In the `CB` mode of a transistor, when the collector voltage is changed by `0.5` volt. The collector current changes by `0.05 mA`. The output resistance will be

Options

A. `10k Omega`
B. `20k Omega`
C. `5k Omega`
D. `2.5k Omega`

Answer

A

Explanation

Here, `DeltaV_(c)=0.5V and DeltaI_(C)=0.05 mA=0.05 xx 10^(-3)A`
Output resistance is given by,
`R_("out") =(DeltaV_(C))/(DeltaI_(C))=(0.5)/(0.05xx10^(-3))`
`=10^(4)Omega=10Omega`