The Electrical Properties of Solids — Question 5

Question 5

A tapered conductor requires integrating local resistance

A conductor of fixed resistivity $\rho$ has length L and cross section $A(x)=A_0(1+x/L)$ for $0\le x\le L$. Steady current I is the same through every slice.

1. Derive its total resistance.

2. Find the field ratio at the narrow and wide ends and compare resistance with a uniform bar of area $A_0$.

Question 5 - Solution

A tapered conductor requires integrating local resistance

(a) Method and calculation

$$dR=\rho\,dx/A(x)\Rightarrow R=\frac{\rho}{A_0}\int_0^L\frac{dx}{1+x/L}.$$$$\boxed{R=(\rho L/A_0)\ln2}.$$

(b) Interpretation and further calculation

$$E(x)=\rho I/A(x)\Rightarrow\boxed{E(0)/E(L)=2}.$$$$R_0=\rho L/A_0\Rightarrow\boxed{R/R_0=\ln2=.6931}.$$

Physical check

Current is constant but current density is larger where the area is smaller. Using an arithmetic average area does not reproduce the exact resistance.