Find the resistance of $1000 m e t e r s$ of copper wire $25 s q . m m$ in cross-section. The resistance of copper is $1 / 58 o h m$ per meter length and $1 s q . m m$ cross-section

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Solution

We know $R = \frac{ρ l}{A} = \frac{1}{58} Ω$

for $l = 1 m$ and $A = 1 s q . m m, R = \frac{1}{58} Ω$

for, $l^{'} = 1000 m = 1000 l$ and $A^{'} = 25 s q . m m = 25 A, R = \frac{ρ 1000 l}{25 A} = \frac{1}{58} \times (\frac{1000}{25}) = \frac{20}{29} Ω$

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Find the resistance of 1000 meters of copper wire 25 sq. mm in cross-section. The resistance of copper is 1/58 ohm per meter length and 1 sq. mm cross-section

We know R=ρlA=158Ω for l=1m and A=1sq.mm,R=158Ωfor, l′=1000m=1000l and A′=25sq.mm=25A,R=ρ1000l25A=158×(100025)=2029Ω

Find the resistance of $1000 m e t e r s$ of copper wire $25 s q . m m$ in cross-section. The resistance of copper is $1 / 58 o h m$ per meter length and $1 s q . m m$ cross-section

We know $R = \frac{ρ l}{A} = \frac{1}{58} Ω$

for $l = 1 m$ and $A = 1 s q . m m, R = \frac{1}{58} Ω$

for, $l^{'} = 1000 m = 1000 l$ and $A^{'} = 25 s q . m m = 25 A, R = \frac{ρ 1000 l}{25 A} = \frac{1}{58} \times (\frac{1000}{25}) = \frac{20}{29} Ω$