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Question

In a common emitter transistor circuit, the base current is 40μA, then VBE is
1381620_6b8e3782463348c9a888034cf7ff3ca8.jpg

A
2V
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B
0.2V
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C
0.8V
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D
zero
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Solution

The correct option is C 0.2V
Given, IB=40μA=40×106A
In the given circuit, the emitter part is grounded, i.e., at zero potential. Therefore, we can use Kirchhoff’s Voltage law-
VCC(245kΩ).IBVBE=0
VBE=10(245×103×40×106)
VBE=(109.8) Volt=0.2 Volt


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