If we can compare thermal currents to electric currents, can we use electric current laws for thermal currrents? I mean, me I take the Kirchoff's laws or other similar laws. Can we find analogous laws for the same in thermal physics?
 A: Most of the good conductors of electricity are good conductors of heat. This is a bit obvious as the cause for both electrical currents and thermal currents are electrons.
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Definitions of the quantities and formulas used in current electricity
$\Delta V = $ potential difference across the electric conductor
$q = $ electric charges
$I_E = \frac{dq}{dt} = $ electric current
$R = $ electrical resistance
$G = \frac{1}{R} = $ electrical conductance
$\rho = $ electrical resistivity
$\sigma = \frac{1}{\rho} = $ electrical conducvity 
We have Ohm's law which states that:
$$I = \frac{\Delta V}{R} \tag{1}$$
From microscopic theory of electrical conductivity, we have:
$$R = \rho \frac{l}{A} \tag{2}$$
where $l$ is the length of the conductor and $A$ is the cross-section area of the conductor.
Kirchhoff's current rule states that the sum of currents leaving a node is equal to zero. This is a consequence of conservation of charge.
Kirchhoff's voltage rule states that the sum of potential drops and potential sources in a circuit is equal to zero. This is a consequence of conservation of energy.
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Definitions of the quantities and formulas used in heat transfer through conduction in steady state
$\Delta T = $ temperature difference across the thermal conductor
$Q = $ heat
$I_T = \frac{dQ}{dt} = $ thermal current
$k = $ conductivity of the thermal conductor
Fourier's law states that:
$$I_T = k\frac{A}{l} \Delta T \tag{3}$$
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Drawing analogies between the electrical quantities/formulas and the thermal equivalents
You can rearrange equation $(3)$ to:
$$I_T = \frac{\Delta T}{\frac{l}{kA}}$$
If you compare the above equation with equation $(1)$, you can 
$$I_T = \frac{\Delta T}{R_T} \tag{4}$$
where $R$ is equal to
$R = \frac{l}{kA} \tag{5}$
Equation $(4)$ looks quite similar to equation $(1)$ and equation $(5)$ looks quite similar to equation $(2)$ except for the $k$ being in the denominator. Aha! You realize that $k$ is thermal 'conductivity' and not resistivity ($\rho$ is the electrical resistivity in equation $(2)$).
Using $(5)$, we can define the quantity thermal resistance. 
The temperature gradient acts like the electric potential difference.
A potential difference across an electrical conductor drives an electric current. Similarly, a temperature difference across an electrical conductor drives a thermal current.
There cannot be heat buildup at a particular node in steady state. Therefore, the sum of thermal currents at a node must be equal to zero.
A point in the thermal conductor cannot have two temperatures. If you sum up the temperature differences as you go around a loop, you must get zero. If it weren't zero, then you'd have two temperatures at one point.
As you can see, that almost every quantity associated with thermal currents has a perfect analogous quantity in electric currents. You can also see that the fundamental laws in electricity work for thermal currents.
Hence, you can conclude that results obtained for simple electrical circuits should work in thermal circuits.

Wheatstone bridge example

The lines in red represent thermal conductors with resistance. The lines in black represent ideal thermal conductors (zero resistance).
Q: Find the thermal current in the middle branch ($QS$).
Instead of solving the problem using equation $(3)$, we will solve the problem as if it were an electrical circuit.
Using $(5)$, you can obtain the resistance for each conducting rods.
$R_1 = \frac{l}{4A}$
$R_2 = \frac{l}{2A}$
$R_3 = \frac{l}{2A}$
$R_4 = \frac{l}{A}$
$R_5 = \frac{l}{16A}$
The temperature difference across the points $P$ and $R$ is
$\Delta T = T_2 - T_1$
Careful observation reveals that the Wheatstone bridge's conditition is satisfied:
$$\frac{R_1}{R_2} = \frac{R_3}{R_4}$$
This would mean that the bridge is balanced and hence the current in the rod $QS$ is zero.
You can solve the above using equation $(3)$ and you'll end up with the same answer.
