From the given current equations, we can cross-multiply:
\[ I_1 (R_1 + r) = E \quad \text{and} \quad I_2 (R_2 + r) = E \]
Equating the two equations for \( E \), we get:
\[ I_1 (R_1 + r) = I_2 (R_2 + r) \]
Expanding both sides:
\[ I_1 R_1 + I_1 r = I_2 R_2 + I_2 r \]
Rearranging the terms:
\[ I_1 r - I_2 r = I_2 R_2 - I_1 R_1 \]
Factor out \( r \):
\[ r (I_1 - I_2) = I_2 R_2 - I_1 R_1 \]
Solving for \( r \), we get:
\[ r = \frac{I_2 R_2 - I_1 R_1}{I_1 - I_2} \]
The internal resistance \( r \) is given by:
\[ r = \frac{I_2 R_2 - I_1 R_1}{I_1 - I_2} \]

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. 

