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Solve for current values in figure. |
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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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