The equation \( \mathbf{E} = \rho \mathbf{J} \) is known as the electrical conductivity equation. Here:
From this equation, we can express the electric field in terms of current density:
\[ \mathbf{E} = \rho \mathbf{J} \]
Now, consider **Ohm's law**, which states that the current density \( \mathbf{J} \) is proportional to the electric field \( \mathbf{E} \) and the material's conductivity \( \sigma \) (the inverse of resistivity). So, we can write:
\[ \mathbf{J} = \sigma \mathbf{E} \]
Since \( \sigma = \frac{1}{\rho} \), we can substitute this into the above equation:
\[ \mathbf{J} = \frac{1}{\rho} \mathbf{E} \]
Rearranging the equation, we get:
\[ \mathbf{E} = \rho \mathbf{J} \]
This is exactly the form of the equation we started with, so we have derived Ohm's law from the equation \( \mathbf{E} = \rho \mathbf{J} \).
Ohm's law assumes that the material has a constant resistivity \( \rho \) and that the current is proportional to the applied voltage (i.e., linear response). However, there are conditions under which Ohm's law does not hold:
Thus, Ohm’s law is not valid in situations where the material’s resistivity is not constant or when extreme conditions like high electric fields or temperatures cause a non-linear relationship between voltage and current.

The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\). 
A Wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances \[ (R_1=R_2=R_3=R_4). \] When \(R_3\) resistance is heated, its resistance value increases by \(10%\). The potential difference \((V_a-V_b)\) after \(R_3\) is heated is _______ V. 
The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9 V with internal resistance of 1 \(\Omega\) is connected across these points is ______ J. 
