Two bulbs of different wattage are connected in series. Which bulb will glow brighter?

Let us suppose that V is the voltage on each bulb and R1 and R2 be the resistance of the two bulbs having power P1 and P2 (P1 < P2) respectively. Then,

\(R_1=\frac{V^2}{P_1}\) and \(R_2=\frac{V^2}{P_2}\)

∴ \(\frac{R_1}{R_2}=\frac{P_2}{P_1}\)

i.e., \(R\propto\frac1P\)

From this relation, we conclude that the resistance of the P1 bulb is more than the P2 bulb. Since in series combination, the same current I flows through each bulb.

Since H = I²R; H is heat developed per unit time, i.e., the power developed

∴ H ∝ R, if the current I is constant.

Therefore power developed across the P1 bulb is more than the P2 bulb. Since the brightness of the bulb is directly proportional to the power developed, the P1 bulb glows more brightly than the P2 bulb.