A copper wire has a diameter of 0.5 mm and a resistivity of $1.6 \times 10^{- 6}$ ohm-cm. How much of this wire would be necessary to make a resistance of 10 ohm?

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Solution

Given, $D i a m e t e r = 0.5 m m, r e s i s t i v i t y = 1.6 \times 10^{- 8} Ω - m, r e s i s t a n c e = 10 Ω$
we know that $R = ρ \frac{1}{A} \Rightarrow 1 = \frac{R A}{ρ}$
Where A is the area of the cross-section,
$= \frac{10 \times 3.14 \times (\frac{0.0005}{2})^{2}}{1.6 \times 10^{- 8}} = 122.7 m$
Thus the length of the wire is $122.7 m$

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A wire has a diameter of 0.5 mm and a resistivity of 1.6 × $10^{- 6}$ ohm cm. How much of this wire would be required to make a 10 ohm coil?

1.A copper wire has diameter 0.5mm & reactivity of 1.6*10 to the power -8 ohm m.what will be the length of this wire to make its resistance 10 ohm ? How much does the resistance charge if the diameter is doubled?

Q. A wire has a diameter of

0.5 m m

and a resistivity of

1.6 \times 10^{- 6} Ω c m

. How much of this wire would be required to make a

10 Ω

coil?

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A copper wire has a diameter of 0.5 mm and a resistivity of 1.6×10−6 ohm-cm. How much of this wire would be necessary to make a resistance of 10 ohm?

Given, Diameter=0.5mm,resistivity=1.6×10−8Ω−m,resistance=10Ωwe know that R=ρ1A⇒1=RAρWhere A is the area of the cross-section,=10×3.14×(0.00052)21.6×10−8=122.7mThus the length of the wire is 122.7m

A copper wire has a diameter of 0.5 mm and a resistivity of $1.6 \times 10^{- 6}$ ohm-cm. How much of this wire would be necessary to make a resistance of 10 ohm?