KVPY SA Exam 2011 Question Paper

Question : 70 of 80

Marks: +1, -0

In the following circuit, the 1 Ω. resistor dissipates power P. If the resistor is replaced by 9 Ω, the power dissipated in it is

Solution:

10=4i⇒i=5/2

P_i=i^2R

=(5/2)^2×1=25/3

P_f=(10/12)^2×9

=100/{12×12}×9

P_f = P_i

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