The transfer ratio of a transistor is $50$. The input resistance of the transistor when used in the common-emitter configuration is $1 \; k\Omega$. The peak value of the collector current for an $A.C.$ input voltage of $0.01 \; V$ peak is:

The graph of the input characteristic of a common emitter amplifier is:

$A$ transistor is operating in the active mode. $v_1$ is the potential barrier across the base-emitter junction and $v_2$ is the potential barrier across the collector-base junction. $b_1$ is the width of the depletion layer of the base-emitter junction and $b_2$ is the width of the collector-base junction.

$A$ 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)

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......$V$.

For a transistor,$\alpha_{dc}$ and $\beta_{dc}$ are the current ratios. Then the value of $\frac{\beta_{dc}-\alpha_{dc}}{\alpha_{dc} \cdot \beta_{dc}}$ is