When the (1) switch S 1 and S3 are closed at the same time, the wires are directly connected to both ends of the power supply, which will lead to short circuit and should be avoided.

(2) Experimental steps:

① Slide the sliding blade of the sliding rheostat to the extreme right;

(2) turn on S2, turn off S 1 and S3, and read out the current representation number I1;

③ Turn off S 1, and read the ammeter instruction I2 when turning off S2 and S3.

(3) The maximum resistance of the sliding rheostat is R = 40Ω, so the voltage of the power supply is u = I1r = 40i1;

When S 1 is off and S2 and S3 are on, the power supply voltage is U = I2Rx.

The supply voltage remains unchanged.

The mathematical expression of ∴Rx is: rx = 40ω× i 1I2.

So the answer is:

(1) No; Power supply is short-circuited (or ammeter is burnt out or power supply is damaged);

(2)① Slide the slider of the sliding rheostat to the extreme right;

(2) turn on S2, turn off S 1 and S3, and read out the current representation number I1;

③ Turn off S 1, and read the ammeter instruction I2 when turning off S2 and S3;

(3)RI = 40ω×I 1i 2。