A) 1.5 mA
B) 1.27 mA
C) 10 mA
D) 12.4 mA
Correct Answer: B
Solution :
We know that, \[A=\frac{\mu {{R}_{L}}}{{{r}_{P}}+{{R}_{L}}}=\frac{14\times 12}{10+12}=\frac{84}{11}\] Peak value of output signal, \[{{V}_{0}}=\frac{84}{11}\times 2\sqrt{2}\,V\] \[{{V}_{rms}}=\frac{{{V}_{0}}}{\sqrt{2}}=\frac{84}{11\times \sqrt{2}}\times 2\sqrt{2}\] \[=\frac{84\times 2}{11}V\] So, the rms value of current through \[{{R}_{L}}=12\,k\Omega \] \[{{i}_{rms}}=\frac{84\times 2}{11\times 12\times {{10}^{3}}}A=1.27\,mA\]
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