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Q.

In the following common emitter circuit if  β=100, VCE=7 V, VBE is negligible, Rc=2 kΩ, then IB =

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a

0.01 mA

b

0.04 mA

c

0.02 mA

d

0.03 mA

answer is B.

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Detailed Solution

VCC=VCE+ICRc15=7+Ic×2×103IC=4 mAβ=IcIBIB=Icβ=4100 mA=0.04 mA
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