A) \[R\]
B) \[4R\]
C) \[\frac{R}{4}\]
D) \[\frac{R}{16}\]
Correct Answer: D
Solution :
\[R=\frac{\rho l}{A}\] or \[R\propto l\] \[\therefore \] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{L}{L\text{/}4}=4\] or \[{{R}_{2}}=\frac{R}{4}\] \[(\because \,{{R}_{1}}=R)\] In parallel combination of such four resistances. \[\frac{1}{R}=\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{2}}}\] or \[\frac{1}{R}=\frac{1}{R\text{/}4}+\frac{1}{R\text{/}4}+\frac{1}{R\text{/}4}+\frac{1}{R\text{/}4}\] or \[\frac{1}{R}=\frac{4}{R}+\frac{4}{R}+\frac{4}{R}+\frac{4}{R}\] or \[\frac{1}{R}=\frac{16}{R}\] or \[R=\frac{R}{16}\]
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