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 |
| \[\text{=}\frac{\text{Change}\,\text{in}\,\text{output}\,\text{power}}{\text{Change}\,\text{in}\,\text{input}\,\text{power}}\] |
| \[=\frac{\Delta {{V}_{c}}\times \Delta {{i}_{c}}}{\Delta {{V}_{i}}\times \Delta {{i}_{b}}}\] |
| \[=\left( \frac{\Delta {{V}_{c}}}{\Delta {{V}_{i}}} \right)\times \left( \frac{\Delta {{i}_{c}}}{\Delta {{i}_{b}}} \right)\] |
| \[={{A}_{V}}\times {{\beta }_{AC}}\] |
| where \[{{A}_{V}}\] 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, \[{{A}_{v}}=50,\text{ }{{R}_{0}}=200\,\Omega ,\,\,{{R}_{i}}=100\,\Omega \] |
| 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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