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Question Number 38967 by rahul 19 last updated on 01/Jul/18

Commented by rahul 19 last updated on 01/Jul/18

$find V_{a} - V_{b} ?$
Commented by rahul 19 last updated on 01/Jul/18
I' m getting correct ans. by Nodal method but when I do by Parallel grouping of cells , I'm getting wrong ans.
Commented by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18

$t h e y c e l l a r e n o t p a r a l l e l g r o u p e d . . t h o u g h i t l o o k$ $l i k e p a r a l l e l b u t p d a c r o s s e a c h c e l l a r e n o t s a m e$
Commented by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18

$o k b u t t o m o r r o w . . . i f y o u w a n t f r e e b o o k s t o$ $d o w n l o a d v i s i t a r c h i e v s . o r g . . .$
Commented by rahul 19 last updated on 01/Jul/18
Assume R1 internal resistance of E1 and R2 internal resistance of E2. Then all the cells have same P.D , right ? ��
Commented by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18

$i h a v e p o s t e f t h e o r e m r e l a t e d t o c e l l s . . . p l s p u r c h a z e$ $r e s n i k a n d h a l i d a y a n d d c p a n d e y b o o k s . . . i h a v$ $r e a d l o t o f p h y s i c s b o o k s b u t r e s n i k h a l i d a y$ $f o r c o n c e p t a n d p a n d e y f o r p r o b l e m s g o o d . .$ $a s p e r y o u r q u e s t i o n i f i n t e r n a l r e s i s t a n c e t h e n$ $p a r a l l e l$
Commented by rahul 19 last updated on 01/Jul/18

$Sir, pls solve this q ., by parallel grouping$
Commented by rahul 19 last updated on 02/Jul/18

$? ? ?$
	Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jul/18

	$s t a r t f r o m b (- I_{1} R_{3} + I_{2} R_{3} + E_{3} - I_{1} R_{2} - E_{2} = 0$ $(- I_{1} \times 4 + I_{2} \times 4 + 10 - I_{1} \times 10 - 8 = 0$ $- 14 I_{1} + 4 I_{2} + 2 = 0$ $s t a r t f r o m E_{1}$ $E_{1} - I_{2} R_{1} - E_{3} - I_{2} R_{3} + I_{1} R_{3} = 0$ $6 - I_{2} \times 5 - 10 - I_{2} \times 4 + I_{1} \times 4 = 0$ $4 I_{1} - 9 I_{2} - 4 = 0$ $s o l v e$ $- 14 I_{1} + 4 I_{2} + 2 = 0 \times 2$ $4 I_{1} - 9 I_{2} - 4 = 0 \times 7$ $- 28 I_{1} + 8 I_{2} + 4 = 0$ $28 I_{1} - 63 I_{2} - 28 = 0$ $- 55 I_{2} + 4 - 28 = 0$ $I_{2} = \frac{24}{- 55} 4 I_{1} + \frac{216}{55} - 4 = 0$ $4 I_{1} + \frac{- 4}{55} = 0$ $I_{1} = \frac{1}{55}$ $c u r r e n t t h r o u g h R_{1} = \frac{24}{55}$ $R_{2} = \frac{1}{55}$ $R_{3} = \frac{1}{55} - (\frac{- 24}{55}) = \frac{25}{55}$ $V_{a} \sim V_{b} = E_{3} - R_{3} (I_{1} - I_{2})$ $= 10 - 4 \times \frac{25}{55}$ $= \frac{550 - 100}{55} = \frac{450}{55} = \frac{90}{11}$ $p l s c h e c k i s i t c o r r e c t . . . i f n o t w h e r e t h e w r o n g$
	Commented by rahul 19 last updated on 04/Jul/18
	Correct! But I asked for parallel grp. method .
	Commented by tanmay.chaudhury50@gmail.com last updated on 05/Jul/18

	$o k p l s$
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