A) \[25\,\Omega \]
B) \[10\,\Omega \]
C) \[5\,\Omega \]
D) None of these
Correct Answer: C
Solution :
Here \[4\,\Omega \] and \[12\,\Omega \]are in parallel. \[\therefore \] \[\frac{1}{R}=\frac{1}{4}+\frac{1}{12}\] \[\Rightarrow \] \[R=\frac{4\times 12}{4+12}=3\Omega \] Similarly, \[6\,\Omega \]and \[3\,\Omega \]are in parallel. \[\therefore \] \[\frac{1}{R}=\frac{1}{6}+\frac{1}{3}\] \[\Rightarrow \] \[R=\frac{6\times 3}{6+3}=2\,\Omega \] R and \[R\] are in series. \[\therefore \]Equivalent resistance between A and B \[={{R}^{}}=R+R=3+2=5\,\Omega \]
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