In the $CB$ mode of a transistor, when the collector voltage is changed by $0.5\,V$, the collector current changes by $0.05\,mA$. The output resistance will be....$k\Omega$.

In $n-p-n$ transistor, in $CE$ configuration:

For a transistor amplifier in a common-emitter configuration,given the load resistance $R_L = 1 \ k\Omega$,$h_{fe} = 50$,and $h_{oe} = 25 \ \mu A/V$,calculate the current gain $(A_i)$.

Assertion : In a common emitter transistor amplifier,the input current is much less than the output current.
Reason : The common emitter transistor amplifier has very high input impedance.

In a $CE$ transistor,when the base current increases by $60 \mu A$,the change in base-emitter voltage is $1.2 \ V$. Then the input resistance is: (in $\Omega$)

For a $CE$ transistor amplifier,the audio signal voltage across the collector resistance of $2.0\,k\Omega$ is $2.0\,V$. Suppose the current amplification factor of the transistor is $100$. What should be the value of $R_B$ (in $k\Omega$) in series with $V_{BB}$ supply of $2.0\,V$ if the $dc$ base current has to be $10$ times the signal current?