Potential Difference, Power and the Fundamentals of Electricity
Learning Objectives
- Recall and use the three power relationships:
P = VI, P = I²R, P = \dfrac{V^{2}}{R}
- Explain electric current, charge carriers and the quantised nature of charge.
- Distinguish between emf, terminal voltage and internal resistance.
- Apply Ohm’s law, the definition of resistance, resistivity and the temperature‑coefficient of resistance.
- Analyse series and parallel circuits using Kirchhoff’s laws.
- Read and draw circuit diagrams using the standard Cambridge symbols.
- Design, carry out and evaluate simple DC‑circuit experiments (AO3).
1. Electric Current and Charge Carriers
- Current (I) – rate of flow of electric charge:
$$I = \frac{Q}{t}\quad\text{(A)}$$
- Charge is quantised; the elementary charge e = 1.60 × 10⁻¹⁹ C.
- Microscopic expression for a conductor containing charge carriers of charge q moving with drift speed v:
$$I = n\,q\,A\,v$$
- n – number of carriers per unit volume (m⁻³)
- A – cross‑sectional area (m²)
- v – drift speed (m s⁻¹)
- Typical charge carriers:
- Metals – electrons (q = –e)
- Electrolytes – positive and negative ions
- Semiconductors – electrons and holes
2. Potential Difference (Voltage), EMF and Internal Resistance
- Potential difference between two points is the work done per unit charge:
$$\Delta V = \frac{W}{Q}\quad\text{(V)}$$
- Electromotive force (emf, 𝓔) – energy supplied per coulomb by a source when no current flows.
- Real sources possess an internal resistance r. When a current I flows, the terminal voltage V is:
$$V = \mathcal{E} - I r$$
- Diagram (battery model): a cell of emf 𝓔 in series with internal resistance r, connected to an external load R. The voltage across the load is the terminal voltage V.
3. Resistance, Resistivity and Temperature Effects
- Ohm’s law (definition of resistance):
$$V = I R$$
- For a uniform conductor of length L and cross‑section A:
$$R = \rho\,\frac{L}{A}$$
where ρ is the resistivity (Ω·m).
- First‑order temperature dependence:
$$R = R_{0}\,[1 + \alpha (T - T_{0})]$$
- α – temperature coefficient (≈ +0.004 K⁻¹ for copper; negative for most semiconductors).
- Special resistive devices:
- Light‑dependent resistor (LDR) – resistance ↓ with light intensity.
- Thermistor – NTC (resistance ↓ with temperature) or PTC (resistance ↑ with temperature).
4. Electrical Power
Power is the rate at which electrical energy is transferred or converted.
$$P = \frac{E}{t}\quad\text{(W)}$$
4.1 Derivation of the Three Forms
- Work done by a charge moving through a potential difference:
$$W = Q\Delta V$$
- Since $I = \dfrac{Q}{t}$, then $Q = I t$. Substituting:
$$W = (I t)\Delta V$$
- Dividing by $t$ gives the power:
$$P = I\Delta V$$
- Using Ohm’s law $\Delta V = I R$:
$$P = I (I R) = I^{2}R$$
- Or replacing $I$ with $\dfrac{\Delta V}{R}$:
$$P = \Delta V\left(\frac{\Delta V}{R}\right)=\frac{\Delta V^{2}}{R}$$
4.2 Summary of Power Relationships
| Form |
Expression |
Most useful when… |
| $P = VI$ |
Power in terms of voltage and current |
Both $V$ and $I$ are known (e.g. mains supply) |
| $P = I^{2}R$ |
Power in terms of current and resistance |
Current and resistance are known (heating of a resistor) |
| $P = \dfrac{V^{2}}{R}$ |
Power in terms of voltage and resistance |
Voltage across a component and its resistance are known |
5. Series and Parallel Circuits & Kirchhoff’s Laws
5.1 Combining Resistors
- Series – same current, total resistance:
$$R_{\text{series}} = R_{1}+R_{2}+ \dots +R_{n}$$
- Parallel – same voltage, total resistance:
$$\frac{1}{R_{\text{parallel}}}= \frac{1}{R_{1}}+\frac{1}{R_{2}}+\dots+\frac{1}{R_{n}}$$
5.2 Kirchhoff’s Laws
- Current Law (KCL): The algebraic sum of currents at a junction is zero.
$$\sum I_{\text{in}} = \sum I_{\text{out}}$$
- Voltage Law (KVL): The algebraic sum of potential differences around any closed loop is zero.
$$\sum V = 0$$
5.3 Worked Example – Mixed Circuit
Given: $R_{1}=4\;\Omega$, $R_{2}=6\;\Omega$ in series; this combination is in parallel with $R_{3}=12\;\Omega$. The circuit is connected to a $24\;\text{V}$ supply. Find the total current drawn.
