question_answer

For the network shown in the figure, the value of the current i is

A) \[\frac{9V}{35}\]             

B)                        \[\frac{5V}{18}\]             

C) \[\frac{5V}{9}\]                

D)        \[\frac{18V}{5}\]

Correct Answer: B

Solution :

The circuit given resembles the balanced Wheatstone Bridge as \[\frac{4}{6}=\frac{2}{3}.\] Thus, middle arm containing 4 Q. resistance will be ineffective and no current flows through it. The equivalent circuit is shown as below: Net resistance of AB and BC \[R=4+2=6\,\Omega \] Net resistance of AD and DC \[R=6+3=9\,\Omega \] Thus, parallel combination of R and R gives \[R=\frac{R\times R}{R+R}\]     \[=\frac{6\times 9}{6+9}=\frac{54}{15}=\frac{18}{5}\,\Omega \] Hence, current \[i=\frac{V}{R}=\frac{V}{18\text{/}5}=\frac{5V}{18}\]