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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