Answer Posted / chaturbhuj poddar

Let us suppose that the first bulbs is marked as 100W-200V and the second bulb as 200W-200V, and both are connected in series to 200V d.c. supply to make easy calculation.
Resistance of 100W bulb R1 = V^2/P = 200*200/100 = 400 ohm,
Resistance of 200W bulb R2 = V^2/200 = 200*200/200 =200 ohm
Thus, Resistance of the combination = =600 ohm. current through each bulb = V/600= 200/600 =1/3 A
Heat produced by 100W bulb per second H1=I^2*R1=(1/9)*200J/sec
Heat produced by 200W bulb per second H2=(1/9)*400 J/s.
H2 is greater than H1. Thus, 200W bulb will glow brighter.

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