A)
B)
C)
D)
Correct Answer: C
Solution :
\[E=\frac{iR}{L}=\frac{i.\rho }{A}=\frac{neA{{v}_{d}}\rho }{A}\] Þ \[{{v}_{d}}\propto E\] (Straight line) \[P={{i}^{2}}R={{\left( \frac{EA}{\rho } \right)}^{2}}R\] Þ \[P\propto {{E}^{2}}\] (Symmetric parabola) Also \[P\propto {{i}^{2}}\] (parabola) Hence all graphs a, b, d are correct and c is incorrect.
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