1. Overview
Practical circuits are governed by the Principle of Conservation of Energy. In any closed loop, the total energy supplied by the source of electromotive force (e.m.f.) must equal the total energy dissipated across the components and the source's own internal resistance. While ideal models assume power sources have no resistance, real-world applications must account for internal resistance ($r$), which causes the terminal potential difference ($V$) to drop as the current ($I$) increases. Mastering this topic requires a precise understanding of energy transfers and the ability to model these relationships both algebraically and graphically.
Key Definitions
To secure full marks in the Cambridge 9702 exam, definitions must include specific keywords often underlined in mark schemes.
- Electromotive Force (e.m.f.): The energy transferred per unit charge in driving charge around a complete circuit. (Unit: Volts, $\text{V}$ or $\text{J},\text{C}^{-1}$).
- Potential Difference (p.d.): The energy transferred per unit charge from electrical energy to other forms (e.g., thermal, light) between two points in a circuit. (Unit: Volts, $\text{V}$ or $\text{J},\text{C}^{-1}$).
- Internal Resistance ($r$): The inherent resistance to the flow of charge within the source of e.m.f. itself, resulting in energy being dissipated as heat inside the battery or cell.
- Terminal Potential Difference ($V$): The potential difference across the external load connected to a power source; it is the energy available to the external circuit per unit charge.
- Lost Volts ($v$ or $Ir$): The potential difference across the internal resistance of a source. It represents the energy per unit charge dissipated as heat within the source and is not available to the external circuit.
Content
3.1 Circuit Symbols and Diagrams
The Cambridge 9702 syllabus requires the use of standardized symbols. Accuracy in drawing these is essential for practical and theory papers.
| Component | Symbol Description | Function in Circuit |
|---|---|---|
| Cell | One long thin line (positive), one short thick line (negative). | Provides e.m.f. via chemical reaction. |
| Battery | Multiple cells connected in series. | Provides higher e.m.f. or capacity. |
| Fixed Resistor | A simple rectangle. | Limits current or provides a specific p.d. |
| Variable Resistor | Rectangle with a diagonal arrow through it. | Allows manual adjustment of circuit current. |
| Potentiometer | Rectangle with an arrow pointing to one side. | Acts as a variable potential divider. |
| Thermistor | Rectangle with a 'hockey stick' line through it. | Resistance decreases as temperature increases (NTC). |
| LDR | Rectangle in a circle with two arrows pointing in. | Resistance decreases as light intensity increases. |
| Diode | Triangle pointing to a vertical bar. | Allows current to flow in one direction only. |
| LED | Diode symbol with two arrows pointing out. | Emits light when current flows in the forward direction. |
| Ammeters | Circle with an 'A'. | Connected in series; ideal resistance is zero. |
| Voltmeters | Circle with a 'V'. | Connected in parallel; ideal resistance is infinite. |
| Galvanometer | Circle with a 'G'. | Detects very small currents and their direction. |
Key Diagram Rules:
- Series Connection: Components are connected end-to-end in a single path. The current is the same through all components.
- Parallel Connection: Components are connected across the same two nodes. The potential difference is the same across all branches.
- Polarity: Ensure the long line of the cell symbol faces the direction of conventional current flow (positive to negative).
3.2 E.M.F. vs Potential Difference (Energy Considerations)
While both are measured in Volts, the distinction lies in the direction of energy transfer.
Electromotive Force ($E$):
- Energy Source: It describes the work done on the charge carriers.
- Conversion: Non-electrical energy (Chemical, Solar, Mechanical) $\rightarrow$ Electrical energy.
- Context: It is a property of the source, independent of the external circuit.
Potential Difference ($V$):
- Energy Sink: It describes the work done by the charge carriers.
- Conversion: Electrical energy $\rightarrow$ Other forms (Thermal, Kinetic, Sound).
- Context: It is a property of the load or component.
Equation Link: $$W = VQ \quad \text{or} \quad W = EQ$$ Where $W$ is the work done (Joules) and $Q$ is the charge (Coulombs).
3.3 Internal Resistance and the Power Source Equation
In a real circuit, the e.m.f. ($E$) provided by the source is shared between the external resistance ($R$) and the internal resistance ($r$).
The Conservation of Energy Derivation: Total Energy Supplied = Energy in External Circuit + Energy lost in Source $$E \cdot I \cdot t = V \cdot I \cdot t + v \cdot I \cdot t$$ Dividing by $(I \cdot t)$ gives the voltage balance: $$E = V + Ir$$ (This equation is NOT on the formula sheet and must be memorized.)
Since $V = IR$ (Ohm's Law for the external circuit): $$E = IR + Ir = I(R + r)$$
The Concept of "Lost Volts": The term $Ir$ represents the "lost volts." As the current $I$ increases (by decreasing the external resistance $R$), the value of $Ir$ increases. Because $E$ is constant for a given source, the terminal p.d. ($V$) must decrease.
3.4 Determining $E$ and $r$ from Graphs
A common practical assessment involves measuring the terminal p.d. ($V$) for various currents ($I$).
Linear Rearrangement: Start with $V = E - Ir$. Rearrange into $y = mx + c$ form: $$V = (-r)I + E$$
Graph Characteristics:
- y-axis: Terminal potential difference ($V$).
- x-axis: Current ($I$).
- Gradient ($m$): The gradient is equal to $-r$. (Internal resistance is the magnitude of the gradient).
- y-intercept ($c$): The intercept on the $V$-axis is the e.m.f. ($E$).
- x-intercept: This represents the short-circuit current ($I_{sc} = E/r$), where the external resistance is zero.
