question_answer

For the network shown in the figure, the value of the current i is:                         [AIPMT (S) 2005]

A) \[\frac{9V}{35}\]            

B)    \[\frac{5V}{18}\]    

C)   \[\frac{5V}{9}\]          

D)  \[\frac{18V}{5}\]

Correct Answer: B

Solution :

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