Resistance
10 flashcards to master Resistance
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A potential difference of 12 V is applied across a resistor. If the current flowing through the resistor is 2 A, calculate the resistance of the resistor.
Resistance (R) = Voltage (V) / Current (I)
R = 12 V / 2 A
R = 6 Ω
Explanation: The formula R=V/I is used to calculate resistance from voltage and current.
A component has a resistance of 10 Ohms. If a potential difference is applied across the component, describe the relationship between the potential difference and the current flowing through it.
According to Ohm's Law (V=IR), the potential difference is directly proportional to the current, assuming the resistance remains constant. This means if you double the potential difference, the current will also double, and vice-versa.
In an experiment to determine the resistance of a resistor, a voltmeter reads 6.0 V when connected across the resistor. The ammeter in the circuit reads 0.50 A. Calculate the resistance of the resistor.
Resistance, R = Voltage, V / Current, I
R = 6.0 V / 0.50 A
R = 12 Ω
The resistance is calculated using Ohm's Law. Voltage is divided by current to find the resistance.
Describe how you would use a voltmeter and an ammeter to determine the resistance of an unknown resistor in a circuit.
1. Connect the resistor in a circuit with a power supply, an ammeter (in series), and a voltmeter (in parallel across the resistor).
2. Adjust the power supply to provide a suitable voltage and record the voltage (V) from the voltmeter and the current (I) from the ammeter.
3. Calculate the resistance (R) using Ohm's Law: R = V/I.
State the relationship between the resistance of a metallic wire and its length.
The resistance of a metallic wire is directly proportional to its length. This means that if the length of the wire increases, the resistance increases proportionally, assuming all other factors remain constant.
Explain, qualitatively, how the cross-sectional area of a metallic wire affects its resistance.
The resistance of a metallic wire is inversely proportional to its cross-sectional area. This means that as the cross-sectional area increases, the resistance decreases, and vice versa, assuming all other factors remain constant. A larger area provides more space for electrons to flow, reducing opposition.
Sketch a current-voltage (I-V) graph for a filament lamp. Explain the shape of the graph.
Sketch should show a curve, starting with a steep slope near the origin, then flattening out at higher voltages/currents.
Explanation: At higher voltages, the filament lamp gets hotter. The increased temperature causes the resistance of the filament to increase. Since resistance = voltage / current (R = V/I), for the same increase in voltage, the increase in current becomes smaller. This results in a less steep slope on the I-V graph at higher voltages.
Describe the shape of the current-voltage (I-V) graph for a diode.
The I-V graph for a diode shows very little current flowing when the voltage is negative (reverse bias). At a specific positive voltage (forward voltage), the current increases rapidly.
A copper wire has a resistance of 2.0 Ω and a length of 5.0 m. What is the resistance of a copper wire of the same material with a length of 15.0 m and the same cross-sectional area?
Resistance is directly proportional to length.
R₁/L₁ = R₂/L₂
2.0 / 5.0 = R₂ / 15.0
R₂ = (2.0 / 5.0) * 15.0
R₂ = 6.0 Ω
Answer: 6.0 Ω
Explain why increasing the cross-sectional area of a wire decreases its resistance.
Increasing the cross-sectional area provides more space for electrons to flow. With more available space, electrons experience fewer collisions with atoms in the wire. Fewer collisions mean less opposition to the current, hence a lower resistance.
(Area ∝ 1/Resistance)
About Resistance (4.2.4)
These 10 flashcards cover everything you need to know about Resistance for your Cambridge IGCSE Physics (0625) exam. Each card is designed based on the official syllabus requirements.
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After mastering Resistance, explore these related topics:
- 4.2.3 Electromotive force and potential difference - 14 flashcards
- 4.2.5 Electrical energy and electrical power - 8 flashcards
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