Determine the current in each branch of the given network.

Explanation

In above circuit, we have to find currents in different branches.
Applying KVL in the closed path ABDA,
`- 10I_(1) - 5I_(3) + 5I_(2) = 0 `
`therefore 10 I_(1) = 5I_(2) - 5I_(3)`
`therefore 2I_(1) = I_(2) - I_(3) " "` ... (1)
Applying KVL in the closed path BCDB,
` - 5 (I_(1) - I_(3)) + 10 (I_(2) + I_(3)) + 5I_(3) = 0 `
`therefore 5I_(1) = 20I_(3) + 10I_(2)`
`therefore I_(1) = 2I_(2) + 4I_(3)`
`therefore 2I_(1) = 4I_(2) + 8I_(3) " " `.... (2)
From equation (1) and (2),
`I_(1) - I_(3) = 4I_(2) + 8I_(3)`
`therefore 3I_(2) = - 9I_(3)`
`therefore I_(2) = - 3I_(3) " " `... (3)
From equation (1) and (3),
`2I_(1) = - 3I_(3) = - 4I_(3) `
`therefore I_(1) = - 2I_(3) " " ` .... (4)
Applying KVL in the closed path ADCEFA , we get,
`- 5I_(2) - 10 (I_(2) + I_(3)) - 10t = - 10`
`therefore 15 I_(2) + 10I_(3) + 10I = 10`
`therefore 15 I_(2) + 10 I_(3) + 10 (I_(1) + I_(2)) = 10 `
`therefore 10I_(1) + 25 I_(2) + 10 I_(3) = 10`
`therefore 10 (-2I_(3)) + 25 (- 3I_(3)) + 10I_(3) = 10`
`therefore - 85I_(3) = 10 `
`therefore I_(3) = - (10)/(85) = - (2)/(17) A " "` ... (5)
From equation (4) and (5).
From equation (4) and (5) ,
`I_(1) = - 2 (- (2)/(17)) = (4)/(17) A ` ... (6)
From equation (3) and (5)
`I_(2) = - 3 (- (2)/(17)) = (6)/(17) A " " `.... (7)
now ,
`I = I_(1) + I_(2) = (4)/(17) + (6)/(17) = (10)/(17) A " " `....(8)
From equation (5) ,
`| I_(3) | = (2)/(17)` A
(Direction of `I_(3)` will be opposite to the direction shown in the figure ).
Final answer :
(i) Current in branch AB is,
`I_(1) = (4)/(17) `A
(ii) Current in branch BC is,
`I_(1) - I_(3) = (4)/(17) - ( - (2)/(17)) = (6)/(17) A `
(iii) Current in branch CD is,
`I_(2) + I_(3) = (6)/(17) + (- (2)/(17)) = (4)/(17)` A
(iv) Current in branch DA is,
`I_(2) = (6)/(17) `A
(v) Current in branch BD is,
`| I_(3)| = (2)/(17) A ` (Direction will be from D to B)
(vi) Current in branch FE is,
I = `(10)/(17) A `