A) 6\[\Omega \]
B) 8\[\Omega \]
C) 24\[\Omega \]
D) 16\[\Omega \]
Correct Answer: B
Solution :
The resistances \[8\Omega ,16\Omega \], and \[16\Omega \]. One connected in parallel (in upper arm) \[\frac{1}{R}=\frac{1}{8}+\frac{1}{16}+\frac{1}{16}=\frac{4}{16}=\frac{1}{4}\] Hence, \[R=4\Omega \] Now \[R=4\Omega \]and \[20\Omega \] are joined in series, their effective resistance is given by \[R=\frac{9\times 18}{9+18}=6\Omega \] ...(i) Now we consider for lower arm, \[9\Omega \] and\[18\Omega \] are connected in parallel so, their equivalent resistance is: \[R=\frac{9\times 18}{9+18}=6\Omega \] Again \[R=6\Omega \] and \[6\Omega \] are connected in series, so their equivalent resistance is \[R=6+6=12\Omega \] ?..(ii) Again equivalent resistance between P and Q \[{{R}_{PQ}}=\frac{R\times R}{R\times R}=\frac{24\times 12}{24+12}=8\Omega \]
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