A) \[\text{25}\]
B) \[\text{50}\]
C) \[75\]
D) \[100\]
Correct Answer: A
Solution :
Since,\[\mu ={{r}_{p}}\cdot {{g}_{m}}\] \[\therefore \] \[{{g}_{{{m}_{1}}}}=\frac{{{\mu }_{1}}}{{{r}_{{{p}_{1}}}}}=\frac{30}{5}=6\] and \[{{g}_{{{m}_{2}}}}=\frac{{{\mu }_{2}}}{{{r}_{{{p}_{2}}}}}=\frac{21}{4}\] Since, triodes are connected in parallel, hence effective plate resistance, \[{{r}_{p}}=\frac{{{r}_{{{p}_{1}}}}{{r}_{{{p}_{2}}}}}{{{r}_{{{p}_{1}}}}+{{r}_{{{p}_{2}}}}}\] \[=\frac{5\times 4}{5+4}=\frac{20}{9}k\Omega \] and trans conductance, \[{{g}_{m}}={{g}_{{{m}_{1}}}}+{{g}_{{{m}_{2}}}}=\left( 6+\frac{21}{4} \right){{(k\Omega )}^{-1}}\] \[\therefore \]Amplification factor \[\mu ={{r}_{p}}\cdot {{g}_{m}}=\frac{20}{9}\left( 6+\frac{21}{4} \right)=25\]
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