question_answer

Resistance of rod is \[1\,\,\Omega \]. It is bent in the form of square. What is the resistance across adjoint corners?

A)  \[1\,\,\Omega \]

B)  \[3\,\,\Omega \]

C)  \[\frac{3}{16}\,\,\Omega \]

D)  \[\frac{3}{4}\,\,\Omega \]

Correct Answer: C

Solution :

When rod is bent in the form of square, then each side has resistance of \[\frac{1}{4}\Omega \]. As shown \[{{R}_{1}},{{R}_{2}}\] and \[{{R}_{3}}\] are connected in series, so their equivalent resistance                                 \[R={{R}_{1}}+{{R}_{2}}+{{R}_{3}}\]                 \[=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\]                 \[=\frac{3}{4}\Omega \] Now, R and \[{{R}_{4}}\] are connected in parallel, so equivalent resistance of the circuit is                 \[R=\frac{R\times {{R}_{4}}}{R+{{R}_{4}}}\]                 \[=\frac{(3/4)\,(1/4)}{(3/4)+(1/4)}\]                 \[=\frac{3/16}{1}\] \[=\frac{3}{16}\Omega \]