Q.

In common emitter configuration, current again of a transistor is 100 and input resistance is  1kΩ. Then trans conductance ∆Ic∆VBE of the transistor is

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a

0.1

b

0.01

c

0.05

d

0.2

answer is A.

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Detailed Solution

gm=∆IC∆VBE=∆IC∆IBx∆IB∆VBE=β x 1Ri=100 x 11000⇒gm=0.1  mho
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In common emitter configuration, current again of a transistor is 100 and input resistance is  1kΩ. Then trans conductance ∆Ic∆VBE of the transistor is