The phase difference between input and output voltages of a $CE$ (common emitter) amplifier circuit is........$^o$.

In the study of a transistor as an amplifier,if $\alpha = \frac{I_{C}}{I_{E}}$ and $\beta = \frac{I_{C}}{I_{B}}$,where $I_{C}$,$I_{B}$,and $I_{E}$ are the collector,base,and emitter currents respectively,then:

$A$ transistor has a current amplification factor of $60$. In a $CE$ amplifier, input resistance is $1 \, k\Omega$ and output voltage is $0.01 \, V$. The transconductance is (in $SI$ units)

$A$ 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:

In common emitter mode of a transistor, the d.c. current gain $(\beta)$ is $20$, and the emitter current $(I_E)$ is $7 \text{ mA}$. The collector current $(I_C)$ is:

An $n-p-n$ transistor is connected in common-emitter configuration as shown in the figure. If the collector current is $5 \text{ mA}$,$V_{BE} = 0.6 \text{ V}$,$V_{CE} = 3 \text{ V}$,and the common-emitter current amplification factor $\beta = 50$,then the values of $R_1$ and $R_2$ are respectively.