Resistivity and Wire-Geometry Scaling
For a uniform wire at fixed temperature, R = rho L/A; therefore wires of the same material obey R2/R1 = (L2/L1)(A1/A2), and circular area scales as diameter squared.
Why this shows up in the exam
Predicting a changed meter-bridge balance · Comparing wire samples · Determining resistivity from measured dimensions
Learn the idea
For the same material, wire resistance grows with length and falls with cross-sectional area. A longer wire gives charge carriers a longer path, while a thicker wire provides more parallel conducting routes; halving diameter reduces area by four, not by two.
🧠 Memory hook: Length counts once; diameter counts inversely twice.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- R = rho L/A = 4 rho L/(pi d²) — uniform circular wire resistance
- R2/R1 = (L2/L1)(d1/d2)² — same-material, same-temperature geometry scaling
How to approach it
- 1Write the resistance ratio
- 2Apply length and diameter changes separately
- 3Feed the new resistance into the bridge relation
Common slip-ups that cost marks
- •Scaling resistance inversely with diameter instead of diameter squared
- •Changing geometry while assuming resistance is unchanged
- •Comparing different materials without retaining rho
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
In a meter bridge, resistance 2 ohm is in the left gap and balance occurs at 40 cm from the left end. Find the right-gap resistance.
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