When a piece of aluminum wire of finite length is drawn through a series of dies to reduce its diameter to half its original value, its resistance will become:
Correct answer: D. Sixteen times
- A. Two times
- B. Four times
- C. Eight times
- D. Sixteen times
Explanation
When the diameter is halved, the cross-sectional area becomes one-fourth its original value. Since the volume of the wire is conserved, the length must become four times longer. The new resistance R' ∝ L'/A' = (4L)/(A/4) = 16(L/A), so the resistance becomes 16 times greater.
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Steady current relates charge flow to potential difference, resistance and resistivity, including how temperature changes a conductor's resistance. Sources with internal resistance have terminal voltage below their emf when delivering current, and maximum power output occurs under a specific load condition rather than at zero resistance.
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