define the potential difference across a component as the energy transferred per unit charge

Potential Difference and Power – Cambridge IGCSE / A‑Level Physics (Syllabus 9.2)

Learning Objective

Define the potential difference across a component as the energy transferred per unit charge and use this definition to derive the expressions for electrical power.

Key Concepts

  • Electric potential (V)

  • Potential difference (ΔV)

  • Energy transferred (ΔE or W)

  • Electric charge (Q)

  • Current (I)

  • Resistance (R) – ohmic vs. non‑ohmic

  • Electromotive force (emf, 𝓔) and internal resistance (r)

  • Electrical power (P)

  • Average power in AC circuits (A‑Level extension)

1. Definition of Potential Difference

The potential difference between two points A and B in a circuit is the amount of energy transferred to a charge as it moves from A to B, divided by the magnitude of the charge.

Key formula (syllabus wording)

\[

\Delta V = \frac{\Delta E}{Q}\;=\;\frac{W}{Q}

\]

  • ΔV – potential difference (volts, V)

  • ΔE (or W) – energy transferred (joules, J)

  • Q – charge (coulombs, C)

Relation to Work Done

If the electric field does work W on the charge, then W = ΔE and the same expression can be written as

\[

\Delta V = \frac{W}{Q}

\]

2. Units

QuantitySymbolSI UnitDefinition
Potential differenceΔVvolt (V)1 V = 1 J · C⁻¹
Energy (or work)ΔE, Wjoule (J)1 J = 1 N·m
ChargeQcoulomb (C)1 C = 1 A·s
CurrentIampere (A)1 A = 1 C · s⁻¹
ResistanceRohm (Ω)1 Ω = 1 V · A⁻¹

3. Electrical Power

Power is the rate at which energy is transferred.

\[

P = \frac{\Delta E}{t}

\]

Using the definition of potential difference (ΔV = ΔE / Q) and the definition of current (I = Q / t):

\[

P = \frac{\Delta V \, Q}{t}= \Delta V \, I

\]

Ohm’s Law (explicit statement)

For an ohmic resistor the relationship between potential difference and current is

\[

\Delta V = I R \qquad\text{(Ohm’s law)}

\]

This statement is required by the syllabus before the power formulae are derived.

Three common power formulae (ohmic components)

Substituting Ohm’s law into \(P = \Delta V I\) gives the three equivalent forms used in the exam:

  1. P = VI – most general expression.

  2. P = I²R – replace ΔV with IR.

  3. P = V²/R – replace I with V/R.

These forms are valid only for components that obey Ohm’s law (linear resistors). For non‑ohmic devices the V‑I relationship must be known before using \(P = VI\).

Real Sources – EMF and Internal Resistance

For a practical source (battery, cell) the terminal potential difference is reduced by the voltage drop across its internal resistance:

\[

\Delta V_{\text{terminal}} = \mathcal{E} - I r

\]

  • 𝓔 – electromotive force (emf)

  • r – internal resistance

  • I – current supplied by the source

A‑Level Extension – Average Power in AC Circuits

For a sinusoidal alternating current the average (real) power is

\[

P{\text{avg}} = V{\text{rms}} I_{\text{rms}} \cos\phi

\]

where \(V{\text{rms}}\) and \(I{\text{rms}}\) are the root‑mean‑square values and φ is the phase angle between voltage and current. This formula is not required for the IGCSE but is useful for A‑Level study.

4. Worked Examples

Example 1 – Calculating Potential Difference

Energy transferred ΔE = 2.5 J to a charge Q = 5.0 × 10⁻³ C. Find ΔV.

\[

\Delta V = \frac{\Delta E}{Q}= \frac{2.5\ \text{J}}{5.0\times10^{-3}\ \text{C}} = 5.0\times10^{2}\ \text{V}

\]

ΔV = 500 V.

Example 2 – Power in a Resistor (three forms)

A 12 Ω resistor carries a current I = 3 A. Find the power dissipated.

  • P = VI V = IR = 3 A × 12 Ω = 36 V → P = 36 V × 3 A = 108 W

  • P = I²R P = (3 A)² × 12 Ω = 9 × 12 = 108 W

  • P = V²/R P = (36 V)² / 12 Ω = 1296 / 12 = 108 W

All three methods give the same result, confirming the consistency of the formulae.

Example 3 – Terminal Voltage of a Real Cell

A 1.5 V cell has internal resistance r = 0.2 Ω and supplies I = 2 A. Find the terminal potential difference.

\[

\Delta V_{\text{terminal}} = \mathcal{E} - I r = 1.5\ \text{V} - (2\ \text{A})(0.2\ \Omega) = 1.1\ \text{V}

\]

5. Practical Investigation (AO3)

Objective: Verify the relationship \(P = VI\) for a resistive element.

  1. Construct the circuit: battery → ammeter → resistor → voltmeter (across the resistor) → back to battery.

  2. Measure the current (I) and the voltage across the resistor (V) for three different resistor values or three different battery voltages.

  3. Calculate the power using \(P = VI\) and compare with the power obtained from \(P = I^{2}R\) (using the measured resistance).

  4. Discuss sources of uncertainty (instrument tolerance, contact resistance, temperature change) and comment on the agreement of the two methods.

6. Common Misconceptions

  • ΔV is not total energy. It is energy per coulomb, not the total joules transferred.

  • Higher voltage does not automatically mean higher power. Power also depends on current: \(P = VI\).

  • Power formulae are for ohmic components only. For non‑linear devices the V‑I relationship must be known before applying \(P = VI\).

  • Internal resistance is often ignored. In real sources the terminal voltage is reduced by the drop \(Ir\).

7. Suggested Diagram

Simple circuit showing battery, ammeter, resistor, and voltmeter with ΔV across the resistor

Simple circuit used in the practical investigation. The arrows indicate conventional current flow; ΔV is measured across the resistor, and W (work done on the charge) is shown conceptually.

8. Summary

  • Potential difference is defined as ΔV = ΔE / Q = W / Q (energy per unit charge).

  • SI unit: 1 V = 1 J · C⁻¹.

  • Electrical power: P = ΔE / t = ΔV I.

  • For an ohmic resistor, Ohm’s law gives ΔV = IR, leading to the three equivalent forms:

    • P = VI

    • P = I²R

    • P = V²/R

  • Real sources: \(\Delta V_{\text{terminal}} = \mathcal{E} - I r\).

  • A‑Level extension: average AC power \(P{\text{avg}} = V{\text{rms}} I_{\text{rms}} \cos\phi\).

  • Practical verification involves measuring V and I, calculating P, and analysing uncertainties.