Photographic lamp u0

A.S2 is closed, S3 is opened, and the ammeter will short-circuit the bulb and the bulb will not emit light; When S2 is open and S3 is closed, the ammeter will short-circuit the resistor; When both are closed or disconnected, the ammeter has no indication. The rated current of the light bulb can't be measured, and the circuit diagram can't measure the power of the small light bulb when it emits light normally, which doesn't meet the question.

B, S2 is closed, S3 is opened, and the voltmeter measures the voltage at both ends of the lamp, with the upper end being a positive electrode and the lower end being a negative electrode; S3 is closed, S2 is opened, and the voltmeter measures the voltage at both ends of the value resistor R, the upper end of which is negative and the lower end is positive; The voltmeter is connected reversely twice, so it can't be measured without changing the circuit connection mode. Figure B can't measure the power of the small bulb when it emits light normally, which doesn't meet the meaning of the question.

C, the switch is in the position of 1, and the voltmeter measures the voltage u across the bulb and resistor after series connection; The switch is in the position of 2, and the voltmeter measures the voltage U0 at both ends of the bulb, so the current of the small bulb is I=U? U0R, the expression of rated power of small bulb is P=U0I=U? U0RU0, this circuit can measure the power of the light bulb when it normally emits light, which is in line with the meaning of the question;

D, closing the switch S3, short-circuiting the bulb and the resistor by the ammeter, and measuring the current of the sliding resistor by the ammeter; The switch S3 is closed, S2 is opened, and the ammeter short-circuits the small bulb to measure the current through the resistor. The rated current of the light bulb can't be measured, and the circuit diagram can't measure the power of the small light bulb when it emits light normally, which doesn't meet the question.

So choose C.