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 \[i=\frac{V}{R}=\frac{V}{18/5}=\frac{5V}{18}\]You need to login to perform this action.
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