For $I_b = 20 \ \mu A$ and $\beta = 100$,the value of the emitter current is ........ $mA$.

In an $N-P-N$ transistor circuit,the common emitter configuration is used. $A$ load resistor $R_L$ is connected to the collector and a base resistor $R_B$ is connected to the base. Given $V_{CC} = 8V$,$V_{CE} = 4V$,$V_{BE} = 0.6V$,collector current $I_C = 4mA$,and current gain $\beta = 100$,find the values of $R_L$ and $R_B$ respectively.

An $npn$ transistor operates as a common emitter amplifier with a power gain of $10^{6}$. The input circuit resistance is $100\, \Omega$ and the output load resistance is $10\, k\Omega$. The common emitter current gain $\beta$ will be (Round off to the nearest integer).

In the following circuit, a voltmeter $V$ is connected across a lamp $L$. What change would occur in the voltmeter reading if the resistance $R$ is reduced in value?

In a common emitter configuration with suitable bias,it is given that $R_L$ is the load resistance and $R_{BE}$ is the small signal dynamic resistance (input side). Then,the voltage gain,current gain,and power gain are given,respectively,by: [$\beta$ is the current gain,$I_B$,$I_C$,$I_E$ are respectively base,collector,and emitter currents]

In a transistor,when the emitter current changes by $9.85 \,mA$,the collector current changes by $9.5 \,mA$. Then the base current is (in $\,mA$)