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प्रश्न
A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10−8Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
उत्तर
Given diameter of the wire,
D = 0.5 × 10−3 m
resistivity of copper, ρ = 1.6 x 10−8 ohm m required resistance, R = 10 ohm
As R = `(ρl)/A, l = (RA)/ρ = (R((piD^2)/4))/ρ = (πRD^2)/(4ρ)` ...`[∵ A = pir^2 = pi(D/2)^2 = (piD^2)/4]`
∴ `l = (3.14 xx 10 xx (0.5 xx 10^-3)^2)/(4 xx 1.6 xx 10^-8)`
m = 122.7 m
Since `R = (ρl)/((πD^2)/4)`
= `(4ρl)/(πD^2), R∝1/D^2`.
When D is doubled, R becomes `1/4` times.
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| Electrical resistivity of some substances at 20°C | ||
| − | Material | Resistivity (Ω m) |
| Conductors |
Silver | 1.60 × 10−8 |
| Copper | 1.62 × 10−8 | |
| Aluminium | 2.63 × 10−8 | |
| Tungsten | 5.20 × 10−8 | |
| Nickel | 6.84 × 10−8 | |
| Iron | 10.0 × 10−8 | |
| Chromium | 12.9 × 10−8 | |
| Mercury | 94.0 × 10−8 | |
| Manganese | 1.84 × 10−6 | |
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49 × 10−6 |
| Manganin (alloy of Cu, Mn and Ni) |
44 × 10−6 | |
| Nichrome (alloy of Ni, Cr, Mn and Fe) |
100 × 10−6 | |
| Insulators | Glass | 1010 − 1014 |
| Hard rubber | 1013 − 1016 | |
| Ebonite | 1015 − 1017 | |
| Diamond | 1012 − 1013 | |
| Paper (dry) | 1012 | |
