CIE IGCSE Physics (0625) Resistance Study Notes - New Syllabus
CIE IGCSE Physics (0625) Resistance Study Notes
LEARNING OBJECTIVE
- Understanding the concepts of Resistance
Key Concepts:
- Resistance and Ohm’s Law
- Experiment to Determine Resistance
- Relationship of Resistance in a Metallic Wire
Resistance and Ohm’s Law
Resistance and Ohm’s Law:
Resistance (R) is the opposition to the flow of electric current in a component.
- Measured in ohms (Ω).
\( R = \frac{V}{I} \)
- \(V\): potential difference (in volts)
- \(I\): current (in amperes)
- \(R\): resistance (in ohms)
Current–Voltage (I–V) Characteristics:
- The I–V graph shows how current changes with voltage for different components.
1. Resistor of Constant Resistance (Ohmic Resistor):
- Follows Ohm’s Law: \(V \propto I\)
- Straight-line graph through the origin
- Resistance is constant (slope = resistance)
2. Filament Lamp:
- As current increases, the filament heats up.
- Resistance increases with temperature.
- Graph is a curve that gets flatter — shows that more voltage is needed to increase current at higher values.
3. Diode:
- Allows current in one direction only (forward bias).
- Very high resistance in reverse direction — no current flows until a threshold voltage is reached.
Example:
A resistor has a voltage of \(6\,\text{V}\) across it and a current of \(2\,\text{A}\) flows through it. Calculate its resistance.
▶️ Answer/Explanation
Using the formula \(R = \frac{V}{I}\):
\(R = \frac{6}{2} = 3\,\Omega\)
Answer: \(\boxed{3\,\Omega}\)
Experiment to Determine Resistance
Experiment to Determine Resistance:
Aim: To measure the resistance of a component using a voltmeter and an ammeter.
Apparatus:
- Power supply (battery)
- Resistor (or other component)
- Ammeter (connected in series)
- Voltmeter (connected in parallel)
- Connecting wires
- Switch
Circuit Setup:
- Connect the resistor in series with the ammeter and power supply.
- Connect the voltmeter in parallel across the resistor.
- Close the switch and record the voltage (V) and current (I).
Calculation:
\( R = \frac{V}{I} \)
- \(R\): Resistance in ohms (Ω)
- \(V\): Potential difference across the component (V)
- \(I\): Current through the component (A)
Example:
In an experiment, a student measures a current of \(0.4\,\text{A}\) through a resistor and a potential difference of \(3.2\,\text{V}\) across it. Calculate the resistance of the resistor.
▶️ Answer/Explanation
Using the formula:
\( R = \frac{V}{I} = \frac{3.2}{0.4} = 8\,\Omega \)
Answer: \(\boxed{8\,\Omega}\)
Relationship of Resistance in a Metallic Wire
Relationship of Resistance in a Metallic Wire:
For a metallic electrical conductor (like copper or nichrome), the resistance \( R \) depends on:
- (a) The length of the wire: \( R \propto L \)
- (b) The cross-sectional area of the wire: \( R \propto \frac{1}{A} \)
Where:
- \( R \) = resistance (ohms, \( \Omega \))
- \( L \) = length of the wire (metres)
- \( A \) = cross-sectional area of the wire (m²)
Combined Equation:
\( R = \rho \frac{L}{A} \)
- \( \rho \) (rho) is the resistivity of the material (Ω·m), a constant depending on the material.
Key Note:
- Longer wires offer more resistance (electrons face more collisions).
- Wider wires (larger area) allow more paths for electrons → lower resistance.
Example:
A copper wire of length 2 m and cross-sectional area \( 0.5 \times 10^{-6} \,\text{m}^2 \) has a resistance of 0.068 Ω. What will the resistance be if the length is doubled and the cross-sectional area is halved?
▶️ Answer/Explanation
Original resistance:
\( R_1 = \rho \frac{L}{A} \)
New resistance with \( L’ = 2L \) and \( A’ = \frac{1}{2}A \):
\( R_2 = \rho \frac{2L}{\frac{1}{2}A} = \rho \frac{4L}{A} = 4 \times R_1 \)
So:
\( R_2 = 4 \times 0.068 = 0.272 \,\Omega \)
Answer: \(\boxed{0.272\,\Omega}\)