A) \[18:1\]
B) \[9:1\]
C) \[6:1\]
D) \[2:3\]
Correct Answer: B
Solution :
Current density\[J=\frac{i}{A}=\frac{i}{\pi {{r}^{2}}}\] \[\Rightarrow \] \[\frac{{{J}_{1}}}{{{J}_{2}}}=\frac{{{i}_{1}}}{{{i}_{2}}}\times \frac{r_{2}^{2}}{r_{1}^{2}}\] But the wires are in series, so they have the same current, hence,\[{{i}_{1}}={{i}_{2}}\] So, \[\frac{{{J}_{1}}}{{{J}_{2}}}=\frac{r_{2}^{2}}{r_{1}^{2}}=9:1\]You need to login to perform this action.
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