Directions: In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
Two resistors when connected in series have an equivalent resistance of \[9\,\Omega \]and when connected in parallel have an equivalent resistance of\[2\,\Omega \]. The value of these resistance of are: (1) \[2\,\Omega \]and\[9\,\Omega \] (2)\[3\,\Omega \]and \[6\,\Omega \] (3) \[6\,\Omega \]and \[3\,\Omega \] (4)\[3\,\Omega \]and \[9\,\Omega \]A) 1 and 2 are correct
B) 2 and 3 are correct
C) 1 and 4 and correct
D) 1, 2 and 4 are correct
Correct Answer: B
Solution :
\[{{R}_{1}}+{{R}_{2}}=9\] ...(i) and \[\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}=2\] or \[\frac{{{R}_{1}}{{R}_{2}}}{9}=2\] or \[{{R}_{1}}{{R}_{2}}=18\] ?.. (ii) \[\therefore \] \[{{({{R}_{1}}+{{R}_{2}})}^{2}}={{({{R}_{1}}+{{R}_{2}})}^{2}}-4{{R}_{2}}{{R}_{2}}\] \[={{(9)}^{2}}-4\times 18=81-72=9\] \[{{R}_{1}}-{{R}_{2}}=3\] Solving Eqs. (i) and (ii), we have \[{{R}_{1}}=6\,\Omega \]and \[{{R}_{2}}=3\,\Omega \] Similarly, we can have \[{{R}_{1}}=3\omega \]and \[{{R}_{2}}=6\,\Omega \]
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