A) \[0.01\,\mu A\]
B) \[0.25\,\mu A\]
C) \[100\,\mu A\]
D) \[500\,\mu A\]
Correct Answer: D
Solution :
Given that the current gain of the transistor is \[\beta =50.\] Input resistance \[=1\,k\Omega ,\]input voltage = 0.01 V Hence, \[50=\frac{\Delta {{i}_{C}}}{\Delta {{i}_{B}}}\Rightarrow \Delta {{i}_{C}}=50\Delta {{i}_{B}}\] Also the change in base current is \[\Delta {{i}_{B}}=\frac{\text{input}\,\text{voltage}}{\text{input}\,\text{resistance}}=\frac{0.01}{1\times {{10}^{3}}}={{10}^{-5}}A\] So, \[\Delta {{i}_{C}}=50\times {{10}^{-5}}=500\mu A\]You need to login to perform this action.
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