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 \[R=\frac{R\times {{R}_{4}}}{R+{{R}_{4}}}\] \[=\frac{(3/4)\,(1/4)}{(3/4)+(1/4)}\] \[=\frac{3/16}{1}\] \[=\frac{3}{16}\Omega \]You need to login to perform this action.
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