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