- Series part: $R_{12}=R_{1}+R_{2}=10\;\Omega$.
- Parallel total:
$$\frac{1}{R_{\text{T}}}= \frac{1}{R_{12}}+\frac{1}{R_{3}}=
\frac{1}{10}+\frac{1}{12}= \frac{11}{60}$$
$$R_{\text{T}}= \frac{60}{11}\approx5.45\;\Omega$$
- Total current: $I = \dfrac{V}{R_{\text{T}}}= \dfrac{24}{5.45}\approx4.4\;\text{A}$.
6. Circuit Symbols (Cambridge Syllabus 6)
| Symbol | Name | Typical Use |
|---|
| ●—|— | Battery (emf) | Provides $\mathcal{E}$ |
| ●—— | Resistor | Ohmic element |
| ●—<|>— | Capacitor | Stores charge (DC circuits – transient analysis) |
| ●—∨— | Switch | Open/close circuit |
| ●—→— | Diode | Allows current in one direction |
| ●—→|— | LED | Light‑emitting diode |
Potential Divider
Two resistors $R_{1}$ and $R_{2}$ in series across a supply $V_{\text{s}}$. The voltage across $R_{2}$ is:
$$V_{R_{2}} = V_{\text{s}}\frac{R_{2}}{R_{1}+R_{2}}$$
Useful for obtaining a required voltage from a higher supply.
7. Practical Skills (AO3)
- Planning an experiment – sketch the circuit, list components, choose measurement devices (ammeter, voltmeter, stopwatch).
- Handling uncertainties – record least counts, calculate absolute and percentage uncertainties, propagate using:
$$\frac{\Delta P}{P}= \frac{\Delta V}{V}+ \frac{\Delta I}{I}\quad\text{(for }P=VI\text{)}$$
$$\frac{\Delta P}{P}= 2\frac{\Delta I}{I}\quad\text{(for }P=I^{2}R\text{)}$$
$$\frac{\Delta P}{P}= 2\frac{\Delta V}{V}+ \frac{\Delta R}{R}\quad\text{(for }P=V^{2}/R\text{)}$$
- Common sources of error
- Internal resistance of the source
- Contact resistance at terminals
- Instrument loading (voltmeter draws current, ammeter adds resistance)
- Temperature change of resistors during prolonged current flow
- Data presentation – clear tables, correct units, graphs (e.g. $V$ vs $I$ to verify Ohm’s law). Determine $R$ from the gradient, include error bars.
8. Worked Examples (Power Focus)
Example 1 – Resistor Heating
A $10\;\Omega$ resistor carries a current of $2\;\text{A}$. Find the power dissipated.
$$P = I^{2}R = (2)^{2}\times10 = 40\;\text{W}$$
Example 2 – Lamp on Mains
A lamp is connected across a $240\;\text{V}$ supply and draws $0.5\;\text{A}$. Determine the power consumed and the lamp’s resistance.
$$P = VI = 240\times0.5 = 120\;\text{W}$$
$$R = \frac{V}{I} = \frac{240}{0.5}=480\;\Omega$$
Example 3 – Heater Rating
A heater is rated at $1500\;\text{W}$ on a $120\;\text{V}$ supply. Find the current and the resistance.
$$I = \frac{P}{V}= \frac{1500}{120}=12.5\;\text{A}$$
$$R = \frac{V^{2}}{P}= \frac{120^{2}}{1500}=9.6\;\Omega$$
Example 4 – Internal Resistance of a Cell
A 1.5 V cell has an internal resistance of $0.5\;\Omega$. When it supplies a current of $2\;\text{A}$, what is the terminal voltage?
$$V = \mathcal{E} - I r = 1.5 - (2)(0.5)=0.5\;\text{V}$$
Example 5 – Potential Divider
Two resistors, $R_{1}=2\;\text{k}\Omega$ and $R_{2}=3\;\text{k}\Omega$, are in series across a $12\;\text{V}$ battery. Find the voltage across $R_{2}$.
$$V_{R_{2}} = 12\;\text{V}\times\frac{3}{2+3}=12\times0.6=7.2\;\text{V}$$
9. Practice Questions
- A $5\;\Omega$ resistor has a current of $3\;\text{A}$ flowing through it. Calculate the voltage across it and the power dissipated.
