In an $n-p-n$ transistor,$10^{10}$ electrons enter the emitter in $10^{-6} \ s$. If $2\%$ of the electrons are lost in the base,the current amplification factor $\beta$ is:

  • A
    $0.02$
  • B
    $7$
  • C
    $33$
  • D
    $49$

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Similar Questions

The output characteristics of an $n-p-n$ transistor represent, ($I_C =$ collector current, $V_{CE} =$ potential difference between collector and emitter, $I_B =$ base current, $V_{BB} =$ voltage given to base, $V_{BE} =$ the potential difference between base and emitter)

Find the value of $I_B$ if $V_{BE} = 0.3\,V$ in the given circuit $(\beta = 100)$. (in $,\mu A$)

In case of a bipolar transistor, $\beta = 45$. The potential drop across the collector resistance of $1 \ k\Omega$ is $5 \ V$. The base current is approximately: (in $\mu A$)

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$A$ 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

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