A battery of voltage V is connected to 30V bulb and 60 W bulb as shown in the figure. (a) Identify brightest bulb (b) which bulb has greater resistance? (c ) Suppose the two bulbs are connected in series, which bulb will glow brighter?

Explanation

(a) The power delivered by the battery `P =VI`. Since the bulbs are connected in parallel, the voltage drop across each bulb is the same. If the voltage is kept fixed, then the power is directly proportional to current `(P prop I)`. So 60 W bulb draws twice as much as current as 30 W and it will glow brighter than others.
(b) To calculate the resistance of the bulbs, we use the relation `P=(V^(2))/(R )`. In both the bulbs, the voltage drop is the same, so the power is inversely proportional to the resistance or resistance is inversely proportional to the power `(R prop (1)/(P))`. It implies that, the 30W has twice as much as resistance as 60 W bulb.
(c ) When these two bulbs are connected in series, the current passing through each bulb is the same. It is equivalent to two resistors connected in series. The bulb which has higher resistance has higher voltage drop. So 30W bulb will glow brighter than 60W bulb. So the higher power rating does not always imply more brightness and it drpends whether bulbs are connected in series or parallel.