Resistance and resistivity

1. Overview

Resistance is the mathematical ratio of the potential difference across a component to the current flowing through it. At a microscopic level, resistance arises from the collisions between drifting charge carriers (usually electrons) and the vibrating ionic lattice of the conductor. These collisions transfer kinetic energy from the electrons to the lattice, manifesting as thermal energy. While Ohmic conductors maintain a constant resistance under constant physical conditions, many practical components—such as filament lamps, diodes, and sensors—exhibit non-linear $I–V$ characteristics where resistance changes in response to temperature, light intensity, or the magnitude of the applied voltage.

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Key Definitions

• Resistance ($R$): The ratio of the potential difference across a component to the current flowing through it ($R = V/I$).
• The Ohm ($\Omega$): The resistance of a component when a potential difference of one volt produces a current of one ampere ($1\ \Omega = 1\text{ V A}^{-1}$).
• Ohm’s Law: The current in a metallic conductor is directly proportional to the potential difference across it, provided that physical conditions (such as temperature) remain constant.
• Resistivity ($\rho$): An intrinsic property of a material, defined as the product of the resistance and the cross-sectional area divided by the length of the specimen ($\rho = RA/L$).
• Negative Temperature Coefficient (NTC): A property of a material where the resistance decreases as the temperature increases. This is characteristic of semiconductor thermistors.

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Content

3.1 Resistance and Ohm’s Law

Resistance is not simply "opposition to current"; it is defined by the relationship:

$$V = IR$$ (Must be memorised)

Where:

• $V$ = Potential Difference (V)
• $I$ = Current (A)
• $R$ = Resistance ($\Omega$)

Ohmic Conductors: A conductor is "Ohmic" only if its resistance remains constant. On an $I–V$ graph, this is represented by a straight line passing through the origin. The gradient of an $I–V$ graph for an Ohmic conductor is $1/R$. If the graph is $V$ against $I$, the gradient is $R$.

3.2 I–V Characteristics

Cambridge 9702 requires you to sketch and explain the $I–V$ curves for three specific components. By convention, $I$ is plotted on the y-axis and $V$ on the x-axis.

1. Metallic Conductor (at constant temperature):

• Graph: A straight line through the origin extending into the third quadrant (negative $V$ and $I$).
• Description: $I \propto V$. The resistance is constant.
• Key Note: The conductor only obeys Ohm's law if the temperature is held constant. If the current is high enough to cause heating, the graph will begin to curve.

2. Filament Lamp:

• Graph: A curve with a decreasing gradient ($dI/dV$) as the magnitude of $V$ increases. It is symmetrical in the positive and negative quadrants.
• Description: As $V$ increases, $I$ increases at a non-linear, decreasing rate.
• The Microscopic Explanation (High-Yield for Exams):
  1. As the potential difference increases, the current increases.
  2. The increased flow of electrons leads to more frequent collisions with the metal ions in the lattice.
  3. These collisions transfer energy to the ions, causing the temperature of the filament to increase.
  4. The metal ions vibrate with greater amplitude.
  5. This increases the frequency of collisions between the charge carriers and the lattice.
  6. Therefore, the resistance increases as the current/temperature increases.

3. Semiconductor Diode:

• Graph:
  • Reverse Bias ($V < 0$): The current is effectively zero (infinite resistance).
  • Forward Bias ($V > 0$): The current remains zero until the threshold voltage (approx. 0.6 V for silicon). Above this voltage, the current increases sharply (resistance drops rapidly).
• Description: A non-ohmic device that allows current to flow in only one direction.

3.3 Resistivity

While resistance depends on the specific dimensions of a sample, resistivity ($\rho$) is a property of the material itself.

$$R = \frac{\rho L}{A}$$ (Found on Formula Sheet)

Where:

• $\rho$ = Resistivity ($\Omega \text{ m}$)
• $L$ = Length of the conductor (m)
• $A$ = Cross-sectional area ($m^2$)

Physical Dependencies:

• Length ($L$): $R \propto L$. Doubling the length of a wire is equivalent to putting two resistors in series; the electrons must travel through twice as many ions, doubling the number of collisions.
• Area ($A$): $R \propto 1/A$. Doubling the cross-sectional area is equivalent to putting two resistors in parallel; there are more available paths for the electrons to flow through, reducing the resistance.

Calculating Area: Most wires are cylindrical. You must be able to calculate $A$ from the diameter ($d$): $$A = \frac{\pi d^2}{4}$$ or $$A = \pi r^2$$

3.4 Sensors: LDRs and Thermistors

These components are made of semiconductors. Unlike metals, semiconductors have a relatively low number of free charge carriers at room temperature.

Light-Dependent Resistor (LDR):

• Behavior: As light intensity increases, resistance decreases.
• Mechanism: Incident photons provide energy to the semiconductor lattice, "liberating" electrons from the valence band to the conduction band. This increase in the number density of charge carriers ($n$) significantly increases conductivity and thus reduces resistance.

NTC Thermistor:

• Behavior: As temperature increases, resistance decreases.
• Mechanism: Thermal energy provides the necessary energy to release more charge carriers (electrons/holes) within the semiconductor. Although the lattice vibrations also increase (which would increase resistance), the increase in the number of charge carriers is the dominant effect, leading to a net decrease in resistance.

