In the given circuit diagram, the currents, $I_{1} = - 0.3 A, I_{4} = 0.8 A$ and $I_{5} = 0.4 A,$ are flowing as shown. The currents $I_{2}, I_{3}$ and $I_{6},$ respectively, are:

Question

In the given circuit diagram, the currents, I 1 = - 0.3 A , I 4 = 0.8 A and I 5 = 0.4 A , are flowing as shown. The currents I 2 , I 3 and I 6 , respectively, are:

TestHub · Accepted Answer

G i v e n I 1 = - 0 . 3 A , I 4 = 0 . 8 A , I 5 = 0 . 4 A , i n s e r i e s c o n n e c t i o n c u r r e n t w i l l b e s a m e , h e n c e I 5 = I 6 = 0 . 4 A , A c c o r d i n g t o k i r c h o f f ' s j u n c i o n r u l e , a l g e b r a i c s u m o f t h e c u r r e n t s m e e t i n g a t t t h e j u c t i o n i s z e r o . i . e c u r r e n t s f l o w i n g i n t o t h e j u n c t i o n i s t a k e n a s p o s i t i v e a n d c u r r e n t f l o w i n g o u t o f t h e j u n c t i o n i s t a k e n a s n e g a t i v e . A t t h e j u n c t i o n R , - I 4 + I 2 + I 1 = 0 ⇒ - 0 . 8 + I 2 - 0 . 3 = 0 ⇒ I 2 = 1 . 1 A At junction Q , I 6 + I 3 - I 2 - I 1 = 0 ⇒ 0 . 4 + I 3 - 1 . 1 + 0 . 3 = 0 ⇒ I 3 = 1 . 1 - 0 . 7 = 0 . 4 A It is problem of kirchhof's current law. I 6 = 0.4 A I 3 = 0.8 A - O . 4 A = 0.4 A I 2 = 0.8 + 0.3 A = 1.1 A

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