two electric bulbs P n Q have their resistances in ratio 1:2.they are connected in series with the battery find the ratio of power dissipated in these bulbs
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1:2
P=v^2/R
2R(p)=R(q)
P=v^2/R
2R(p)=R(q)
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as P=i²R. and current in series is same for both bulbs. hence P1:P2=R1:R2 =1:2 ans.
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