A) \[{{I}_{B}}=20\mu A,{{I}_{C}}=5mA,\beta =250\]
B) \[{{I}_{B}}=25\mu A,{{I}_{C}}=5mA,\beta =200\]
C) \[{{I}_{B}}=40\mu A,\,\,{{I}_{C}}=10mA,\,\,\beta =250\]
D) \[{{I}_{B}}=40\mu A,\,\,{{I}_{C}}=5mA,\,\,\beta =125\]
Correct Answer: D
Solution :
\[{{\text{V}}_{\text{BE}}}\text{=0}\] \[{{\text{V}}_{\text{CE}}}\text{=0}\] \[{{\text{V}}_{\text{b}}}\text{=0}\] \[{{\text{I}}_{C}}=\frac{(20-0)}{4\times {{10}^{3}}}\] \[{{I}_{C}}=5\times {{10}^{-3}}=5mA\] \[{{\text{V}}_{\text{i}}}\text{=}{{\text{V}}_{\text{BE}}}\text{+}{{\text{I}}_{\text{B}}}{{\text{R}}_{\text{B}}}\] \[{{\text{V}}_{\text{i}}}\text{=0+}{{\text{I}}_{\text{B}}}{{\text{R}}_{\text{B}}}\] \[\text{20=}{{\text{I}}_{B}}\times 500\times {{10}^{3}}\] \[{{I}_{B}}=\frac{20}{500\times {{10}^{3}}}=40\mu A\] \[\beta \text{=}\frac{{{\text{l}}_{\text{c}}}}{{{\text{l}}_{\text{b}}}}\text{=}\frac{\text{25 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}}{\text{40 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-6}}}}\text{=125}\]You need to login to perform this action.
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