: \(V = \varepsilon - I r = 12\ \text{V} - (2.86\ \text{A})(0.2\ \Omega)=11.43\ \text{V}\).
  4. Power to load: \(P_{\text{load}} = V I = 11.43\ \text{V} \times 2.86\ \text{A}=32.7\ \text{W}\).

7. Common Mistakes to Avoid

  • Writing the voltage formula as \(V = Q/W\); the correct order is energy **over** charge.
  • Confusing the symbols V (voltage) and v (speed) in the same problem – keep them distinct.
  • Forgetting the time factor when converting between energy (J) and power (W). Remember \(1\ \text{W}=1\ \text{J s}^{-1}\).
  • Mixing up charge (C) with current (A). Current is charge per unit time: \(I = Q/t\).
  • Neglecting internal resistance when a source is asked to “provide a voltage”. Use \(V = \varepsilon - I r\).
  • Applying Ohm’s law to a non‑linear component (e.g., a filament lamp). The syllabus specifies the law holds only for linear resistors.

8. Diagram (Suggested)

Simple circuit showing a source with emf ε and internal resistance r connected to an external resistor R. An ammeter (I) and a voltmeter (V) indicate the current and the terminal voltage.
Battery (ε, r) → ammeter → resistor R → voltmeter across R; arrows indicate direction of charge flow Q and work done W across the resistor.

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