- A device operates at $12\;\text{V}$ and consumes $24\;\text{W}$. Determine the current drawn and the equivalent resistance.
- Two resistors, $R_{1}=8\;\Omega$ and $R_{2}=12\;\Omega$, are connected in series across a $24\;\text{V}$ battery. Find the total power supplied by the battery.
- A heating element is designed to produce $2000\;\text{W}$ when connected to a $240\;\text{V}$ supply. What resistance must the element have?
- If a current of $0.2\;\text{A}$ flows through a $50\;\Omega$ resistor, what is the rate of energy conversion (power) in kilojoules per hour?
- A 9 V battery has an internal resistance of $1\;\Omega$. When a $10\;\Omega$ lamp is connected:
- Find the terminal voltage across the lamp.
- Find the current through the circuit.
- Find the power dissipated in the lamp.
- In the circuit below (series‑parallel combination), the supply voltage is $15\;\text{V}$. Determine the total current drawn.
Resistors: $R_{1}=2\;\Omega$ (in series with the parallel pair), $R_{2}=3\;\Omega$ and $R_{3}=6\;\Omega$ in parallel.
Answers (for self‑checking)
- $V = IR = 3\times5 = 15\;\text{V}$; $P = I^{2}R = 3^{2}\times5 = 45\;\text{W}$.
- $I = P/V = 24/12 = 2\;\text{A}$; $R = V/I = 12/2 = 6\;\Omega$.
- Total resistance $R_{\text{T}} = 8+12 = 20\;\Omega$; $P = V^{2}/R_{\text{T}} = 24^{2}/20 = 28.8\;\text{W}$.
- $R = V^{2}/P = 240^{2}/2000 = 28.8\;\Omega$.
- $P = I^{2}R = 0.2^{2}\times50 = 2\;\text{W}$.
$2\;\text{W}=2\;\text{J s}^{-1}$ → in one hour: $2\times3600 = 7200\;\text{J}=7.2\;\text{kJ}$.
- Total resistance $R_{\text{T}} = 10 + 1 = 11\;\Omega$.
$I = \mathcal{E}/R_{\text{T}} = 9/11 = 0.818\;\text{A}$.
Terminal voltage $V = I\times10 = 8.18\;\text{V}$.
Power $P = V I = 8.18\times0.818 \approx 6.7\;\text{W}$.
- Parallel pair: $1/R_{\text{p}} = 1/3 + 1/6 = 1/2$ → $R_{\text{p}} = 2\;\Omega$.
Total $R_{\text{T}} = R_{1}+R_{\text{p}} = 2+2 = 4\;\Omega$.
$I_{\text{total}} = V/R_{\text{T}} = 15/4 = 3.75\;\text{A}$.
10. Key Points to Remember
- Power can be written as
P = VI, P = I²R or P = V²/R. Choose the form that matches the quantities you know.
- Current is the flow of quantised charge; $I = nqAv$ links microscopic motion to macroscopic current.
- EMF is the ideal voltage of a source; the terminal voltage is reduced by internal resistance: $V = \mathcal{E} - I r$.
- Resistance depends on material, dimensions and temperature: $R = \rho L/A$ and $R = R_{0}[1+\alpha(T-T_{0})]$.
- Series: $R_{\text{total}} = \sum R$; Parallel: $\displaystyle \frac{1}{R_{\text{total}}}= \sum \frac{1}{R}$.
- KCL – algebraic sum of currents at a junction is zero; KVL – algebraic sum of voltages round any closed loop is zero.
- When analysing circuits, start by reducing series/parallel groups, then apply KVL/KCL as needed.
- In experiments, always record uncertainties, propagate them correctly, and comment on possible systematic errors.