The resistance of a conductor at a temperature \( T \) is given by: \[ R_T = R_0 \left( 1 + \alpha (T - T_0) \right), \] where: \( R_T = 1.25R_0 \) (final resistance is 25% greater than the initial resistance), \( R_0 \) is the resistance at \( T_0 = 27^\circ \text{C} \), \( \alpha = 2.0 \times 10^{-4} \, \text{C}^{-1} \) (temperature coefficient of resistance). Substitute \( R_T = 1.25R_0 \) into the equation: \[ 1.25R_0 = R_0 \left( 1 + \alpha (T - T_0) \right). \] Simplify: \[ 1.25 = 1 + \alpha (T - 27). \] Rearrange: \[ \alpha (T - 27) = 0.25. \] Substitute \( \alpha = 2.0 \times 10^{-4} \): \[ (2.0 \times 10^{-4}) (T - 27) = 0.25. \] Solve for \( T \): \[ T - 27 = \frac{0.25}{2.0 \times 10^{-4}} = 1250. \] \[ T = 27 + 1250 = 1277^\circ \text{C}. \] Answer: The temperature is \( 1277^\circ \text{C} \).

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. 
