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Question Number 40918 by Tawa1 last updated on 29/Jul/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 30/Jul/18

	$i a m u s i n g a n o t h e r m d t h o d t o f i n d c u r r e n t t h r o u g h R_{L}$ $l e t i_{1} c u r r e n t c o n f i n e d i n f i r s t m e s h$ $i_{2} i n s e c o n d m e s h$ $i_{3} i n t h i r d m e s h$ $l$ $10 - 6 i_{1} + 6 i_{2} - 3 i_{1} = 0$ $9 i_{1} - 6 i_{2} = 10 . . . . e q n 1$ $9 - 4 i_{2} + 4 i_{3} - 6 i_{2} + 6 i_{1} - 10 = 0$ $6 i_{1} - 10 i_{2} + 4 i_{3} = 1 . . . e q n 2$ $4 - 3 i_{3} - 4 i_{3} + 4 i_{2} - 9 = 0$ $4 i_{2} - 7 i_{3} = 5 . . . e q n 3$ $i_{3} = \frac{4 i_{2} - 5}{7} p u t t i n g t h i s v a l u e i n e q n 2$ $6 i_{1} - 10 i_{2} + 4 (\frac{4 i_{2} - 5}{7}) = 1$ $42 i_{1} - 70 i_{2} + 16 i_{2} - 20 = 7$ $42 i_{1} - 54 i_{2} = 27$ $14 i_{1} - 18 i_{2} = 9 . . . e q n 4$ $9 i_{1} - 6 i_{2} = 10 . . . e q n 1 \times 3$ $14 i_{1} - 18 i_{2} = 9$ $27 i_{1} - 18 i_{2} = 30$ $s u b s t r u c t . . .$ $- 13 i_{1} = - 21$ $i_{1} = \frac{21}{13} A i s t h e r e q u i r e d c u r r e n t p a s s i n g t h r l u g h$ $R_{L} . . . . p l s c h e c k$
	Commented by Tawa1 last updated on 30/Jul/18

	$God bless you sir . thanks .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 30/Jul/18

	$t h a n k y o u G o d s h o w e r i t s b l e s s i n g t o a l l . . .$
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