A) 1.27 mA
B) 10 mA
C) 1.5 mA
D) 12.4 mA
Correct Answer: A
Solution :
\[A=\frac{\mu {{R}_{L}}}{{{r}_{p}}+{{R}_{L}}}=\frac{14\times 12}{10+12}=\frac{84}{11}\]. Peak value of output signal \[{{V}_{0}}=\frac{84}{11}\times 2\sqrt{2}V\] Þ \[{{V}_{rms}}=\frac{{{V}_{0}}}{\sqrt{2}}=\frac{84\times 2}{11}V\] Þ r.m.s. value of current through the load \[=\frac{84\times 2}{11\times 12\times {{10}^{3}}}A=1.27\,mA\]
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