P28_023.PDF

be the emf of the battery. When the bulbs are connected in parallel, the potential diﬀerence

across them is the same and is also the same as the emf of the battery. The power dissipated by

bulb 1 is P 1 =

2 /R 2 .Since R 1 is greater than

R 2 , bulb 2 dissipates energy at a greater rate than bulb 1 and is the brighter of the two.

(b) When the bulbs are connected in series the current in them is the same. The power dissipated by

bulb 1 is now P 1 = i 2 R 1 and the power dissipated by bulb 2 is P 2 = i 2 R 2 .Since R 1 is greater than

R 2 greater power is dissipated by bulb 1 than by bulb 2 and bulb 1 is the brighter of the two.

2 /R 1 , and the power dissipated by bulb 2 is P 2 =

23. (a) Let