recall and use P = VI, P = I 2R and P = V 2 / R

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

1. Work done by a charge moving through a potential difference:
   

   \$W = Q\Delta V\$

2. Since \$I = \dfrac{Q}{t}\$, then \$Q = I t\$. Substituting:
   

   \$W = (I t)\Delta V\$

3. Dividing by \$t\$ gives the power:
   

   \$P = I\Delta V\$

4. Using Ohm’s law \$\Delta V = I R\$:
   

   \$P = I (I R) = I^{2}R\$

5. 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

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

1. Current Law (KCL): The algebraic sum of currents at a junction is zero.
   

   \$\sum I{\text{in}} = \sum I{\text{out}}\$

2. 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.

1. Series part: \$R{12}=R{1}+R_{2}=10\;\Omega\$.
2. 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\$

3. Total current: \$I = \dfrac{V}{R_{\text{T}}}= \dfrac{24}{5.45}\approx4.4\;\text{A}\$.

6. Circuit Symbols (Cambridge Syllabus 6)

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

1. A \$5\;\Omega\$ resistor has a current of \$3\;\text{A}\$ flowing through it. Calculate the voltage across it and the power dissipated.
2. A device operates at \$12\;\text{V}\$ and consumes \$24\;\text{W}\$. Determine the current drawn and the equivalent resistance.
3. 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.
4. A heating element is designed to produce \$2000\;\text{W}\$ when connected to a \$240\;\text{V}\$ supply. What resistance must the element have?
5. 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?
6. 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.

7. 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)

1. \$V = IR = 3\times5 = 15\;\text{V}\$; \$P = I^{2}R = 3^{2}\times5 = 45\;\text{W}\$.
2. \$I = P/V = 24/12 = 2\;\text{A}\$; \$R = V/I = 12/2 = 6\;\Omega\$.
3. Total resistance \$R{\text{T}} = 8+12 = 20\;\Omega\$; \$P = V^{2}/R{\text{T}} = 24^{2}/20 = 28.8\;\text{W}\$.
4. \$R = V^{2}/P = 240^{2}/2000 = 28.8\;\Omega\$.
5. \$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}\$.

6. 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}\$.

7. 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.