Answer :

Let


Resistance, power and current of bulb P = RP, PP, IP


Resistance, power and current of bulb Q = RQ ,PQ, IQ


It is given in question that,



We know that power dissipated across any resistance is given as


P = I2R


Where P = power dissipated


I = current through the resistance


According to question both the resistance is in series. Therefore, current through both the resistance will be same.


IP = IQ …..(ii)



From equation(ii),



From equation (i),



the ratio of power dissipation in bulbs P and Q is 1:2.


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