Orientation of resistor when connected in a cicuit Consider a cuboid shaped resistor.
Now
Case1: Connect the resistor along is length to a battery.
Case2: Connect the resistor along is breadth to a battery.
In the two cases the length and cross-sectional area along and through which current flows are different so resistance should change right? 
Is this correct or the resistance is same in both cases?

 A: More generally, the resistance depends on the position of the connections and the shape of the conductor. 
A: In my answer, I assume the material of the resistor to be homogeneous and isotropic and described by a local resistivity $\rho$ within Drude's theory of linear conductance.
I further assume, the contacts to the resistive material do not contribute an appreciable contact resistance at the surface of the material. They cover the surface areas of the cuboid which are normal to the current flow.
Let the dimensions of the cuboid be $L_x$, $L_y$, and $L_z$.
Under all these assumptions, the resistance of the cuboid is given by
$$ R = \rho \frac{L}{A},$$
where $L$ is the cuboid's length along the direction of current flow, and $A$ is the cross sectional area normal to the direction of current flow in each of the individual cases mentioned below.
So, driving the current along $x$ leads to
$$ R_x = \rho \frac{L_x}{L_yL_z},$$
driving the current along $y$ gives
$$ R_y = \rho \frac{L_y}{L_xL_z},$$
and driving it along $z$ gives
$$ R_z = \rho \frac{L_z}{L_xL_y}.$$
A: The resistance of the cuboid will be
$$R=ρ\frac{L}{A}$$
Where $ρ$ is the resistivity of the material (Ohm-m) and is assumed the same through out the material and in every direction. $L$ is the length (m) along the current path. $A$ is the area perpendicular to the current path ($m^2$).
The resistance is therefore higher in case 1 than case 2 due to both the length being greater and the area being less than case 2. 
Hope this helps
