A) \[2\,\Omega \]
B) \[4\,\Omega \]
C) \[32\,\Omega \]
D) \[16\,\Omega \]
Correct Answer: B
Solution :
The given circuit is reccrawn and solve as shown in the figure. Clearly it is a balanced Wheatstone bridge hence \[10\Omega \]resistance is ineffective. In the upper arm \[4\Omega \] and \[8\Omega \]. resistance are in series their effective resistance is. \[{{R}_{u}}=4+8=12\Omega .\] Similarly, in the lower arm, resistance \[2\Omega .\] and \[4\Omega \] are also in series their effective resistance is \[{{R}_{L}}=2+4=6\Omega \] Now, resistance \[{{R}_{u}}\] and \[{{R}_{L}}\] are in parallel. So, resultant resistance between points A and B is \[{{R}_{AB}}=\frac{{{R}_{u}}{{R}_{L}}}{{{R}_{u}}+{{R}_{L}}}=\frac{12\times 6}{12+6}=4\Omega \]You need to login to perform this action.
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