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In a p-n-p transistor in CB circuit, the...

In a p-n-p transistor in CB circuit, the emitter current changes from 0.6 mA to 0.4 mA , when the bae emitter voltage is changed from 0.68 V to 0.64 V. The input resistance of the transistor is

A

`100 Omega`

B

`200 Omega`

C

`300 Omega`

D

`400 Omega`

Text Solution

Verified by Experts

The correct Answer is:
B
`R_(i)=(deltaV_("BE))/(deltaI_(e))=(0.68-0.64)/((0.6-0.4)xx10^(-3))`
`=(0.04xx10^(3))/(0.2)=200 Omega`
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