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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
Expert Answer
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 = I12 (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 = I22 (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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