Principle of the analogy
Thermal conductance, $\displaystyle{\frac{KA}{L}}$, is the amount of heat that passes through a slab of area ($A$) and thickness ($L$) in unit time, when the opposite faces of the surface differ in tmperature by $1K$. It's reciprocal is the thermal resistance $R_T$,
$$R_T=\frac{L}{KA}\longrightarrow (1)$$
Hence, the rate of heat transfer from one face to the other is given by:
$$\frac{Q}{t}=\frac{KA\Delta T}{L}\longrightarrow(2)$$
which follows from the definition of thermal conductance, when the opposite faces are having a difference in temperature by $\Delta T$. From equation $(1)$, we have
$$\frac{Q}{t}=\frac{\Delta T}{R_T}\longrightarrow(3)$$
Let, the heat current (amount of heat transferred in unit time) be denoted by $I_H$. Hence
$$I_H=\frac{Q}{t}=\frac{\Delta T}{R_T}\longrightarrow(4)$$
Now, comparing this equation with the Ohm's law, we have the electric current, which is the amount of charge flowing per second
$$I=\frac{V}{R}\longrightarrow(5)$$
Comparing both equations, we can make an analogy. We have in equation $(4)$, $\Delta T$, the difference in temperature in analogy with the electric potential $V$, and $R_T$, the thermal resistance in analogy with the electrical resistance $R$.
Just as like a voltage (or more precisely, potential difference) makes a current flow through a metallic conductor, the temperature difference between the two ends of a metallic rod makes the heat flow to happen from a point of higher temperature to that of lower temperature. This makes sense. Hence this analogy is useful.
Hence it can be extended to Kirchhoff's law in electricity also, as the only difference in the two cases are that in one heat is the energy transferred, while in the other, electrical energy is being transferred.
How to work out the problem using this analogy?
One of the ends of the three rods start from a common point, say a heat source maintained at a constant temperature $T$. The direction of current flow states that $T$ is the high temperature region.
The current starts from the junction. Hence at the junction, by the conservation of energy, the currents should be added to zero, just as in Kirchhoff's rule. How's that? Simple: For current flow to take place, there should be temperature difference between the two points. If you take the junction point alone, which is at a single temperature, there is no current flow. Hence the currents add up to zero at the junction. But it flows to the other ends, as obviously there is some temperature difference.
First, use the principle that at the junction $T$,
$$i_{Al}+i_{S}+i_{Cu}=0\longrightarrow(6)$$
If the temperature at the junction is $T$ and that at the end points of the rods are $T_{Al}$, $T_{S}$, and $T_{Cu}$, then the corresponding temperature differences become $(\Delta T)_{Al}=(T-T_{Al})$; $(\Delta T)_{S}=(T-T_{S})$; $(\Delta T)_{Cu}=(T-T_{Cu})$. Now using equation $(4)$, $(6)$ reads
$$\frac{(T-T_{Al})}{R_{Al}}+\frac{(T-T_{S})}{R_{S}}+\frac{(T-T_{Cu})}{R_{Cu}}=0$$
$$T\left[\frac{1}{R_{Al}}+\frac{1}{R_{S}}+\frac{1}{R_{Cu}}\right]=\frac{T_{Al}}{R_{Al}}+\frac{T_{S}}{R_{S}}+\frac{T_{Cu}}{R_{Cu}}\longrightarrow(7)$$.
Now, knowing the resistances, we can calculate $T$.
Best Answer
Okay, the "cylinder" thing means that we can think of this as a 2D system; we can assume that the temperature of the cylinder has reached a steady state. Then there are three constraints at play here:
You are right that if you have two boxes separated by a sheet, then the linear transfer equilibrium is solved for a linear temperature gradient. That is effectively a 1D problem.
But when you generalize this to 2D and higher dimensions, the linear temperature gradient doesn't give you this linear transfer equilibrium. Why not? Well think about the absolute simplest case, two boxes at T=0 and T=1 which are at opposite ends of a 3x3 matrix separated by walls. It's not too hard to work out that the temperature solution of this:
does not look linear:
So the way to think of this example is that, say, focusing on the 1/3 cell, you see a flow of 1/3 "out" to the 0 cell exactly balanced by a flow of 2 * 1/6 "in" from the 1/2 cells next to it. Notice that we're missing two values, 1/6 and 5/6, needed for a "linear progression" across the diagonals. We instead see a graph which has a little bit of a curve when we look along the diagonal. There's an even-simpler case where we force the 3 cells in the top corner to be 0, and that's solved by:
Again, the progression is not linear! {0, 1/4, 1/2, 1} is missing a value 3/4 which is nowhere to be found.
The bottom line is: surface area matters. When you've got this increasing/decreasing surface area, even in the "transfer region" you don't necessarily see linear gradients, because it's not just how much energy is in the box that matters -- it's how many boxes are nearby, that this one is flowing into.
OKAY. Returning to your case, how do we solve the conjunction of (2) and (3) for a circular symmetry? The answer is actually really simple. First we notice that the "local" heat flux at any point $J$ must be dropping like $J = k / r$ because, by (1), it is constant along the circles we're drawing, so the integral of it is simply $2 \pi r J$. If that's constant with respect to radius then we have $J \propto 1/r$.
Now we use (3). The gradient in the radial direction is actually still just d/dr, so we just have to integrate $k/r$ to get: $T(r) ~=~ A ~-~ B~ \operatorname{ln} (k / r) $. Those look like three independent constants but because of log-laws we know that they are not, so we can just set $k = R_0$ or so. Solve for A and B such that $T(R_0) = T_0$ and $T(R_1) = R_1$, if you like.
Finally notice that your intuition is right if the cylinder gets super big and thin, in which case we can write $r = R + \delta r$ and expand to first order, getting a linear temperature gradient.