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 \[R=\frac{R\times {{R}_{4}}}{R+{{R}_{4}}}\] \[=\frac{(3/4)(1/4)}{(3/4)+(1/4)}\] \[=\frac{3/16}{1}\] \[=\frac{3}{16}\,\Omega \]