(1) When the slider of the sliding rheostat moves to the right, the resistance of the access circuit increases, and the total resistance of the circuit increases;

According to ohm's law, the current in the circuit becomes smaller, that is, the pointer of the ammeter becomes smaller;

According to the characteristics of resistance voltage division in series circuit, the voltage divided at both ends of sliding rheostat becomes larger, that is, the indication of voltmeter becomes larger.

(2) When the current representation number I=0.4A,

The voltage across R 1 is u1= IR1= 0.4a×10ω = 4v,

The voltage U2 on R2 = u-u1=18v-4v =14v,

The electric power consumed by the sliding rheostat R2 is P2 = U2i =14v× 0.4a = 5.6W. 。

(3) According to the problem, the maximum current in the circuit should be iMax =0.6A,

That is, I max = ur1+R2 min =18v10ω? +R2 minimum =0.6A,

∴ Minimum value of sliding rheostat connected to the circuit r Minimum value = 20 Ω,

R 1 maximum electric power consumed: p1max = imax2× r1= (0.6a) 2×10ω = 3.6w. 。

(4) In order to protect the voltmeter, the maximum voltage across the sliding rheostat is 15V. At this time, the resistance of the sliding rheostat connection is the largest and the current in the circuit is the smallest.

∫U = U 1′+U2′，

∴u 1′=u-u2′= 18v- 15v=3v，

At this point, the minimum current in the circuit is:

I′= U′ 1r 1 = 3v 10ω= 0.3A，

The maximum resistance of sliding rheostat is:

r max = u2i ' = 15v 0.3a = 50ω；

R 1 Minimum power consumption:

p 1min = u 1′I′= 3v×0.3A = 0.9W；

It can be seen that the allowable adjustment range of sliding rheostat is 20 Ω ~ 50 Ω,

R 1 the ratio of the maximum and minimum electric power consumed is P 1 maximum: P 1 minimum = 3.6W: 0.9W = 4: 1.

So choose BCD. ..