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 \[={{I}_{1}}+{{I}_{2}}=\frac{10}{17}A\]
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