Determine the impedance of the circuit phase of current
.

Explanation

The currents through the various arms of the circuit.
According to Kirchoff's second law,
In a closed circuit EABCE ,
`10 = 10 (i_(1) + i_(2)) + 10 i_(1) + 5 (i_(1) - i_(3))`
or `10 = 25 i_(1) + 10 i_(2) - 5 i_(3)`
or ` 2 = 5 i_(1) + 2 i_(2) - i_(3)`
In closed circuit ABDA,
`10 i_(1) + 5 i_(3) - 5 i_(2)=0`
or `2 i_(1) + i_(3) - i_(2) = 0`
or `i_(2)=2 i_(1) + i_(3)`
In a closed circuit BCDB ,
`5 (i_(1) - i_(3)) - 10 (i_(2) + i_(3)) - 5 i_(3) = 0`
or `5 i_(1) - 10 i_(2) - 20 i_(3) = 0 or i_(1) = 2 i_(2) + 4 i_(3)`
From (ii) and (iii) ,
`i_(2) = 2(2 i_(1) + i_(3)) + 4 i_(3) = 4 i_(1) + 6 i_(3) or 3 i_(1) = - 6 i_(3) or i_(1) = - 2 i_(3)`
Putting this value in (ii), `i_(2) = 2 (- 2 i_(3)) + i_(3) = - 3 i_(3)`
Putting this values in (i) , `2 = 5 (- 2 i_(3)) + 2 ( - 3 i_(3)) - i_(3) or 2 = - 17 i_(3) or i_(3) = - 2//17A`
From (iv), `i_(1) = - 2 (- 2//17) = 4//17A`,
From (v), `i_(2) = - 3(- 2 //17) = (6//17)A`
`:. i_(1) + i_(2) = (4 //17) + (6//17)=(10//17)A`
`i_(1) - i_(3) = 4//17 - (- 2//17) = (6//17)A`
`i_(2) + i_(3) = 6//17 + (- 2//17) = 4//17A`