How to plot solutions for simultaneous equations

I have 2 simultaneous equation of $x$ and $y$. Here is how I solve it, and it works. But I want to vary the parameters n and imax, and plot the result of seeb*(xx - yy) vs imax for different values of n (say imax from 1 to 10 and n=17,30,100,1000). How can I do that? Thanks.

n = 17;
imax = 0.5;

seeb = 2*n*0.0002;
dTmax = 300 - ((Sqrt[1 + 2*0.72] - 1)/(0.72/300));
vmax = seeb*300;
qmax = seebimax*300 - 0.5*imax*seeb*(300 - dTmax);
r = (300 - dTmax)*vmax/(300*imax);
rth = (2*300*dTmax)/((300 - dTmax) vmax*imax);
length = r/(0.001/100)*0.65*0.65/1000/2/n;

equation1[x_, y_] := -(308 - x)/0.01 + (x - y)/rth +
seeb^2*x*(x - y)/2/r - 0.5 seeb^2*(x - y)*(x - y)/4/r
equation2[x_, y_] := (y - 300)/100 - (x - y)/rth -
seeb^2*y*(x - y)/2/r - 0.5 seeb^2*(x - y)*(x - y)/4/r
{xx, yy} = {x, y} /.
FindRoot[{equation1[x, y], equation2[x, y]}, {{x, 308}, {y, 300}}]

Print["V="]; seeb*(xx - yy)

You can do it by wrapping your code by Table. Something like this:

Clear[imax];
n=17;

lst = Table[{imax, seeb = 2*n*0.0002;
dTmax = 300 - ((Sqrt[1 + 2*0.72] - 1)/(0.72/300));
vmax = seeb*300;

r = (300 - dTmax)*vmax/(300*imax);
rth = (2*300*dTmax)/((300 - dTmax) vmax*imax);
length = r/(0.001/100)*0.65*0.65/1000/2/n;

equation1[x_, y_] := -(308 - x)/0.01 + (x - y)/rth +
seeb^2*x*(x - y)/2/r - 0.5 seeb^2*(x - y)*(x - y)/4/r;
equation2[x_, y_] := (y - 300)/100 - (x - y)/rth -
seeb^2*y*(x - y)/2/r - 0.5 seeb^2*(x - y)*(x - y)/4/r;
fr = FindRoot[{equation1[x, y],
equation2[x, y]}, {{x, 308}, {y, 300}}];
{xx, yy} = {x, y} /. fr}, {imax, 1, 10, 0.5}];

Then you get the list each element of which has the structure {imax, {solution1, solution2}}. It can be printed as follows:

ListPlot[{lst /. {x_, {y_, z_}} -> {x, y}, lst /. {x_, {y_, z_}} -> {x, z}}, PlotStyle -> {Blue, Red}]

yielding the plot: Since you need to do all this with different n values you can wrap it by table once more:

Clear[n, imax]];

lst2=Table[thePreviousCodeWithoutnAssignment,{n,{17,30,100,1000}}]

However, your plot will be overcrowded in such a case.

Have fun!