NTA Abhyas JEE Main2020PhysicsSemiconductorsPractice
In the circuit shows in the figure, the input voltage V i i s 20 V , V B E = 0 a n d V C E = 0 . The values of I B , I C a n d β are given by
Options
- AI B = 20 μA , I C = 5 mA , β = 250
- BI B = 25 μA , I C = 5 mA , β = 200
- CI B = 40 μA , I C = 10 mA , β = 250
- DI B = 40 μA , I C = 5 mA , β = 125
Correct answer
D. I B = 40 μA , I C = 5 mA , β = 125
Step-by-step solution
V i = I B R B + V B E (By Kirchoff's Voltage Law) 20 = I B × 500 × 10 3 + 0 I B = 20 500 × 10 3 = 40 μA V C C = I C R C + V C E (By Kirchoff's Voltage Law) 20 = I C × 4 × 10 3 + 0 I C = 5 × 10 - 3 = 5 mA β = I C I B = 5 × 10 - 3 40 × 10 - 6 = 125