question_answer

Resistance of rod is 1\[\Omega \]. It is bent in form of square. What is 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 \[R=\frac{R\times {{R}_{4}}}{R+{{R}_{4}}}\]    \[=\frac{(3/4)(1/4)}{(3/4)+(1/4)}\]    \[=\frac{(3/16)}{1}=\frac{3}{16}\Omega \]