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In the circuit shown in the figure, the input voltage ViV_i is 20 V, VBE=0V_{BE} = 0 and VCE=0V_{CE} = 0. The values of IBI_B, ICI_C and β\beta are given by

1

IB=20μA,IC=5mA,β=250I_B = 20 \mu A, I_C = 5 mA, \beta = 250

2

IB=25μA,IC=5mA,β=200I_B = 25 \mu A, I_C = 5 mA, \beta = 200

3

IB=40μA,IC=10mA,β=250I_B = 40 \mu A, I_C = 10 mA, \beta = 250

4

IB=40μA,IC=5mA,β=125I_B = 40 \mu A, I_C = 5 mA, \beta = 125

Step-by-Step Solution

Given VCE=0V_{CE} = 0, IC=VCC−VCERC=20−04×103=5mAI_C = \frac{V_{CC} - V_{CE}}{R_C} = \frac{20 - 0}{4 \times 10^3} = 5 mA. Given VBE=0V_{BE} = 0, Vi=VBE+IBRB  ⟹  20=0+IB(500×103)  ⟹  IB=20500×103=40μAV_i = V_{BE} + I_B R_B \implies 20 = 0 + I_B(500 \times 10^3) \implies I_B = \frac{20}{500 \times 10^3} = 40 \mu A. Then β=ICIB=5×10−340×10−6=125\beta = \frac{I_C}{I_B} = \frac{5 \times 10^{-3}}{40 \times 10^{-6}} = 125.

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