The current when the lamps of Party A and Party B work normally is: I a = P/U = 3/3 =1a; I b =P/U= 12/6=2A。
After connecting them in series, the total resistance is R=3+3=6 ohms.
The current in the series circuit is I = u/r = 9/6 =1.5a.
After series connection, due to the same lighting resistance, the partial pressure is also equal. That is, UA = UB = 9V/2 = 4.5V.
From this, it can be compared that the actual voltage of lamp A is greater than the rated voltage, while the actual voltage of lamp B is less than the rated voltage, so lamp A is bright when it emits light normally, and lamp B is dark when it emits light normally, as if D is right. However, the actual voltage of a lamp exceeds the allowable voltage of the rated voltage, which soon burns out a lamp, so D does not hold. Since the A lamp is burnt out, we must choose the A lamp. ..
Then from the rated current, the rated current of a lamp is 1A, and the actual current is 1.5A, which is half higher. The rated current of lamp B is 2A, but the actual current is only1.5A. Working for a long time may burn out a lamp, so the answer A is correct.
Based on the above analysis, the correct answer is: A.
When the light bulb A burns out, an open circuit is formed, that is, an open circuit. The second lamp in series with it has no current and will not emit light because the circuit is broken. Because the characteristic of series circuit is that somewhere in the series circuit is broken, the whole circuit cannot work.