Answer
$7.50\Omega$
Work Step by Step
We can find the known resistance of Wheatstone bridge as follows:
$\frac{R_2}{R_1}=\frac{R_x}{R_3}$
This can be rearranged as:
$R_x=\frac{R_2}{R_1}.R_3$
We plug in the known values to obtain:
$R_x=\frac{15.0\Omega}{25.0\Omega}(12.5\Omega)$
$R_x=7.50\Omega$
