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Two electric bulbs rated `P_(1)` and `P_(2)` watt at `V` volt are connected in series across `V` volt mains then their total power consumption `P` isA. `(P_(1)+P_(2))`B. `sqrt(P_(1)P_(2))`C. `P_(1)P_(2)//(P_(1)+P_(2))`D. `(P_(1)+P_(2))//P_(1)P_(2)` |
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Answer» Resistance of `I^(st)=` bulb `R_(1)=(V^(2))/(P_(1))` Resistance of `II^(nd)` bulb `R_(2)=(V^(2))/(P_(2))` When both bulb are connected in series `R_(eq)=V^(2)[(1)/(P_(1))+(1)/(P_(2))]=(V^(2)(P_(1)+P_(2)))/(P_(1)P_(2))` Hence power consumed `P=(V^(2))/(R )=(V^(2))/(V^(2)((P_(1)+P_(2))/(P_(1)P_(2))))=(P_(1)P_(2))/(P_(1)+P_(2))` |
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