Two electric bulbs P and Q have their resistances in the ratio 1:2 they are connected in series find the ratio of power dissipation
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Power dissipation, P= V × I or
I^2×R
When connected in series current is same in resistor P & Q.
So, ratio of power dissipation =
P(p)/P(q) = I^2×R(p)/I^2×R(q)
= 1/2 ans
I^2×R
When connected in series current is same in resistor P & Q.
So, ratio of power dissipation =
P(p)/P(q) = I^2×R(p)/I^2×R(q)
= 1/2 ans
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