In a common base amplifier circuit,calculate the change in base current if the change in emitter current is $2\, mA$ and $\alpha = 0.98$.

In case of a bipolar transistor, $\beta = 45$. The potential drop across the collector resistance of $1 \ k\Omega$ is $5 \ V$. The base current is approximately: (in $\mu A$)

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 $(A_i)$.

In an $NPN$ transistor,$10^{10}$ electrons enter the emitter in $10^{-6} \ s$ and $2 \%$ of electrons recombine with holes in the base. The current ratios $\alpha$ and $\beta$ of the transistor are respectively (nearly):

An $N-P-N$ transistor conducts when:

The current gain of a common emitter amplifier is $69$. If the emitter current is $7.0\,mA$,the collector current is.....$mA$.