A wire is pulled at a rate of 3 cm/hr. What is the change in its resistance?

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justaguide | College Teacher | (Level 2) Distinguished Educator

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It is given that the wire is being pulled at the rate of 3 cm/hr. So, we can take the mass of the substance making up the wire and its volume as constant.

Due to the pull, the length increases and there is a corresponding decrease in the area of cross-section.

Let’s denote the initial volume, length and area of cross-section by V, L and A resp.

We have V = L*A

dV/dt = d/dt (L*A)

=> 0 = (dL/dt)*A + L *(dA/dt)

=> 0 = 0.03A + L*(dA / dt)

=> dA/dt = -0.03A/L

Now resistance = R = rho*L/A, where rho is the resistivity, L is the length of the wire and A is the area of cross-section.

The rate of change in resistance is given by

dR/dt = d/dt (rho*L/A)

=> dR / dt = rho*[(A*dL/dt – L*dA/dt)/A^2]

=> rho*[(0.03A – L*dA/dt)/A^2]

substitute dA/dt = -0.03A/L

=> rho*[(0.03A – L*(-0.03A/L))/A^2]

=> rho*[(0.03A + 0.03A)/A^2]

=> rho*0.06/A ohm/hr

Therefore the rate of change of resistance is rho*0.06/A ohm/hr

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