physics electric circuit

Let E be the emf of the power source.

When S1 and S2 are closed and S3 is open, the first resistance R is being by-passed, leaving only the second resistance (the one drawn vertically on the diagram) connected to the power source.

Hence, P = E^2/R

When both S1 and S2 are open and S3 is closed, the two resistances are connected in series. The total resistance = (R + R) = 2R

Hence, new power P' = E^2/(2R) P/2

The answer is option B

If you wish to find out the current, when S1 and S2 are closed and S3 is open,
i = E/R
thus P = i^2.R = (E/R)^2.R = E^2/R

When S1 and S2 are open and S3 is closed,
new current i' = E/2R
New power P' = (i')^2.(2R) = (E/2R)^2.(2R) = E^2/2R = P/2

Obviously, you get the same result.


2013-07-17 22:41:54 補充：
Sorry, an "equal sign" ( = ) is missing in the 8th line. It should be:
"Hence, new power P' = E^2/(2R) = P/2"