What is the resistance, under normal working conditions, of an electric lamp rated at '240 V, 60 W'? If two such lamps are connected in series across a 240 V mains supply, explain why each one appears less bright.

Given,

Voltage (V) = 240 V

Power (P) = 60 W

Resistance (R) = ?

From relation

$P = \dfrac{V^2}{R} \\[0.5em]$

We get,

$60 = \dfrac{240^2}{R} \\[0.5em] \Rightarrow R = \dfrac{240^2}{60} \\[0.5em] \Rightarrow R = 960 Ω$

Hence, resistance of the electric lamp = 960 Ω

From the relation for current through the element, we know,

$I = \dfrac{P}{V}$

Substituting the values in the formula above, we get,

$I = \dfrac{60}{240} \\[0.5em] \Rightarrow I = 0.25 A$

When one lamp is connected across the mains, it draws 0.25 A current. If two such lamps are connected in series across the mains, current through each bulb becomes —

$\dfrac{240 V}{(960 + 960) Ω} \\[0.5em] = 0.125 A$

i.e., current is halved. Hence, heating (= I²Rt) in each bulb becomes one fourth, so each bulb appears less bright.