A) \[516\,k\Omega \]
B) \[216\,k\Omega \]
C) \[300\,k\Omega \]
D) \[416\,k\Omega \]
Correct Answer: D
Solution :
We know that Voltage gain, \[\Delta V=\beta \frac{{{R}_{o}}}{{{R}_{i}}}\] Also, \[\beta =\frac{\alpha }{1-\alpha }=\frac{0.98}{1-0.98}=49\] \[\Delta V=(49)\left( \frac{{{R}_{0}}}{200} \right)\] Power gain AP\[\Delta P\] = Current gain \[(\beta )\times \]voltage gain \[(\Delta V)\] \[\Rightarrow \] \[5\times {{10}^{6}}=49\times 49\times \frac{{{R}_{o}}}{200}\] \[\Rightarrow \]\[{{R}_{e}}=\frac{5\times {{10}^{6}}\times 200}{49\times 49}=0.416\times {{10}^{6}}\simeq 416\,k\Omega \]
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