In an $NPN$ transistor circuit,the collector current is $10 \,mA$. If $90\%$ of the electrons emitted reach the collector,the emitter current $(i_E)$ and base current $(i_B)$ are given by

In a transistor,the forward bias is always kept smaller than the reverse bias. The correct reason is:

The collector supply voltage is $6 \text{ V}$ and a voltage drop across a resistor of $600 \Omega$ in the collector circuit is $0.6 \text{ V}$,in a circuit of a transistor connected in common emitter mode. If the current gain is $20$,then the base current is: (in $\text{ mA}$)

$A$ $n-p-n$ transistor is connected in common-emitter configuration in which the collector supply is $8\, V$ and the voltage drop across the load resistance of $800\,\Omega$ connected in the collector circuit is $0.8\, V$. If the current amplification factor $\alpha$ is $25/26$,the collector-emitter voltage $V_{CE}$ and the base current $I_B$ will be:

$A$ $npn$ transistor is connected in common emitter configuration in a given amplifier. $A$ load resistance of $800 \,\,\Omega$ is connected in the collector circuit and the voltage drop across it is $0.8 \,\, V$. If the current amplification factor is $0.96$ and the input resistance of the circuit is $192 \,\,\Omega$,the voltage gain and the power gain of the amplifier will respectively be

$A$ transistor is used as a common emitter amplifier with a load resistance $2 \ k\Omega$. The input resistance is $150 \ \Omega$. The base current is changed by $20 \ \mu A$,which results in a change in collector current by $1.5 \ mA$. The voltage gain of the amplifier is