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 \[=\frac{Change\,in\,output\,power}{Change\,in\,input\,power}\] \[=\frac{\Delta {{V}_{c}}\times \Delta {{i}_{c}}}{\Delta {{V}_{i}}\times \Delta {{i}_{b}}}\] \[=\left( \frac{\Delta {{V}_{c}}}{\Delta {{V}_{i}}} \right)\times \left( \frac{\Delta {{i}_{c}}}{\Delta {{i}_{b}}} \right)\] \[={{A}_{V}}\times {{\beta }_{AC}}\] where AV is voltage gain and \[{{(\beta )}_{AC}}\] is AC current gain. Also \[{{A}_{v}}={{\beta }_{AC}}\times \]resistance gain \[\left( \,=\frac{{{R}_{o}}}{{{R}_{i}}} \right)\] Given, \[{{A}_{v}}=50,\,\,{{R}_{0}}=200\,\,\Omega ,\,\,{{R}_{i}}=100\,\Omega \] 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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