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Determine the current in each brance of ...

Determine the current in each brance of the network showin in

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Applying KVL `( sum IR = sum epsilon)` In the closed loop ADCA, we get
-4 `(I_(1) - I_(2) ) + 2 (I_(2) + I_(3) - I_(1) )= -10`
`therefore 4 (I_(1) - I_(2) ) - 2 (I_(2) + I_(3) - I_(1)) + I_(1) = 10`
`therefore 7 I_(1) - 6I_(2) - 2I_(3) = 10 " "` .... (1)
Similarly for closed loop ABCDA,
` -4 I_(2) -2 (I_(2) + I_(3)) - I_(1) = - 10 `
`therefore 4I_(2) + I_(2) -2 (I_(2) + I_(3)) + I_(1) = 10`
`therefore I_(1) + 6I_(2) + 2I_(3) = 10 " "` ... (2)
For closed loop BCDEB,
` - 2 (I_(2) + I_(3)) - 2 (I_(2) + I_(3) - I_(1)) = - 5 `
` therefore 2 (I_(2) + I_(3)) + 2 (I_(2) + I_(3) - I_(1)) = 5 `
`therefore 4 I_(2) + 4I_(3) - 2I_(1) = 5 `
`therefore 2I_(1) - 4I_(2) - 4I_(3) = - 5`
`therefore I_(1) - 2I_(2) - 2I_(3) = - 2.5 " "` .... (3)
Adding equations (1) and (2),
`8I_(1) = 20`
`therefore I_(1) = (5)/(2) A " "` .... (4)
Adding equations (2) and (3)
`2I_(1) + 4I_(2) = 7.5`
`therefore 2 ((5)/(2)) + 4I_(2) = 7.5 `
`therefore 5 + 4I_(2) = 7.5`
`therefore I_(2) = (2.5)/(4) = (25)/(40) = (5)/(8) A " " `... (5)
Substituting values of `I_(1) and I_(2)` in equation (3),
2.5 - 2 `((5)/(8)) - 2I_(3) = - 2.5 `
`therefore 2.5 - (5)/(4) + 2.5 = 2 I_(3)`
`therefore 10 - 5 + 10 = 8I_(3)`
`therefore 15 = 8I_(3)`
`therefore I_(3) = (15)/(8) A " "` .... (6)
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