Subelement A: Principles— Topic 8: Conduction
Question 3-8A3
Element 3 (GROL)Halving the cross-sectional area of a conductor will:
Explanation
The resistance of a conductor is inversely proportional to its cross-sectional area. This means that if the area decreases, the resistance increases proportionally, and vice-versa.
Imagine a conductor as a pipe carrying water (current). If you make the pipe narrower (reduce its cross-sectional area), it becomes harder for the water to flow through, increasing the "resistance" to flow.
Mathematically, resistance $R = \rho \times (L/A)$, where $\rho$ is resistivity, $L$ is length, and $A$ is the cross-sectional area. If $A$ is halved, the denominator becomes half its original value, effectively doubling the resistance. Therefore, halving the cross-sectional area of a conductor will double its resistance.
Related Questions
3-8A1 What is the term used to identify an AC voltage that would cause the same heating in a resistor as a corresponding value of DC voltage?3-8A2 What happens to reactive power in a circuit that has both inductors and capacitors?3-8A4 Which of the following groups is correct for listing common materials in order of descending conductivity?3-8A5 How do you compute true power (power dissipated in the circuit) in a circuit where AC voltage and current are out of phase?3-8A6 Assuming a power source to have a fixed value of internal resistance, maximum power will be transferred to the load when: