Solution: The equivalent circuit diagram when (1)S is closed and S 1 and S2 are disconnected is shown in figure 1.

IL = I = u 1r 1 = 2vr 1，

Bulb power PL=I2RL=(2VR 1)2RL=4W,

Then RL=R 12, R2=0.5R 1,

Then RL=R 12=4R22,

(2) The equivalent circuit diagram when switches S, S 1 and S2 are all closed is shown in Figure 2.

The ratio of the current through the bulb to the resistor is:

I2IL′= U′R2U′RL = rlr 2 = 4 r22r 2 = 4 R2，I2 = 4IL′R2，

Circuit current IL+I2=2.5A, IL'+4IL'R2=2.5A ①,

PL'=IL'2RL= 1W, because RL=4R22,

Therefore, 4IL ′ 2R22 =1w, IL ′ R2 = 0.5②,

Substituting ② into ① gives the following results: IL ′ = 0.5a, I2=2A,

Substitute IL'= 0.5a into ②, R2 = 1ω,

∵rl=r 12=4r22,∴r 1=2ω,rl=4ω，

IL = I = u 1r 1 = 2vr 1 = 2v2ω= 1A，

u = UV+IRL = 2V+ 1a×4ω= 6，

u′= IL′RL = 0.5a×4ω= 2V；

The ratio of two power supply voltages used continuously is u:u'= 3: 1,

When switches S, S 1 and S2 are all closed, the total power of the circuit is p ′ = u ′ i ′ = 2v× 2.5a = 5w;

Therefore, ABC is correct and does not meet the meaning of the question; D's statement is wrong and in line with the meaning of the question.

So choose D.