(1) resistors R 1 and R2 are connected in series with the bulb L, and a voltmeter V 1 measures the voltage between the resistor R 1 and the bulb L;

Voltmeter resistor V2 measures the voltage between resistor R2 and both ends of bulb L;

(2) The current I flowing through each circuit element of the series circuit is equal, and U1U2 = IR1IR2 = R1R2 = 8Ω 4Ω = 2, then U1= 2U2 (1);

According to the characteristics of series circuit, UV 1=U 1+UL=7V? ②,

UV2=UL+U2=5V？ ③; U 1=4V, U2=2V, and UL = 3V come from ① ② ③ solution;

(3) The circuit current I = u1r1= 4V8Ω = 0.5A, and the bulb resistance RL = ULI = 3V0.5A = 6Ω;

Then the rated power of the bulb PL=U2L, RL = (6V) 26ω = 6W.

So choose B.