A transistor has a current amplification factor (current gain) of \[50\]. In a common emitter amplifier circuit the collector resistance is chosen as \[5{\text{ }}\Omega \] and input resistance is \[{\text{1 }}\Omega \]. The output voltage when input voltage is \[{\text{0}}{\text{.01 V}}\]
A. \[{\text{ - 5 V}}\]
B. \[{\text{ - 2 V}}\]
C. \[{\text{ - 1 V}}\]
D. \[{\text{ 2}}{\text{.5 V}}\]

Hint: We will use the expression of voltage gain to find the value of output voltage of a common emitter amplifier when the input voltage applied is \[{\text{0}}{\text{.01 V}}\]. Voltage gain of a common emitter amplifier can be expressed in terms of input and output voltage. Also it can be expressed in terms of amplification factor, collector resistance and input resistance.

Formula Used:
$A_v = \dfrac{V_o}{V_i}$
$A_v = \beta \dfrac{R_c}{R_i}$
Here, $A_v$= voltage gain,$V_o$= output voltage,$V_i$= input voltage, $R_o$= collector resistance and $R_i$=input resistance.

Complete step by step answer:
Let a transistor whose current amplification factor \[\beta \] is given as \[50\]. In Common emitter amplifier mode its collector resistance \[{R_c}\] is given as \[5{\text{ }}\Omega \] and input resistance \[{R_i}\] is given as \[{\text{1 }}\Omega \]. If its input voltage\[{V_i}\] is given as \[{\text{0}}{\text{.01 V}}\] then output voltage \[{V_o}\] can be find by using the voltage gain \[{A_v}\].

The voltage gain can be expressed as, ratio of output voltage and input voltage,
\[{A_v}{\text{ = }}\dfrac{{{V_o}}}{{{V_i}}}\] _____________________\[(1)\]
Also the voltage gain can be expressed in terms of \[\beta \], \[{R_c}\] and \[{R_i}\] as,
\[{A_v}{\text{ = }}\beta \dfrac{{{R_c}}}{{{R_i}}}\] ____________________\[(2)\]
On comparing equations \[(1)\] and \[(2)\] we get the result as,
\[\dfrac{{{V_o}}}{{{V_i}}}{\text{ = }}\beta \dfrac{{{R_c}}}{{{R_i}}}\]
On substituting the values we get the result as,
\[\dfrac{{{V_o}}}{{0.01}}{\text{ = 50 }} \times {\text{ }}\dfrac{5}{1}\]
We get the output voltage as,
\[{{\text{V}}_o}{\text{ = }}\left( {{\text{250 }} \times {\text{ 0}}{\text{.01}}} \right){\text{ V}}\]
\[\therefore {{\text{V}}_o}{\text{ = 2}}{\text{.5 V}}\]
Therefore the output voltage of the transistor will be equal to \[{\text{2}}{\text{.5 V}}\].Current amplification factor is also known as current gain of circuit.

Hence the correct option is D.

Note: In common emitter configuration the input voltage is given between the base and emitter terminals while output voltage is obtained between the collector and emitter terminals. The current amplification factor has no units since it is a ratio of output and input.Similarly voltage gain has no units as it is also a ratio of output voltage and input voltage.