According to the power formula P=UI and the rated current of the bulb = P U = 36V = 0.5A, then the current in the circuit cannot exceed 0.5A;

According to Ohm's law, the resistance of the bulb RL=U, I = 6V0.5A =12ω;

When the resistance of the sliding rheostat is the minimum and the current in the circuit should be 0.5A, it can be obtained from Ohm's law: UI quantity = r1+RL+R2;

Then r1= 24V0.5A-12ω-12ω = 24ω;

The maximum electric power consumed by the resistor R2 is p2 = I = 2R2 = (0.5a) 2×12ω = 3W;

The total power of the circuit is p = UI = 24V× 0.5A =12w;

When the voltage representation number is U 1= 15V, the voltage U2 on the bulb and resistor R2 is u-u1= 24v-15v = 9v;

Then the current in the circuit is I1= u2rl+R2 = 9v12ω+12ω = 0.375a;

The total power of the circuit is p ′ = UI1= 24V× 0.375A = 9W;

The total power of the circuit varies from 9W to 12W;

Then the access resistance of the sliding rheostat r1'= u1I1=15v 0.375a = 40ω;

That is, the allowable resistance range of sliding rheostat is 24Ω ~ 40Ω;

The total voltage of the series circuit is equal to the sum of the partial voltages everywhere,

The voltage across the sliding rheostat is U 1=U-U2-UL=U-I(R2+RL),

Therefore, the ratio of voltmeter to absolute value of current change is expressed as R2+RL = 24ω.

So choose D.