A npn transistor is connected in common emitter configuration in a given amplifier. A load resistance of $800\Omega$ is connected in the collector circuit and the voltage drop across it is \[\text{0}.8V\]. If the current amplification factor is \[\text{0}.96\] and the input resistance of the circuit is $192\Omega$ , the voltage gain and the power gain of the amplifier will respectively be:
\[\begin{align}
  & \text{A}\text{. }4,3.84 \\
 & \text{B}\text{. 3}\text{.69,3}\text{.84} \\
 & \text{C}\text{. 4,4} \\
 & \text{D}\text{. 4,3}\text{.69} \\
\end{align}\]

Hint: A transistor is a semiconductor device which is used to amplify or switch electrical signals. To calculate the power gain we need current and voltage gain, since the current gain is given, we can calculate the individual current and multiply it with resistance, to get individual voltage.

Formula used:
$A_{i}=\dfrac{I_{out}}{I_{in}}= \dfrac{I_{c}}{I_{b}}=\beta$
$V_{i}=\dfrac{V_{out}}{V_{in}}= \dfrac{V_{l}}{V_{b}}$
$A_{P}=A_{i}\times A_{v}$

Complete step by step answer:
A transistor is a semiconductor which is used to amplify the input, or acts as a switch. It has three terminals, namely collector, emitter, and base. The common transistors are bipolar transistors, which are of two types i.e. are npn and pnp.
In an NPN or negative-positive-negative transistor, can be biased in three ways: common base, common emitter, and common collector.
Here we have a NPN transistor with a common emitter configuration, then the input terminal of the transistor is the base and the output is the collector.
It is given that the, resistance due to collector $R_{c}=800\Omega$, voltage in the circuit is $V_{l}=0.8V$, resistance of the base is $R_{b}=192\Omega$ and the current gain $\beta=0.96$
We know that current gain $\beta=\dfrac{I_{c}}{I_{b}}=0.96$
If the transistor is forward biased, i.e. BE is forward biased, we know that$V_{BE}=0$
Then the collector current $I_{c}=\dfrac{V_{l}}{R_{c}}=\dfrac{0.8}{800}=10^{-3}A$ ,
Using the current gain, the base current $I_{b}=\dfrac{I_{c}}{\beta}=\dfrac{10^{-3}}{0.96}A$
From KVL or Kirchhoff voltage law, $V_{b}=I_{b}R_{b}$
Then $V_{b}=\dfrac{10^{-3}}{0.96}\times 192=0.2V$
We know that, current gain $A_{i}=\dfrac{I_{out}}{I_{in}}= \dfrac{I_{c}}{I_{b}}=\beta=0.96$
Similarly, voltage gain is $V_{i}=\dfrac{V_{out}}{V_{in}}= \dfrac{V_{l}}{V_{b}}=\dfrac{0.8}{0.2}=4$
We also know that the power gain $A_{P}=A_{i}\times A_{v}$ where $A_{v}$ is the voltage gain and $A_{i}$ is the current gain.
Then $A_{P}= 0.96 \times 4= 3.84$
Thus, the voltage gain $A_{v}$ and power gain$A_{P}$ is \[4,3.84\] respectively.
Hence the answer is \[A.4,3.84\].

Note:
Current gain, voltage gain, and power gain, defined as input quantity by output quantity do not have units, since these are ratios. Also, note that we use ohm's law to find $V_{c}$. It is important to know that here, $\beta=\dfrac{I_{c}}{I_{b}}$, which in turn gives $I_{b}$.