Using kirchhoff law in the given network calculate the values of I1, I2 and I3

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Solution

Dear Student,

$I 1 + I 2 = I 3 f r o m j u n c t i o r u l e . i n l o o p 1 12 = 5 I 1 + 2 I 3 12 = 5 I 1 + 2 (I 1 + I 2) 12 = 7 I 1 + 2 I 2 i n l o o p 2 6 = 3 I 2 + 2 I 3 6 = 3 I 2 + 2 (I 1 + I 2) 6 = 5 I 2 + 2 I 1 s o l v e b o t h t h e e q n 60 = 42 I 1 + 10 I 2 12 = 4 I 1 + 10 I 2 48 = 38 I 1 I 1 = \frac{48}{38} = 1.26 A I 2 = 3.16 A R e g a r d s$

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Similar questions

I_{1}, I_{2}, I_{3}, I_{4}

I_{1} = π / 2 \int 0 cos (sin x) d x; I_{2} = π / 2 \int 0 sin (cos x) d x

I_{3} = π / 2 \int 0 cos x d x,

I_{1} = \int_{0}^{\frac{π}{4}} x^{2008} (t a n x)^{2008} d x, I_{2} = \int_{0}^{\frac{π}{4}} x^{2009} (t a n x)^{2009} d x a n d I_{3} = \int_{0}^{\frac{π}{4}} x^{2010} (t a n x)^{2010} d x

I_{1} = \int_{0}^{\frac{π}{2}} c o s (π s i n^{2} x) d x, I_{2} = \int_{0}^{\frac{π}{2}} c o s (2 π s i n^{2} x) d x a n d I_{3} = \int_{0}^{\frac{π}{2}} c o s (π s i n x) d x,

Now what should be the correct relation between $I_{1}$ , $I_{2}$ and $I_{3}$ .

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