question_answer

In the adjoining circuit diagram each resistance is of 10 W. The current in the arm AD will be

A)            \[\frac{2i}{5}\]

B)            \[\frac{3i}{5}\]

C)            \[\frac{4i}{5}\]

D)            \[\frac{i}{5}\]

Correct Answer: A

Solution :

Applying Kirchoff's law in mesh ABCDA      \[-10(i-{{i}_{1}})-10{{i}_{2}}+20{{i}_{1}}=0\]   Þ \[\,3{{i}_{1}}-{{i}_{2}}=i\] .......(i) and in mesh BEFCB        \[-\,20\,(i-{{i}_{1}}-{{i}_{2}})+\,10\,({{i}_{1}}+{{i}_{2}})+10{{i}_{2}}=0\] Þ    \[3{{i}_{1}}+4{{i}_{2}}=2i\]                    ......(ii) From equation (i) and (ii)  \[{{i}_{1}}=\frac{2i}{5},\,{{i}_{2}}=\frac{i}{5}\]Þ \[{{i}_{AD}}=\frac{2i}{5}\]