recall and use V = W / Q
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.
- Total resistance: \(R_{\text{tot}} = r + R = 0.2\ \Omega + 4\ \Omega = 4.2\ \Omega\).
- Current: \(I = \dfrac{\varepsilon}{R_{\text{tot}}}= \dfrac{12\ \text{V}}{4.2\ \Omega}=2.86\ \text{A}\).
- Terminal voltage: \(V = \varepsilon - I r = 12\ \text{V} - (2.86\ \text{A})(0.2\ \Omega)=11.43\ \text{V}\).
- 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)
