question_answer

Determine the current in each branch of the network shown in figure.

 

Answer:

            Applying Kirchhoff's second law to mesh ABDA,             \[-10{{I}_{1}}-5{{I}_{3}}+5{{I}_{2}}=0\] \[2{{I}_{1}}-{{I}_{2}}+{{I}_{3}}=0\]                         ?(i) Applying Kirchhoff's second law to mesh BCDB,             \[-5({{I}_{1}}-{{I}_{3}})+10({{I}_{2}}+{{I}_{3}})+5{{I}_{3}}=0\]             \[5{{I}_{1}}-10{{I}_{2}}-20{{I}_{3}}=0\]             \[{{I}_{1}}-2{{I}_{2}}-4{{I}_{3}}=0\]             ?(ii) Applying Kirchhoff's second law to mesh ADCA,             \[-5{{I}_{2}}-10({{I}_{2}}+{{I}_{3}})+10-10({{I}_{1}}+{{I}_{2}})=10\]             \[2{{I}_{1}}+5{{I}_{2}}+2{{I}_{3}}=2\]                    ?(iii) From Eq. (i),      \[{{I}_{2}}=2{{I}_{1}}+{{I}_{3}}\]                ?(iv) From Eq. (ii),      \[{{I}_{1}}=2{{I}_{2}}+4{{I}_{3}}\]              ?(v) \[\therefore \] Substituting \[{{I}_{1}}\] in Eq. (iv), we get             \[{{I}_{2}}=2(2{{I}_{2}}+4{{I}_{3}})+{{I}_{3}}\]             \[{{I}_{2}}=-3{{I}_{3}}\] From Eq. (v)             \[{{I}_{1}}=-6{{I}_{3}}+4{{I}_{3}}\Rightarrow {{I}_{1}}=-2{{I}_{3}}\] Now, from Eq. (iii)             \[-4{{I}_{3}}-15{{I}_{3}}+2{{I}_{3}}=2\]       \[\Rightarrow {{I}_{3}}=\frac{-2}{17}A\] \[\therefore \]      \[{{I}_{1}}=\frac{+4}{17}A,\]  \[{{I}_{2}}=\frac{6}{17}A,\]            \[I={{I}_{1}}+{{I}_{2}}=\frac{10}{17}A\] So, current in branch AB \[={{I}_{1}}=\frac{4}{17}A\] Current in branch BC \[={{I}_{1}}-{{I}_{3}}=\frac{6}{17}A\] Current in branch AD \[={{I}_{2}}=\frac{6}{17}A\] Current in branch DC \[={{I}_{2}}+{{I}_{3}}=\frac{4}{17}A\] Current In branch BD \[={{I}_{3}}=-\frac{2}{17}A\] (Direction of current is from D to B) Current in cell \[={{I}_{1}}+{{I}_{2}}=\frac{10}{17}A\]