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Question

In the circuit shown in the figure, the input voltage V1 is 20 V, VBE=0 and VCE=0. The values of IB, IC and β are given by


A
IB=401 μA, IC=10 mA, β=250
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B
IB=20 μA, IC=5 mA, β=250
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C
IB=40 μA, IC=5 mA, β=125
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D
IB=25 μA, IC=5 mA, β=200
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Solution

The correct option is C IB=40 μA, IC=5 mA, β=125
For Emitter-Base circuit,
Vi=IBRB+VBE

20=IB×(500×103)+0

IB=20500×103=40 μA

For CE circuit
VCC=ICRC+VCE

IC=5×103=5 mA

β=ICIB=5×10340×106=125.

Hence (D) is the correct answer.

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