Finite cable

There are a variety of possible end conditions we can apply to the cable. Among them are the (a) sealed end where no current can pass and so dV/dx=0, (b) short circuit or open end where the voltage is clamped to 0, (c) leaky ends which is a mixture of the two, some current escapes but not an infinite amount. Let's revert to the dimensionless equations $X=x/\lambda$ and $L=l/\lambda.$Assume that the voltage at X=0 is V₀. Then the general solution to the steady-state equation is:

$$V(X) = V_0\frac{\cosh(L-X) + B_L\sinh(L-X)}{\cosh L + B_L \sinh L}$$

where B_L is an arbitrary constant. This general solution is equivalent to asserting that the boundary condition at X=0 is V₀ and that at X=L

$$B_L V(L) + \frac{dV}{dX}(L) = 0$$

The free parameter, B_L is the ratio of the input conductance for the cable, G_L to that of the semi-infinite cable, $G_\infty.$ That is, $B_L=G_L/G_\infty.$

For example, if we want the sealed end condition at X=L we take B_L=0 so that

$$V(X) = V_0 \frac{\cosh(L-X)}{\cosh L}$$

If we want the open end conditions, we take $B_L=\infty$ so that

$$V(X) = V_0 \frac{\sinh(L-X)}{\sinh L}$$

If we choose B_L=1 then

V(X) = V₀ e^-X

which is precisely the solution to the semi-infinite cable. In figure 3 we plot the steady state voltages for a variety of different cables and at different electronic lengths. These could be solved analytically, but the plots were in fact generated by using XPPAUT.

  

Figure 3: Steady state voltages for a variety of electrotonic lengths and for different end conditions