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 fig. It is balanced Wheatstone bridge, therefore \[10\Omega \] resistance is ineffective. In the upper arm \[4\Omega \] and \[8\Omega \] are in series, their effective resistance is \[{{R}_{U}}=4+8=12\Omega \] Similarly, in the lower arm resistances \[2\Omega \]and\[4\Omega \] are also in series their effective resistance is \[{{R}_{L}}=2+4=6\Omega \] Now, resistances R,- and K, 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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