| In the circuit shown in the figure, the input voltage \[{{\text{V}}_{\text{i}}}\] is 20 V, \[{{V}_{BE}}=0\]and\[{{V}_{CE}}=0\]. The values of \[{{\text{I}}_{\text{B}}}\text{,}{{\text{I}}_{\text{C}}}\] and \[\beta \] are given by [NEET - 2018] |
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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 :
| [d] \[{{\text{V}}_{\text{BE}}}\text{=0}\] |
| \[{{\text{V}}_{\text{CE}}}\text{=0}\] |
| \[{{\text{V}}_{\text{b}}}\text{=0}\] |
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| \[{{\text{I}}_{C}}=\frac{(20-0)}{4\times {{10}^{3}}}\] |
| \[{{I}_{C}}=5\times {{10}^{-3}}=5mA\] |
| \[{{\text{V}}_{\text{i}}}\text{=}{{\text{V}}_{\text{BE}}}\text{+}{{\text{I}}_{\text{B}}}{{\text{R}}_{\text{B}}}\] |
| \[{{\text{V}}_{\text{i}}}\text{=0+}{{\text{I}}_{\text{B}}}{{\text{R}}_{\text{B}}}\] |
| \[\text{20=}{{\text{I}}_{B}}\times 500\times {{10}^{3}}\] |
| \[{{I}_{B}}=\frac{20}{500\times {{10}^{3}}}=40\mu A\] |
| \[\beta \text{=}\frac{{{\text{l}}_{\text{c}}}}{{{\text{l}}_{\text{b}}}}\text{=}\frac{\text{25 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}}{\text{40 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-6}}}}\text{=125}\] |
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