A) \[\frac{2}{3}G\]
B) \[1.5G\]
C) \[\frac{1}{3}G\]
D) \[\frac{5}{4}G\]
Correct Answer: A
Solution :
As\[{{A}_{v}}=\beta \frac{{{R}_{L}}}{{{R}_{i}}}\] \[\left[ \because {{g}_{m}}=\frac{\Delta {{l}_{c}}}{\Delta {{V}_{B}}}=\frac{\Delta {{l}_{c}}}{\Delta {{l}_{B}}{{R}_{i}}} \right]\] or \[G=\left( \frac{\beta }{{{R}_{i}}} \right){{R}_{L}}\] \[\left[ \because {{g}_{m}}=\frac{\beta }{{{R}_{i}}} \right]\] \[\Rightarrow \] \[G={{g}_{m}}{{R}_{L}}\] \[\Rightarrow \] \[G\propto {{g}_{m}}\] \[\therefore \] \[\frac{{{G}_{2}}}{{{G}_{1}}}=\frac{{{g}_{{{m}_{1}}}}}{{{g}_{{{m}_{2}}}}}\] \[\Rightarrow \] \[{{G}_{2}}=\frac{0.02}{0.03}\times G\] \[\therefore \]Voltage gain \[{{G}_{2}}=\frac{2}{3}G\]
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