For the transistor circuit shown below,if $\beta = 100$ and the voltage drop between the emitter and base is $0.7 \, V$,then the value of $V_{CE}$ will be......$V$.

$10^{10}$ electrons enter the emitter of a junction transistor in a time of $0.4 \mu s$. If $5 \%$ of the electrons are lost in the base, then the collector current is (in $\text{ mA}$)

For a common emitter configuration,if $\alpha$ and $\beta$ have their usual meanings,the incorrect relation between $\alpha$ and $\beta$ is

$A$ transistor is connected in a common emitter configuration with $R_o = 4 \text{ k}\Omega$,$R_i = 1 \text{ k}\Omega$,$I_C = 1 \text{ mA}$,and $I_b = 20 \text{ }\mu\text{A}$. The voltage gain is:

An $npn$ transistor operates as a common emitter amplifier,with a power gain of $60\, dB$. The input circuit resistance is $100\,\Omega$ and the output load resistance is $10\, k\Omega$. The common emitter current gain $\beta$ is:

In an $n-p-n$ transistor circuit,the collector current is $20 \, mA$. If $90 \%$ of the emitted electrons reach the collector,calculate the emitter current and base current.