Resistance

1. Overview

Resistance is a measure of how much a component opposes the flow of electrical current. Understanding resistance is crucial because it allows engineers to control the amount of current in a circuit, ensuring that electronic devices operate safely and efficiently.

Key Definitions

• Resistance: The ratio of the potential difference across a component to the current flowing through it.
• Ohm ($\Omega$): The unit of electrical resistance.
• Ohmic Conductor: A conductor that follows Ohm’s Law, where current is directly proportional to voltage (at a constant temperature).
• Potential Difference (Voltage): The energy transferred per unit charge between two points in a circuit.
• Current: The rate of flow of electrical charge.

Core Content

The Resistance Equation

To calculate the resistance of a component, we use the following formula: $$Resistance (R) = \frac{Potential Difference (V)}{Current (I)}$$

Worked Example: A light bulb has a potential difference of $12\text{ V}$ across it and a current of $3\text{ A}$ flowing through it. Calculate the resistance.

• $R = V / I$
• $R = 12 / 3$
• $R = 4\text{ }\Omega$

Determining Resistance Experimentally

To find the resistance of an unknown component (e.g., a wire or a resistor), use an ammeter and a voltmeter.

Method:

1. Connect the component in series with a power supply and an ammeter.
2. Connect a voltmeter in parallel specifically across the component being tested.
3. Close the switch and record the reading on the ammeter ($I$) and the voltmeter ($V$).
4. (Optional) Use a variable resistor to change the current and take multiple readings to find an average.
5. Calculate resistance using $R = V / I$.

Factors Affecting Resistance of a Wire (Qualitative)

The resistance of a metallic wire depends on its physical dimensions:

• Length: The longer the wire, the higher the resistance (electrons collide with more ions).
• Cross-sectional Area (Thickness): The thicker the wire, the lower the resistance (there is more space for electrons to flow).

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Extended Content (Extended Curriculum Only)

Current-Voltage ($I-V$) Graphs

The relationship between current and voltage is not the same for all components.

1. Fixed Resistor (Ohmic Conductor): At a constant temperature, the graph is a straight line through the origin. Resistance is constant.
   • 📊A straight diagonal line passing through (0,0)
2. Filament Lamp: As current increases, the temperature of the filament increases, causing resistance to increase. The graph curves, becoming flatter at higher voltages.
   • 📊An S-shaped curve passing through the origin
3. Diode: A diode only allows current to flow in one direction. It has very high resistance in the reverse direction and low resistance in the forward direction after a certain voltage.
   • 📊Horizontal line on the x-axis for negative voltage, then a sharp upward curve for positive voltage

Quantitative Relationships

For a metallic conductor at a constant temperature:

• Resistance is directly proportional to length ($R \propto L$): Doubling the length will double the resistance.
• Resistance is inversely proportional to cross-sectional area ($R \propto 1/A$): Doubling the area will halve the resistance.

Worked Example (Extended): Wire A has a resistance of $10\text{ }\Omega$. Wire B is made of the same metal but is twice as long and has twice the cross-sectional area. What is the resistance of Wire B?

1. Length is doubled $\rightarrow$ Resistance doubles ($10 \times 2 = 20\text{ }\Omega$).
2. Area is doubled $\rightarrow$ Resistance halves ($20 / 2 = 10\text{ }\Omega$).
3. Final Resistance = $10\text{ }\Omega$.

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Key Equations

Equation	Symbols	Units
$R = \frac{V}{I}$	$R$ = Resistance, $V$ = Voltage, $I$ = Current	$\Omega$ (Ohms), $V$ (Volts), $A$ (Amps)
$R \propto L$	$L$ = Length	$m$ (Meters)
$R \propto \frac{1}{A}$	$A$ = Cross-sectional area	$m^2$

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Common Mistakes to Avoid

• ❌ Wrong: Placing the voltmeter in series with the circuit or across the power supply.
  • ✅ Right: Always place the voltmeter in parallel across the specific component you are measuring.
• ❌ Wrong: Subtracting current from voltage ($12 - 4$) to find resistance.
  • ✅ Right: Always use division ($V / I$).
• ❌ Wrong: Using the symbol for a variable resistor (rectangle with a diagonal arrow) when asked for a fixed resistor.
  • ✅ Right: A fixed resistor is a simple plain rectangle.
• ❌ Wrong: Forgetting to convert time units if calculating charge or energy (e.g., leaving 2 minutes as "2").
  • ✅ Right: Always convert time to seconds ($2\text{ minutes} = 120\text{ seconds}$).

