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\[\Omega \] 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 \[i=\frac{V}{R}=\frac{V}{18/5}=\frac{5V}{18}\]