1.

Solve for current values in figure.

Answer»

Solution :APPLYING Kirchhoff.s first law at the junction B we have `i_(1)+i_(2)=i_(3)"….(1)"`
Applying Kirchhoff. second law to loop ABEFA
`-12+i_(2)xx1.5-i_(1)xx1+8=0`
`i_(1)=1.5i_(2)=-4"…..(2)"`
From loop BCDEB `-(i_(2)xx1.5)-(i_(3)xx9)+12=0`
`1.5i_(2)+9i_(3)=12".....(3)"`
`"on SOLVING "i_(1)=-1A and i_(3)=1A`


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