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Exam Tips

1. Check Units: Ensure current is in Amps (A), not milliamps (mA). If you see $50\text{ mA}$, divide by $1000$ to get $0.05\text{ A}$ before calculating resistance.
2. Identify the Component: If an exam question asks you to identify a component from an $I-V$ graph, look at the shape. A straight line is always a fixed resistor/ohmic conductor; a curve that levels off is a filament lamp.
3. Proportionality: If the question says the wire is "thicker" or "wider," it refers to the cross-sectional area. If they give you the diameter or radius, remember that area is proportional to the square of the radius ($A = \pi r^2$).

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

A student investigates the resistance of a length of nichrome wire. She connects the wire to a power supply, ammeter, and voltmeter.

(a) Draw a circuit diagram to show how she should connect the components to measure the resistance of the nichrome wire. [2]

(b) State two things the student should do to ensure an accurate measurement of the resistance. [2]

(c) The student finds the resistance of the wire is $4.5 , \Omega$. She then doubles the length of the wire. Predict the new resistance of the wire. [1]

Worked Solution:

(a)

1. 📊A circuit diagram showing a power supply connected in series to an ammeter and a nichrome wire. A voltmeter is connected in parallel with the nichrome wire. The circuit should include connecting wires.
   *Correct placement of ammeter and voltmeter.*

How to earn full marks:

• Ammeter in series with the wire
• Voltmeter in parallel with the wire

(b)

1. Use a low current to prevent the wire from heating up. Heating changes the resistance.
2. Take multiple readings of current and voltage and calculate the average resistance. Averaging reduces random errors.

How to earn full marks:

• Mention using a low current
• Mention taking multiple readings and averaging

(c)

1. Resistance is directly proportional to length. If the length doubles, the resistance doubles. $R_{new} = 2 \times 4.5 , \Omega = 9.0 , \Omega$ Applying the proportionality.

How to earn full marks:

• Correctly doubling the resistance value
• Correct unit

Final Answer: $\boxed{9.0 , \Omega}$

Common Pitfall: Many students struggle with drawing circuit diagrams, especially placing the ammeter and voltmeter correctly. Remember that an ammeter must be in series to measure the current flowing through the component, while a voltmeter must be in parallel to measure the potential difference across the component.

Exam-Style Question 2 — Calculation [6 marks]

Question:

A copper wire has a length of 20.0 m and a cross-sectional area of $1.70 \times 10^{-6} , \text{m}^2$. The resistance of the wire is measured to be $0.20 , \Omega$.

(a) Calculate the current flowing through the wire when a potential difference of 1.5 V is applied across it. [2]

(b) Calculate the resistivity of the copper wire. [3]

(c) The wire is then replaced with another copper wire of the same length but with twice the cross-sectional area. Without calculation, state and explain how the current changes when the same potential difference of 1.5 V is applied. [1]

Worked Solution:

(a)

1. Use Ohm's Law: $V = IR$, so $I = \frac{V}{R}$ Recalling Ohm's Law.
2. Substitute the given values: $I = \frac{1.5 , \text{V}}{0.20 , \Omega} = 7.5 , \text{A}$ Substituting and calculating.

How to earn full marks:

• Correct formula and substitution
• Correct answer and unit

(b)

1. Use the formula for resistance: $R = \frac{\rho L}{A}$, so $\rho = \frac{RA}{L}$ Recalling the resistance formula.
2. Substitute the given values: $\rho = \frac{0.20 , \Omega \times 1.70 \times 10^{-6} , \text{m}^2}{20.0 , \text{m}}$ Substituting the values.
3. Calculate the resistivity: $\rho = 1.70 \times 10^{-8} , \Omega \cdot \text{m}$ Calculating the resistivity.

How to earn full marks:

• Correct formula and rearrangement
• Correct substitution
• Correct answer and unit

(c)

1. The current will double. Stating the change.
2. Since the cross-sectional area doubles, the resistance halves. Linking area and resistance.
3. Since $I = \frac{V}{R}$ and $R$ is halved, the current doubles for the same potential difference. Explaining the effect on current.

