A) 500
B) 1000
C) 1250
D) 100
Correct Answer: C
Solution :
Key Idea AC power gain is ratio of change in output power to the change in input power, AC power gain \[=\frac{\text{Change}\,\text{in}\,\text{output}\,\text{power}}{\text{Change}\,\text{in}\,\text{input}\,\text{power}}\] \[=\frac{\Delta {{V}_{c}}\times \Delta {{i}_{c}}}{\Delta {{V}_{i}}\times \Delta {{i}_{b}}}\] \[=\left( \frac{\Delta {{V}_{c}}}{\Delta {{V}_{i}}} \right)\times \left( \frac{\Delta {{i}_{c}}}{\Delta {{i}_{b}}} \right)={{A}_{v}}\times {{\beta }_{AC}}\] where Ay is voltage gain and \[{{(\beta )}_{AC}}\]is AC current gain. Also, \[{{A}_{V}}={{\beta }_{AC}}\times \text{resistance}\,\text{gain}\left( =\frac{{{R}_{o}}}{{{R}_{i}}} \right)\] Given, \[{{A}_{V}}=50,\]\[{{R}_{o}}=200\,\Omega ,\,{{R}_{i}}=100\,\Omega \] Hence, \[50={{\beta }_{AC}}\times \frac{200}{100}\] \[\therefore \] \[{{\beta }_{AC}}=25\] Now, AC power gain\[{{A}_{V}}\times {{\beta }_{AC}}\] \[=50\times 25=1250\]
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