A) \[7.5\times {{10}^{-5}}A\]
B) \[7.5\times {{10}^{-4}}A\]
C) \[7.5\times {{10}^{-3}}A\]
D) \[7.5\times {{10}^{-2}}A\]
Correct Answer: D
Solution :
Here, the resistance for 10\[=2\times {{10}^{-14}}\left( W/{{m}^{2}} \right)\], 60\[=\frac{{{l}_{ear}}}{{{l}_{eye}}}=\frac{{{10}^{-13}}}{2\times {{10}^{-14}}}=5\] and 100\[\mu =\frac{\Delta {{V}_{p}}}{\Delta {{V}_{g}}}\] are in series and they together are in parallel to 200\[\Rightarrow \] resistance when a potential difference of 15 V is applied across 200\[\Delta {{V}_{p}}=-\mu \times \Delta {{V}_{s}}\], then current through it is \[=-50\left( -20 \right)=10V\]You need to login to perform this action.
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