Easy INTEGER +4 / -1 PYQ · JEE Mains 2022

An electrical bulb rated 220 V, 100 W, is connected in series with another bulb rated 220 V, 60 W. If the voltage across combination is 220 V, the power consumed by the 100 W bulb will be about _______ W.

Answer (integer) 14

Solution

<p>$${P_{100}} = {{{V^2}} \over {{R_{100}}}} \Rightarrow {R_{100}} = {{{V^2}} \over {{P_{100}}}}$$</p> <p>$${P_{60}} = {{{V^2}} \over {{R_{60}}}} \Rightarrow {R_{60}} = {{{V^2}} \over {{P_{60}}}}$$</p> <p>$${P_{net}} = {{{V^2}} \over {{R_{60}} + {R_{100}}}} = {{{P_{60}}{P_{100}}} \over {{P_{60}} + {P_{100}}}} = {{60 \times 100} \over {160}} = 37.5$$</p> <p>This power developed is proportional to resistance.</p> <p>So, $$P{'_{60}} = {P_{net}} \times {{60} \over {160}} = 37.5 \times {{60} \over {160}} \simeq 14\,W$$</p>

About this question

Subject: Physics · Chapter: Current Electricity · Topic: EMF and Internal Resistance

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