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# JEE Class main Answered

three bulbs are connected in series after two bulbs s key is connected in parallel then what will be the total power p is?
Asked by avneetsingh09 | 09 Mar, 2024, 01:18: PM

Initially two bulbs are connected in series with battery because switch shorted the third bulb.

If R is the equal resistance of each bulb , then current I1 drawn from battery is

I1 = V / (2R)

Power delivered by battery to two bulbs, P = I1(2R) ..................(1)

When switch is open , all three bulbs are in series and connected to battery.

Hence current I2 drawn from battery is

I2 = V/(3R) = (2/3) V(2R) = (2/3) I1

Power delivered by battery to two bulbs, P2 = I2(3R)  = (2/3)2 I12 (3R)

P2 = (3/2) (2/3)2 I12 (2R) = (2/3) I12 (2R) = (2/3) P

As the switch S is opened , then power output becomes (2/3)P

Answered by Thiyagarajan K | 09 Mar, 2024, 05:29: PM

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