A) \[\frac{5}{8}\,\,ohms\]
B) \[\frac{8}{5}\,\,ohms\]
C) \[\frac{3}{8}\,\,ohms\]
D) \[\frac{8}{3}\,\,ohms\]
Correct Answer: B
Solution :
\[\rho -\text{same, }\,l-\text{same, }\,{{A}_{2}}=\frac{1}{4}{{A}_{1}}\] (as \[{{r}_{2}}=\frac{{{r}_{1}}}{2}\]) By using \[R=\rho \frac{l}{A}\Rightarrow \frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{A}_{2}}}{{{A}_{1}}}\Rightarrow \frac{{{R}_{1}}}{8}=\frac{1}{4}\Rightarrow {{R}_{1}}=2\Omega \] Hence, \[{{R}_{eq}}=\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}=\frac{2\times 8}{(2+8)}=\frac{8}{5}\Omega .\]
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