[DIAGRAM DESCRIPTION: A graph of $V$ against $I$. The line is straight with a negative slope. The point where the line hits the vertical axis is $E$. The slope of the line is steepness $r$. A steeper line indicates a higher internal resistance.]
3.5 Power in Practical Circuits
The power delivered to the external load is: $$P = VI = I^2R$$ The power dissipated internally is: $$P_{lost} = I^2r$$ The total power produced by the source is: $$P_{total} = EI$$
Worked Example 1 — Calculating Internal Resistance
A battery has an e.m.f. of $9.00\ \text{V}$. When a $10.0\ \Omega$ resistor is connected across its terminals, the terminal potential difference is measured to be $8.20\ \text{V}$. Calculate the internal resistance of the battery.
Step 1: Identify the knowns $E = 9.00\ \text{V}$ $V = 8.20\ \text{V}$ $R = 10.0\ \Omega$
Step 2: Calculate the circuit current ($I$) using the external load $$V = IR \implies I = \frac{V}{R}$$ $$I = \frac{8.20}{10.0} = 0.820\ \text{A}$$
Step 3: Use the power source equation to find $r$ $$E = V + Ir$$ $$9.00 = 8.20 + (0.820 \times r)$$ $$0.80 = 0.820r$$ $$r = \frac{0.80}{0.820} = 0.9756... \Omega$$
Step 4: Final answer with units and sig figs Answer: $r = 0.98\ \Omega$ (to 2 s.f., matching the precision of the subtraction $9.00 - 8.20$).
Worked Example 2 — Comparing Two Sources
Source A has $E = 12\ \text{V}, r = 0.5\ \Omega$. Source B has $E = 12\ \text{V}, r = 5.0\ \Omega$. Both are connected to a $2.0\ \text{A}$ load. Calculate the terminal p.d. for both and explain which is more efficient.
Step 1: Calculate $V$ for Source A $$V_A = E - Ir = 12 - (2.0 \times 0.5) = 12 - 1.0 = 11\ \text{V}$$
Step 2: Calculate $V$ for Source B $$V_B = E - Ir = 12 - (2.0 \times 5.0) = 12 - 10 = 2\ \text{V}$$
Step 3: Compare efficiency Efficiency can be thought of as $\frac{V}{E} \times 100%$. Efficiency A: $(11/12) \times 100 = 91.7%$ Efficiency B: $(2/12) \times 100 = 16.7%$
Answer: Source A is significantly more efficient. Source B loses most of its energy as heat internally ($10\ \text{V}$ of "lost volts").
Key Equations
| Quantity | Equation | Data Sheet? | Notes |
|---|---|---|---|
| e.m.f. / p.d. | $V = \frac{W}{Q}$ | No | $W$ is work done, $Q$ is charge. |
| Power Source | $E = V + Ir$ | No | Fundamental for internal resistance. |
| Total Resistance | $R_{total} = R + r$ | No | For a simple series circuit. |
| Terminal p.d. | $V = E - Ir$ | No | Used for $V$-$I$ graph analysis. |
| Ohm's Law | $V = IR$ | No | Applies to the external circuit only. |
Common Mistakes to Avoid
❌ Wrong: Thinking e.m.f. is a "force."
✓ Right: e.m.f. is energy per unit charge (Potential). The name is a historical artifact.
❌ Wrong: Using $E$ in the formula $P = I^2R$ to find power in the external resistor.
✓ Right: Use $V$ (terminal p.d.) for external power ($P = VI$) or use the total resistance ($P_{total} = I^2(R+r)$) for total power.
❌ Wrong: Assuming $V$ is constant when $R$ changes.
✓ Right: As $R$ decreases, $I$ increases, which increases "lost volts" ($Ir$), therefore $V$ must decrease.
❌ Wrong: Drawing a voltmeter in series.
✓ Right: Voltmeters have very high resistance; if placed in series, they will effectively block the current, and the reading will simply be the e.m.f. of the source.
❌ Wrong: Forgetting that $r$ is part of the total circuit resistance.
✓ Right: When calculating current, always use $I = \frac{E}{R + r}$.
Exam Tips
- The "Open Circuit" Condition: If an exam question mentions an "open circuit" or a "voltmeter of infinite resistance" connected directly across a battery, the current $I = 0$. In this specific case, $V = E$. This is how e.m.f. is measured in practice.
- Internal Resistance in High Voltage Supplies: Some exam questions ask why Extra High Tension (EHT) supplies have very high internal resistance (e.g., $5\ \text{M}\Omega$). This is a safety feature: if the terminals are short-circuited, the high internal resistance limits the current to a non-lethal level.
- Graph Gradient: When calculating the gradient for internal resistance from a $V$-$I$ graph, use the largest possible triangle to minimize percentage uncertainty. Ensure you state $r = - \text{gradient}$.
- Significant Figures: In Paper 2 and Paper 3, pay close attention to the precision of the given data. If the e.m.f. is $1.5\ \text{V}$ (2 s.f.) and current is $0.70\ \text{A}$ (2 s.f.), your internal resistance should be given to 2 s.f.
- Multiple Cells: If cells are in series, add their e.m.f.s ($E_{total} = E_1 + E_2$) and their internal resistances ($r_{total} = r_1 + r_2$). If a cell is connected with reverse polarity, subtract its e.m.f.
- "Show that" Questions: If asked to "show that $V = E - Ir$", always start from the principle of conservation of energy: $E = I(R+r)$, then expand to $E = IR + Ir$, and substitute $V = IR$.