The correct way to think about the internal forces in the ropes is in terms of stress and strain (eg http://en.wikipedia.org/wiki/Stress%E2%80%93strain_curve).
The stress, $\sigma$, is the force per unit area in the rope, for this system we expect the stress to be constant across the cross section of either of the given ropes so, for a given rope, $\sigma = \frac{F}{A}$ where $F$ is the force in that rope and $A$ is that rope's cross sectional area.
The strain, $\epsilon$, is the ratio of extension ($x$) to natural length ($l$) (ie the length of the rope when no force is applied), that is $\epsilon = \frac{x}{l}$.
In general the stress is a function of strain, $\sigma = f(\epsilon)$; because both our ropes are made of the same material they have the same relation between stress and strain that is: (where subscripts label the rope to which the given property belongs)
$$ \sigma_A = \frac{F_A}{A} = f(\epsilon_A) = f\left(\frac{x_A}{l_A}\right) $$
$$ \sigma_B = \frac{F_B}{A} = f(\epsilon_B) = f\left(\frac{x_B}{l_B}\right) $$
A special case of this is called Hookean elasticity, where the stress and strain are proportional, their ratio being a constant, $E$, called the Young's Modulus:
$$ \sigma = E \epsilon $$
A rope will break when it's stress reaches a point known as the ultimate strength of that material, $\sigma_{ult}$
From now on I will assume that both ropes are attached to some kind of a rig such that their attachment points are always parallel, the system cannot twist or shear, and the only way that the system can move is by the bars, to which the ropes are attached, being separated by a force, $F_{app}$, being applied.
In this problem, there are two regions whose behaviour is qualitatively different.
The first region is where the total separation of the bars is less than the natural length of the longer rope, in this case the longer rope is slack, ie it has no tension, and the applied force is entirely balanced by the force from the shorter rope:
$$ F_{app} = F_B = Af\left(\frac{x_B}{l_B}\right) $$
This region continues until the separation of the bars, $x_A+l_A$, is equal to the natural length of the longer rope. If the ultimate strength of the ropes is reached before that, the shorter rope will break and (at least in the instant of the breakage) the tension in the longer rope will be zero.
When the separation of the bars is equal to the natural length of the longer rope:
$$ x_B+l_B = l_A$$
$$ \implies \frac{x_b}{l_A} = \frac{l_A-l_B}{l_B} $$
$$ \implies f\left(\frac{x_b}{l_A}\right) = f\left(\frac{l_A-l_B}{l_B}\right) $$
So the tension in the longer rope will be zero when the shorter rope breaks if the ultimate stress is less than the stress at this separation, ie:
$$ \sigma_{ult} < f\left(\frac{l_A-l_B}{l_B}\right) $$
Otherwise we move into the second region, where the external force is balanced by (possibly different) tensions in each of the ropes. For some given separation of the bars, $d$, we have:
$$ F_{app} = F_A + F_B = Af\left(\frac{d-l_A}{l_A}\right) + Af\left(\frac{d-l_B}{l_B}\right) $$
Now, the shorter rope will still break first, but the tension in the longer rope when this happens will be non-zero, to find this tension: set the stress in the shorter rope to be the material's ultimate strength, find the bar separation for which this occurs, and substitute this into the the equation for the stress in the longer rope.
To do this you need to know the exact form of the stress-strain relationship, for Hookean behaviour:
$$ \sigma_{ult} = E \frac{d-l_B}{l_B} $$
$$ \implies d = \frac{l_B\sigma_{ult}}{E} + l_B $$
$$ \implies F_A = A\sigma_A(d) = A\sigma_{ult}\frac{l_B}{l_A}+AE\left(\frac{l_B}{l_A}-1\right) $$
EDIT:
In general the cross sectional area of the ropes will also be a function of strain, ie $A=g(\epsilon)$. Hence:
$$ \sigma = \frac{F}{A} = f(\epsilon) \implies F = g(\epsilon)f(\epsilon) $$
A common form for the dependence of the area on strain is:
$$ A = A_0\epsilon^{-\nu} $$
where $\nu$ is the Poisson's ratio. http://en.wikipedia.org/wiki/Poisson's_ratio
If the volume is conserved during deformation, Poisson's ratio is equal to exactly one half, but there is no fundamental reason that volume should be conserved.
The results above are easy to generalise using this new formula, although the algebra will be complicated. Start with:
$$ F_{app} = F_A + F_B = g\left(\frac{d-l_A}{l_A}\right)f\left(\frac{d-l_A}{l_A}\right) + g\left(\frac{d-l_B}{l_B}\right)f\left(\frac{d-l_B}{l_B}\right) $$
then substitute whatever functional dependence you want the area and stress to have for $f$ and $g$, eg $f(\epsilon)=E\epsilon$, $g(\epsilon)=A_0\epsilon^{-\nu}$, find the value of $d$ for which the shorter rope breaks using $\sigma_{ult} = f\left(\frac{d-l_B}{l_B}\right)$, then substitute this into the equation for the tension in rope A. The algebra will be complicated but it's not difficult.
