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