Answer
a. $R_x= R_2\frac{R_3}{R_1}$.
b. $129\Omega$.
Work Step by Step
a. When the switch is closed, the ammeter shows zero current. Points B and D are at the same electric potential, because if there were a potential difference between points B and D, current would flow through the ammeter.
The potential drop between A and B equals the drop from A to D.
$$I_3R_3=I_1R_1$$
$$\frac{R_3}{R_1}=\frac{I_1}{I_3}$$
Also, the potential drop between B and C must be the same as the drop between D and C.
$$I_3R_x=I_1R_2$$
$$\frac{R_x}{R_2}=\frac{I_1}{I_3}$$
Equate the left hand sides of the equations to find the unknown resistance.
$$\frac{R_x}{R_2}=\frac{R_3}{R_1}$$
$$R_x= R_2\frac{R_3}{R_1}$$
b. Plug in numbers and evaluate.
$$R_x= R_2\frac{R_3}{R_1}= (972\Omega)\frac{78.6\Omega}{590\Omega}=129\Omega$$
