As these are identical lamps we can assume that their brightness depends either on the current through them or on the voltage across them, whichever is easier to find. (Note that if they had been non-identical lamps, then we would have had to find the power, VI, to detect the brightness).

Opening the switch will increase the total resistance of the circuit, reducing the current through W. Hence B and D can be eliminated. And opening the switch will also increase the voltage across Y – from about V/3 to V/2. Hence C.