A) 3
B) 1/3
C) 8/9
D) 2
Correct Answer: B
Solution :
Since, voltage remains same in parallel, so, \[i\propto \frac{1}{R}\] \[\Rightarrow \] \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{{{R}_{2}}}{{{R}_{1}}}\] \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{\rho {{l}_{2}}/{{A}_{2}}}{\rho {{l}_{2}}/{{A}_{1}}}\] \[\left( \because R=\frac{\rho l}{A} \right)\] \[\Rightarrow \] \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{{{l}_{2}}}{{{l}_{1}}}\times {{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}\] \[(\because A=\pi {{r}^{2}})\] \[\Rightarrow \] \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{3}{4}\times {{\left( \frac{2}{3} \right)}^{2}}\] Hence, \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{1}{3}\]
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