How to earn full marks:

• Correct statement of change (current doubles)
• Correct explanation relating area, resistance, and current

Final Answer: (a) $\boxed{7.5 , \text{A}}$ (b) $\boxed{1.70 \times 10^{-8} , \Omega \cdot \text{m}}$ (c) The current doubles because resistance is halved.

Common Pitfall: Students often forget the correct formula for resistance in terms of resistivity, length, and cross-sectional area. Make sure you know that $R = \frac{\rho L}{A}$ and can rearrange it correctly. Also, pay close attention to units, especially when dealing with area (m$^2$) and resistivity ($\Omega \cdot$m).

Exam-Style Question 3 — Graph Interpretation [5 marks]

Question:

A student investigates the current-voltage (I-V) characteristics of three different electrical components: a resistor, a filament lamp, and a diode. The graphs of their I-V characteristics are shown below.

(a)

Describe the shape of the I-V graph for the resistor. [1]

(b) Explain how the resistance of the filament lamp changes as the voltage increases. [2]

(c) Describe the I-V characteristic of the diode and explain its function. [2]

Worked Solution:

(a)

1. The I-V graph for the resistor is a straight line through the origin. Describing the graph.

How to earn full marks:

• Correct description of the graph

(b)

1. As the voltage increases, the current increases, but not proportionally. Describing the non-linear relationship.
2. The resistance of the filament lamp increases as the voltage increases because the lamp gets hotter. Linking temperature to resistance.

How to earn full marks:

• Mentioning the non-linear relationship between voltage and current
• Explaining that resistance increases due to increasing temperature

(c)

1. The diode allows current to flow easily in one direction (forward bias) but blocks current in the opposite direction (reverse bias). Describing the diode's behavior.
2. The diode acts as a one-way valve for current, allowing current to flow only when the voltage is above a certain threshold in the forward direction. Explaining the function.

How to earn full marks:

• Describing the one-way current flow
• Explaining the diode's function as a one-way valve

Final Answer: (a) Straight line through the origin. (b) Resistance increases as voltage increases. (c) Allows current in one direction only.

Common Pitfall: Students often confuse the I-V characteristics of different components. Remember that a resistor has a linear I-V relationship, a filament lamp's resistance increases with temperature (non-linear), and a diode allows current to flow in only one direction. Don't say the gradient is the resistance; it's actually the inverse of resistance.

Exam-Style Question 4 — Resistance and Dimensions [4 marks]

Question:

A cylindrical aluminum wire has a resistance of $2.0 , \Omega$. The wire has a length of 10.0 m and a radius of 1.0 mm.

(a) Calculate the cross-sectional area of the wire. [2]

(b) Calculate the resistivity of the aluminum wire. [2]

Worked Solution:

(a)

1. The cross-sectional area of a cylindrical wire is given by $A = \pi r^2$. Recalling the area formula.
2. Convert the radius to meters: $r = 1.0 , \text{mm} = 1.0 \times 10^{-3} , \text{m}$ Converting units.
3. Calculate the area: $A = \pi (1.0 \times 10^{-3} , \text{m})^2 = \pi \times 10^{-6} , \text{m}^2 = 3.14 \times 10^{-6} , \text{m}^2$ Substituting and calculating.

How to earn full marks:

• Correct formula for the area of a circle
• Correct answer and unit

(b)

1. Use the formula for resistance: $R = \frac{\rho L}{A}$, so $\rho = \frac{RA}{L}$ Recalling the resistance formula.
2. Substitute the given values: $\rho = \frac{2.0 , \Omega \times 3.14 \times 10^{-6} , \text{m}^2}{10.0 , \text{m}}$ Substituting the values.
3. Calculate the resistivity: $\rho = 6.28 \times 10^{-7} , \Omega \cdot \text{m}$ Calculating the resistivity.

How to earn full marks:

• Correct formula and rearrangement
• Correct answer and unit

Final Answer: (a) $\boxed{3.14 \times 10^{-6} , \text{m}^2}$ (b) $\boxed{6.28 \times 10^{-7} , \Omega \cdot \text{m}}$

Common Pitfall: A very common mistake is forgetting to convert the radius from millimeters to meters before calculating the area. Always double-check your units and ensure they are in the standard SI units (meters, kilograms, seconds, etc.) before plugging them into formulas. Also, remember that resistance is inversely proportional to the area, not the radius directly.