Question Number 36700 by rahul 19 last updated on 04/Jun/18
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Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jun/18

Commented by rahul 19 last updated on 04/Jun/18
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Commented by tanmay.chaudhury50@gmail.com last updated on 04/Jun/18

Answered by ajfour last updated on 15/Jun/18
![let R_x =x P_x =i_x ^2 x P_x = [(V^( 2) /((20+((30x)/(30+x)))^2 ))×(((30)/(30+x)))^2 ]x =((900V^( 2) x)/((1800+50x)^2 )) = ((9V^( 2) )/(25))×(x/((x+36)^2 )) for small variations in x if power doesn′t change much then for such a value of x (dP_x /dx)=0 ; ⇒ (x+36)^2 =2x(x+36) ⇒ x = 18 Ω .](https://www.tinkutara.com/question/Q37578.png)