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$\begin{align}

& A.1.57mA,49 \\

& B.1.92mA,70 \\

& C.2mA,25 \\

& D.2.25mA,100 \\

\end{align}$

Answer

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${{I}_{c}}=\alpha {{I}_{e}}$

$\alpha $ is given by how much current from the emitter reaches the collector. So using this $\alpha $ amplification factor can be found out.

Current amplification factor in a bi junction transistor is described as the ratio of output current to its input current. In the common base configuration, current amplification factor is given as the ratio of collector current to the emitter current. The low current travels from the voltage source into the transistor base. The current at the base will turn on the transistor. Then the current is amplified and travels from the emitter of the transistor to the collector. This is how current amplification takes place in a bi junction transistor. The amplifier is a very complex circuit which is using properties of a transistor.

Here let us first write what all are given in the question.

Number of electrons entering emitter,

${{n}_{e}}={{10}^{10}}$

Time taken=${{10}^{-6}}s$

Therefore the current in the emitter is

${{I}_{e}}=\dfrac{{{q}_{e}}\times {{n}_{e}}}{t}$

${{I}_{e}}=\dfrac{{{10}^{10}}\times 1.6\times {{10}^{-19}}}{{{10}^{-6}}}$

$=1.6mA$

Since only 2% current is lost, therefore 98% will reach the collector

So

\[\alpha =0.98\]

And also we know that

${{I}_{c}}=\alpha {{I}_{e}}$

Therefore the collector current will be,

\[\begin{align}

& =\alpha \times {{I}_{e}} \\

& =0.98\times 1.6 \\

& {{I}_{c}}\approx 1.57mA \\

\end{align}\]

Also the current amplification factor will be,

\[\beta =\dfrac{\alpha }{1-\alpha }=49\]

MOSFETs are controlled by voltage not on their own. It is used to amplify current using a MOSFET by passing the input current through a resistor there by converting the current signal to a voltage. This signal voltage is controlling the MOSFET, which gives an output signal current.