Power in Electric Circuits — Question 4

Question 4

A tapered conductor dissipates more power where its cross-section is smaller

A wire has $A(x)=A_0(1+x/L)$ on $0<x<L$, resistivity rho and constant current I. Use volumetric heating $p_v=\vec J\cdot\vec E$.

1. Find total power by integrating the local heating density.

2. Find the position x that divides the dissipated power into equal halves.

Question 4 - Solution

A tapered conductor dissipates more power where its cross-section is smaller

(a) Method and calculation

$$J=I/A(x),\quad p_v=\rho I^2/A(x)^2,\quad dP=p_vA(x)dx.$$$$P=\frac{\rho I^2L}{A_0}\ln2=I^2R.$$

(b) Interpretation and further calculation

$$\frac{P(0,x)}{P(0,L)}=\frac{\ln(1+x/L)}{\ln2}.$$$$\ln(1+x/L)=\tfrac12\ln2\Rightarrow x=(\sqrt2-1)L.$$

Physical check

Integrating power density requires the local volume element A(x) dx, not just dx.