Vi=IBRB+VBE (By Kirchoff's Voltage Law)
20=IB×(500×103)+0
IB=500×10320=40μA
VCC=ICRC+VCE(By Kirchoff's Voltage Law)
20=IC×(4×103)+0
IC=5×10−3=5mA
β=IBIC=40×10−65×10−3=125
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,I_C and β are given by NEET 2018 Physics,…
NEET UG 2018 — Physics Modern Physics
2018mcqmedium
In the circuit shown in the figure, the input voltage Viis20V,VBE=0andVCE=0. The values of IB,ICandβ are given by
Official previous-year question
Held on 30 Apr 2018 · Verified 9 Jul 2026.
Options
- A
IB=20μA,IC=5mA,β=250
- B
IB=25μA,IC=5mA,β=200
- C
IB=40μA,IC=10mA,β=250
- D
IB=40μA,IC=5mA,β=125
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