Transmission Line Model, Symmetric vs Unsymmetric
There's absolutely no difference, in fact four different circuits could be used, the L and the T depicted in the question and the Pi and the inverted L.
All of them are equivalent.
You have two constitutive equations, one for the the voltage drop and the other for the shunt current.
Taking for instance this L diagram



simulate this circuit &ndash; Schematic created using CircuitLab
and writing  the current change along the x direction we have
$$ i(x)-i(x+\mathrm{d}x) = G\,\mathrm{d}x\;v(x+\mathrm{d}x)+C\,\mathrm{d}x\;\frac{\partial}{\partial t}v(x+\mathrm{d}x)$$
then we substitute for  removing v(x+dx)
$$v(x+\mathrm{d}x)=v(x)-R\,\mathrm{d}x\;i(x)-L\,\mathrm{d}x\;\frac{\partial}{\partial t}i(x)$$
we get
$$ 
\begin{align} 
i(x)-i(x+\mathrm{d}x) &amp;= G\,\mathrm{d}x\left(v(x)-R\,\mathrm{d}x\;i(x)-L\,\mathrm{d}x\;\frac{\partial}{\partial t}i(x)\right)+\\ &amp;+C\,\mathrm{d}x\;\frac{\partial}{\partial t}\left(v(x)-R\,\mathrm{d}x\;i(x)-L\,\mathrm{d}x\;\frac{\partial}{\partial t}i(x)\right)
\end{align}$$
$$ 
\begin{align} 
i(x)-i(x+\mathrm{d}x) &amp;= G\,\mathrm{d}x \;v(x)
-\left(RG\;i(x)+LG\;\frac{\partial}{\partial t}i(x)\right)(\mathrm{d}x)^2+
\\ &amp;+C\,\mathrm{d}x\;\frac{\partial}{\partial t}v(x)-\frac{\partial}{\partial t}\left(RC\;i(x)+LC\;\frac{\partial}{\partial t}i(x)\right)(\mathrm{d}x)^2
\end{align}$$
so, in turn, we have terms proportional to dx while those proportional to dx squared, are more infinitesimal and has then to be neglected.
$$(\mathrm{d}x)^2=\omicron(\mathrm{d}x) $$
and finally
$$i(x)-i(x+\mathrm{d}x)=G\,\mathrm{d}x \;v(x)+C\,\mathrm{d}x\;\frac{\partial}{\partial t}v(x)$$
$$\frac{\partial}{\partial x}i(x,t)=G \,v(x,t)+C\,\frac{\partial}{\partial t}v(x,t)$$
The same applies to the voltage drop and shunt current for thr L and inverted L circuits and, possibly with some more boring algebra, to the T and Pi versions as well.