A wire has a resistance R. If the length of a wire is doubled and the radius is doubled, what is the new resistance in terms of R?

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The resistance of a resistor is given by the formula `R = rho*l/A ` where `rho` is the resistivity of the material the resistor is made of, l is the length of the resistor and A is the cross-sectional area.

In the problem, the resistance of the wire is initially equal to R. `R = rho*l/(pi*r^2)` . The length of the wire is doubled and so is the radius of its cross-section. This increases the cross sectional area to four times the initial area. `R' = rho*(2*l)/(pi*(2*r)^2)` = `rho*(2*l)/(pi*(4*r^2))` = `rho*l/(pi*r^2)*2/4`

The new resistance of the wire is `R*(2/4) = R/2` .

The change in the dimensions of the wire reduces the resistance to half the initial value.

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