Question-36700

Question Number 36700 by rahul 19 last updated on 04/Jun/18

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jun/18

R_{2} {a n d R}_{x} p a r r a l a l s o e q u i v a l a n t r e s i s t a n c e

= \frac{30 x}{30 + x}

t o t a l r e s i s t a n c e = R = 20 + \frac{30 x}{30 + x}

= \frac{600 + 50 x}{30 + x}

t h e r m a l p o w e r = \frac{V^{2}}{R}

p = \frac{V^{2} \times (30 + x)}{600 + 50 x}

n o w x c h a n g e s t o x + △ x

p c h a n g e s t o p + △ p

t o s h o w w h e n △ x \to 0

△ p \to 0

p + △ p = V^{2} \times \frac{(30 + x + △ x)}{600 + 50 x + 50 △ x}

△ p = V^{2} \times {\frac{30 + x + △ x}{600 + 50 x + 50 △ x} - \frac{30 + x}{600 + 50 x}}

= V^{2} {\frac{18000 + 1500 x + 600 x + 50 x^{2} + △ x (600 + 50 x)}{}

\frac{- 18000 - 1500 x - 1500 △ x - 600 x - 50 x^{2} - x 50 △ x}{}

= \frac{N_{r}}{D_{r}}

w h e n △ x \to 0 t h e v a l u e o f N_{r} \to 0

t h a t m e a n s △ p \to 0

h e n c e p r o v e d

Commented by rahul 19 last updated on 04/Jun/18

So, what is the final value of R_{x} ? ?

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

d a t a i n s u f f i c i e n t . . b e c a u s e p = \frac{v^{2}}{R}

t o f i n d w e n e e d v a n d p

Answered by ajfour last updated on 15/Jun/18

l e t R_{x} = x

P_{x} = i_{x}^{2} x

P_{x} = [\frac{V^{2}}{{(20 + \frac{30 x}{30 + x})}^{2}} \times {(\frac{30}{30 + x})}^{2}] x

= \frac{900 V^{2} x}{{(1800 + 50 x)}^{2}} = \frac{9 V^{2}}{25} \times \frac{x}{{(x + 36)}^{2}}

f o r s m a l l v a r i a t i o n s i n x

i f p o w e r {d o e s n}^{'} t c h a n g e m u c h

t h e n f o r s u c h a v a l u e o f x

\frac{{d P}_{x}}{d x} = 0; \Rightarrow {(x + 36)}^{2} = 2 x (x + 36)

\Rightarrow x = 18 Ω .

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