phy electricity

1. A two-way swtich must be used for such purpose. Thus (1) is not correct.
The difference between (2) and (3) is only the lamps are connected in series and parallel, which doesn't affect the switching of the lamps.

2. As you can see from the given circuit that the two sockets are indeed connected in series instead of connected in parallel. Hence, the lamps could only glow when both sockets are plugged. Without any one of the socket be plugged with a lamp, the series circuit would not be complete, and no current flows through it.

Therefore, in situation of statement (1), both lamps would glow.
In statement (2), the lamp would not glow, as the circuit is not complete.
Statement (3) is correct, as the circuit is not complete.

3. When bulb X burns out, the voltage across bulb Y remains unchanged (which equals to the battery voltage). Thus the current through Y is the same as before (Ohm's Law, as the resistance of bulb Y would not be changed). As a result, the brightness of bulb Y doesn't change.

The decrease of current from the battery (as recorded by the ammeter) is that no current needs to be supplied to bulb X.

Your explanation is wrong because you have mixed up the "current delivered by the battery (ammeter reading)" and the "current consumed by bulb Y". The two are not the same before bulb X burns out.

2014-04-09 17:00:41 補充：
Q:could you please explain again why circuit 1 is not correct?
A: In circuit 1, the 2 switches are connected in parallel. Hence, any one of the switch is closed will make the circuit complete and the lamps glow....

2014-04-09 17:02:27 補充：
(cont'd)... Thus, no matter what the position of the other switch is (open or close), the circuit is still complete and you cannot switch off the lamps.