A) \[2\Omega \]
B) \[4\Omega \]
C) \[8\Omega \]
D) \[16\Omega \]
Correct Answer: B
Solution :
The given circuit is redrawn as shown in the figure. It is balanced Wheatstone bridge, therefore \[10\text{ }\Omega ,\]resistance is ineffective. In the upper arm\[\text{4 }\Omega ,\]and\[\text{8 }\Omega ,\]arc in series, their effective resistance is \[{{R}_{U}}=4+8=12\,\Omega \] Similarly, in the lower arm resistances\[2\,\Omega \]. and\[4\,\Omega \]arc also in series, their effective resistance is \[{{R}_{L}}=2+4=6\Omega \] Now, resistances\[{{R}_{U}}\]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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