(2) The voltage across the sliding rheostat is 2V, which can be obtained from the voltage law of series circuit. The voltage across R 1 is = 15v-2v = 13v, and the current through R 1 is 0.2A, so the electric power consumed by R 1 is P65438+.

(3) The maximum value that the voltmeter can reach in the problem, if the range of 0 ~ 15V is selected, because the voltmeter is connected in parallel at both ends of R2, the total voltage is 15V, no matter which resistor is selected, the voltage value cannot reach 15V, so the range of the voltmeter can only be 0 ~ 3V.

The ammeter should also be able to reach the maximum current. If a15Ω resistor is connected, the maximum current in the circuit is IM1= ur =15v15Ω =1a;

If 24Ω is connected, the current in the circuit IM2 = ur ′ =15V24Ω = 0.625a; That is, the maximum value that can be achieved in the circuit can only be 0.6A, so the range of ammeter can only be 0 ~ 0.6a

If R 1 5Ω is replaced by r1,when the maximum current is reached, the total resistance r in the circuit is always = uim2 =15v0.6a = 25Ω; The sliding rheostat should be connected with resistance R' = 25Ω-15Ω =10Ω;

The voltage across the sliding rheostat is u' = im2r' = 0.6a×10Ω = 6v > 3V, which exceeds the maximum voltage of 3v, so we can only choose the resistance of 24Ω instead of r 1.

(2) the maximum current of ammeter I max =0.6A, which can be obtained from ohm's law:

The minimum resistance r in the circuit is the minimum value = the maximum value of UI =15v0.6a = 25Ω;

The access resistance of the sliding rheostat is R2 = 25 Ω-24 Ω =1Ω;

The maximum voltage of the voltmeter is u max =3V, so the voltage across the resistor is U2= 15V-3V= 12V.

Then the current in the circuit is I2 = u2r =12v24ω = 0.5a;

Sliding rheostat access resistance R2' = u maxi2 = 3v0.5a = 6ω;

Therefore, the access range of sliding rheostat is1ω ~ 6ω;

Answer: (1) The access resistance is10Ω; (2) The power consumption of R1is 2.6W；; (3)①24, ② The access range of sliding rheostat is 1ω~ 6ω.