NEET Solved Paper 2018 Question 15
Question: In the circuit shown in the figure, the input voltage $ {V_i} $ is 20 V, $ V _{BE}=0 $ and $ V _{CE}=0 $ . The values of $ {I_B}\text{,}{I_C} $ and $ \beta $ are given by [NEET - 2018]
Options:
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 $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ {V _{BE}}\text{=0} $ $ {V _{CE}}\text{=0} $ $ {V_b}\text{=0} $ $ {I_C}=\frac{(20-0)}{4\times 10^{3}} $ $ I _{C}=5\times {10^{-3}}=5mA $ $ {V_i}\text{=}{V _{BE}}\text{+}{I_B}{R_B} $ $ {V_i}\text{=0+}{I_B}{R_B} $ $ \text{20=}{I_B}\times 500\times 10^{3} $ $ I _{B}=\frac{20}{500\times 10^{3}}=40\mu A $ $ \beta \text{=}\frac{{l_c}}{{l_b}}\text{=}\frac{\text{25 }\times\text{ 1}{0^{\text{-3}}}}{\text{40 }\times\text{ 1}{0^{\text{-6}}}}\text{=125} $
