The wattage of a bulb is 24W when its is connected to a 12V battery. Calculate its effective wattage if it operate on a 6V battery (Neglect the change in resistance due to unequal heating of the filament in the two cases)
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Answers
Answered by
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A) POWER (P) = Potential Difference (V) × Current (l) OR P = Vl
P = 24W
V = 12V
Therefore,
24 = 12 × l
OR l = 2 Amperes
_______________________________
B ) V = 6V
l = 2 Amperes (concluding from the 1st case)
Therefore,
P = 6 × 2 = 12
Therefore The Effective Wattage = 12W
Hope this helps !
P = 24W
V = 12V
Therefore,
24 = 12 × l
OR l = 2 Amperes
_______________________________
B ) V = 6V
l = 2 Amperes (concluding from the 1st case)
Therefore,
P = 6 × 2 = 12
Therefore The Effective Wattage = 12W
Hope this helps !
rajeshshady9:
P1=24W
p2=6W*
How can it be possible ?!
For Me It's Correct !
Even idk
p1/p2=v1/v2. p2=(v2/v1)•p1=(6/12)^2•24=1/4•24=6W
How Did You Come To Know?
it is given in the book with solution bro..
Okay
But I think this is Correct
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