Two bulbs, P and Q, have their resistance in ratio 1 :2. They are connected in a series across a battery. What is the the ratio of power dissipation of this bulb?
Zenishtha:
I remains d same. So using the formula P= I^2 R, we get power of first bulb as 1a and power of second bulb as 2a (where a is some number). Upon dividing them and making 'em into a ratio we get 1:2. But I think this is wrong.
Answers
Answered by
86
Hello Dear.
Given :-
Ratio of the Resistance of two bulbs P and Q = 1 : 2.
Let the Resistance of the bulbs P and Q be x and 2x respectively.
∵ Current remains same in the series,
∴ Let the current in the Resistors P and Q be I.
Now,
For Bulbs P,
Power = P₁
Current = I
Resistance = x
∵ Power = I² × R
∴ P₁ = I² × x
For Bulb Q,
Power = P₂
Current = I
Resistance = 2x
∴ Power(P₂) = I² × 2x
∴ Ratio of the Powers,
P₁/P₂ = I²x/I²2x
P₁/P₂ = 1/2
⇒ P₁ : P₂ = 1 : 2
Hence, the Ratio between the Power Dissipation of the bulb P and bulb Q is 1 : 2.
Hope it helps.
Given :-
Ratio of the Resistance of two bulbs P and Q = 1 : 2.
Let the Resistance of the bulbs P and Q be x and 2x respectively.
∵ Current remains same in the series,
∴ Let the current in the Resistors P and Q be I.
Now,
For Bulbs P,
Power = P₁
Current = I
Resistance = x
∵ Power = I² × R
∴ P₁ = I² × x
For Bulb Q,
Power = P₂
Current = I
Resistance = 2x
∴ Power(P₂) = I² × 2x
∴ Ratio of the Powers,
P₁/P₂ = I²x/I²2x
P₁/P₂ = 1/2
⇒ P₁ : P₂ = 1 : 2
Hence, the Ratio between the Power Dissipation of the bulb P and bulb Q is 1 : 2.
Hope it helps.
Answered by
14
Sir why can we solve it using formula P=Vsqaure/R
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