question_answer

The equivalent resistance of the arrangement of resistances shown in the given figure between points P and Q is

A)  60                                         

B)  80

C)  240                                       

D)  160

Correct Answer: B

Solution :

                 In the given circuit, the three resistors of \[3\,\,\Omega ,\,\,16\,\,\Omega \] and \[16\,\,\Omega \] are connected in parallel, hence equivalent resistance is                                 \[\frac{1}{R}=\frac{1}{8}+\frac{1}{16}+\frac{1}{16}=\frac{4}{16}=\frac{1}{4}\] \[\Rightarrow \]               \[F=4\,\Omega \] This \[4\,\Omega \] resistor is connected in series with \[20\,\,\Omega \] resistor.                 \[R\,=20\,\,\Omega +4\Omega =24\,\,\Omega \] Similarly in the lower arm, the \[9\,\,\Omega \] and \[18\,\,\Omega \]resistances are connected in parallel, hence equivalent resistance is                 \[\frac{1}{R\,\,}=\frac{1}{9}+\frac{1}{18}=\frac{18+9}{18\times 9}\] \[\Rightarrow \]               \[R\,\,=6\,\Omega \] The two \[6\,\Omega \] resistors are connected in series, hence equivalent resistance is                 \[R\,\,=6\,\,\Omega +6\,\,\Omega =120\,\Omega \] Resultant equivalent resistance between P and Q is                 \[\frac{1}{R}=\frac{1}{24}+\frac{1}{12}\] \[\Rightarrow \]               \[R=8\,\,\Omega \]