An $n-p-n$ transistor has three leads $A, B$ and $C$. Connecting $B$ and $C$ by moist fingers,$A$ to the positive lead of an ammeter,and $C$ to the negative lead of the ammeter,one finds large deflection. Then,$A, B$ and $C$ refer respectively to

  • A
    Emitter,base and collector
  • B
    Base,emitter and collector
  • C
    Base,collector and emitter
  • D
    Collector,emitter and base

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In a transistor in $CE$ configuration,the ratio of power gain to voltage gain is

In the study of a transistor as an amplifier,if $\alpha = \frac{I_C}{I_E} = 0.98$ and $\beta = \frac{I_C}{I_B} = 49$,where $I_C, I_B,$ and $I_E$ are collector,base,and emitter currents respectively,then $\left(\frac{1}{\alpha} - \frac{1}{\beta}\right)$ is equal to:

$A$ change of $0.04 \ V$ takes place between the base and the emitter when an input signal is connected to the common emitter transistor amplifier. As a result,a $20 \ \mu A$ change takes place in the base current and a change of $2 \ mA$ takes place in the collector current. The input resistance and $AC$ current gain are:

In a common emitter configuration,a transistor has $\beta=50$ and input resistance $1 \text{ k}\Omega$. If the peak value of a.c. input is $0.01 \text{ V}$,then the peak value of collector current is: (in $\mu\text{A}$)

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.

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