Cylindrical capacitor in an electric circuit I've come across a tricky question and would appreciate some hints or explanations as to why the given solution is the way it is. The question reads as follows:

A coaxial cable consists of a wire with radius $a$ (the core of the cable), which is wrapped with insulating material with dielectric constant $\varepsilon$, until radius $b$ (called the insulator). Around the cable there is a layer of conducting material (radius $c$ from the center of the cable and is called the wrapper).
The wire's length is $d$ such that $d \gg a,b,c$. At one side of the cable, a voltage source $V_0$ with inner resistance $R_0$ is connected to both the wire and the wrapper, and at the other side, a resistor $R$ is connected instead of a voltage source.


It asks to find the magnetic and electric fields $B(r)$,  $E(r)$, where $r$ is the distance from the center of the cable (from $z$-axis in the picture), when $t\rightarrow+\infty $.
In the solution, they said that when $t\rightarrow +\infty$, no current will pass through the cylindrical capacitor so: $I=\frac{V_0}{R_0+R}$ therefore $V\left(\text{final}\right)=V_0 \frac{R}{R_0+R}$.
I do not get this, how can one imagine how this circuit works? Is there an equivalent and more simple circuit? According to what they said, after infinite time, no current passes through the capacitor, but the wires are connected to the wrapper so how can there be current at all in the circuit? All I know is when an uncharged capacitor is charged, it will act as an open switch in the circuit after a long time.
Possible equivalent Circuit?:

 A: Your "possible equivalent circuit" is correct and you have to just understand that the capacitor, in this case, happens to be particularly long, so that it happens to incorporate into its body both the wire up-top and the wire beneath.
As for "how can current flow?" the answer is "for long time scales a capacitor looks like a break in the circuit, that's why we draw it with this space in the middle. Eventually the capacitor gets charged up and no charge flows over the capacitor. However charge does still flow across the wire up-top and the wire on the bottom; it just doesn't flow by means of the capacitor: it follows the wires on the top and the bottom." 
A: There is only one connection to the wrapper, at the power supply end.  One can read the problem to imply that the resistor at the other end is connected to the wrapper, but you shouldn't.  If there were a connection there, the resistor would be shorted out.  As the wrapper has only one connection, at long $t$ it will have come to the proper equilibrium potential and there will be no current in the wrapper.  Your basic circuit is then an ideal battery in series with two resistors, which is where the final current calculation comes from.
A: I think the solution is incorrect. 
If the current becomes zero after an infinite time, the potential of the rod and the wrapper would become equal ( because they are connected by a wire of $R$ resistance, and if $I=0$, the $V$ across the wire $=0$. ) 
Also, your diagram seems incorrect. You assume that there are different wires connecting the capacitor and the resistance $R$ but, in the question, the wire IS a part of the capacitor.