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Worked Example 1 — Resistivity and Ratios

A wire $X$ has resistance $R$. Wire $Y$ is made of the same material but has twice the length and one-third the diameter of wire $X$. Determine the resistance of wire $Y$ in terms of $R$.

Step 1: Write the formula for both wires. For Wire $X$: $R_X = \frac{\rho L}{A_X}$ where $A_X = \frac{\pi d^2}{4}$ For Wire $Y$: $R_Y = \frac{\rho L_Y}{A_Y}$

Step 2: Express Wire $Y$ dimensions in terms of Wire $X$. $L_Y = 2L$ $d_Y = \frac{1}{3}d$ $A_Y = \frac{\pi (d/3)^2}{4} = \frac{1}{9} \left( \frac{\pi d^2}{4} \right) = \frac{1}{9} A_X$

Step 3: Substitute into the resistance formula for $Y$. $R_Y = \frac{\rho (2L)}{(1/9) A_X}$ $R_Y = 18 \left( \frac{\rho L}{A_X} \right)$

Step 4: Relate back to $R$. Since $R = \frac{\rho L}{A_X}$, then $R_Y = 18R$.

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Worked Example 2 — I-V Graph Analysis

A filament lamp is connected to a variable power supply. When the potential difference across the lamp is 6.0 V, the current is 0.50 A. When the potential difference is increased to 12.0 V, the current becomes 0.80 A. (a) Calculate the resistance at 6.0 V and 12.0 V. (b) Explain why the resistance has changed.

Part (a) Solution:

1. At 6.0 V: $R = V / I$ $R = 6.0 / 0.50 = \mathbf{12\ \Omega}$
2. At 12.0 V: $R = V / I$ $R = 12.0 / 0.80 = \mathbf{15\ \Omega}$

Part (b) Solution: As the potential difference increases, the current increases, which leads to a higher temperature in the filament. This causes the metal ions in the lattice to vibrate with larger amplitudes, increasing the frequency of collisions between the electrons and the ions. Consequently, the resistance increases from $12\ \Omega$ to $15\ \Omega$.

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Key Equations

Equation	Description	Status
$V = IR$	Definition of Resistance	Memorise
$R = \frac{\rho L}{A}$	Resistivity formula	Formula Sheet
$A = \frac{\pi d^2}{4}$	Cross-sectional area from diameter	Memorise
$P = VI = I^2 R = \frac{V^2}{R}$	Electrical power dissipated	Formula Sheet
$G = \frac{1}{R}$	Conductance (occasionally used)	Memorise

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Common Mistakes to Avoid

• ❌ Wrong: Defining resistance as the gradient of an $I–V$ graph.
• ✅ Right: Resistance is the ratio $V/I$ at a specific point. For a curve (like a lamp), the gradient is $dI/dV$, which is not $1/R$. Always use the coordinates $(V, I)$ of the point, not the slope of the tangent.
• ❌ Wrong: Using diameter instead of radius in $\pi r^2$, or forgetting to square the radius.
• ✅ Right: Use $A = \pi (d/2)^2$ or $A = \frac{\pi d^2}{4}$. Double-check that you have squared the value.
• ❌ Wrong: Incorrect unit conversions for area (e.g., thinking $1\text{ mm}^2 = 10^{-3}\text{ m}^2$).
• ✅ Right: $1\text{ mm} = 10^{-3}\text{ m}$, so $1\text{ mm}^2 = (10^{-3}\text{ m})^2 = \mathbf{10^{-6}\text{ m}^2}$.
• ❌ Wrong: Assuming all resistors obey Ohm's Law.
• ✅ Right: Only "Ohmic" conductors have a constant resistance. Filament lamps and diodes are explicitly non-Ohmic.
• ❌ Wrong: Stating that resistance of a thermistor increases with temperature.
• ✅ Right: In the 9702 syllabus, thermistors are assumed to be NTC (Negative Temperature Coefficient). Resistance decreases as temperature increases.

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Exam Tips

1. The "Explain" Mark: When explaining why resistance increases in a metal, you must use the word "ions" or "lattice". Referring to "atoms" is often not precise enough for the mark scheme, as the electrons have been stripped to form the sea of delocalised charge carriers.
2. Resistivity Units: Always check that resistivity is in $\Omega \text{ m}$. If a question gives you $\Omega \text{ cm}$, convert it immediately ($1\ \Omega \text{ cm} = 10^{-2}\ \Omega \text{ m}$).
3. Significant Figures: Cambridge 9702 is strict. If the input data is $2.0\text{ V}$ and $0.50\text{ A}$, your answer must be $4.0\ \Omega$ (2 s.f.), not $4\ \Omega$.
4. Graph Sketching:
   • Diode: Ensure the line is perfectly flat on the x-axis for negative voltages.
   • Filament Lamp: Ensure the curve passes through $(0,0)$ and the "flattening" is visible in both the positive and negative quadrants.
5. Command Words: If a question asks to "State Ohm's Law," you must include the condition "provided temperature remains constant." Omitting this condition is a common reason for losing the mark.