A) e
B) \[\frac{1}{e}\]
C) \[{{e}^{2}}\]
D) \[{{e}^{3}}\]
Correct Answer: A
Solution :
Let \[=q\times {{n}^{2/3}}\] On differentiating w.r.t. x, we get \[R=1k\Omega \] For maxima and minima, put f?(x) = 0. Log x ? 1 = 0 \[{{i}_{1}}=\frac{15}{1}mA=15mA\] x = e Now, \[R=250\Omega \] \[{{i}_{250}}=\frac{20-15}{250}=\frac{5}{250}\] \[=\frac{20}{1000}=20mA\] \[{{i}_{Zener}}=20-15\] f(x) is minimum at x = e. Hence, minimum value of f(x) at x = e is \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}}={{\left( \frac{1}{2} \right)}^{t/7}}\]
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