
A common emitter amplifier has a voltage gain of $50V$, an input impedance of ${\text{100 Units}}$ and an output impedance of ${\text{200 Units}}$. The power gain of the amplifier is
A. $500$
B. $1000$
C. $1250$
D. $50$
Answer
574.2k+ views
Hint: Power gain in any circuit is defined as the ratio of the output power to the input power. In order to find the solution of the given question write down the provided physical quantities and then apply the formula of power gain to find the required correct answer.
Formula used: ${\text{Power gain}} = \dfrac{{{\text{o/p power}}}}{{{\text{i/p power}}}}$
Power gain $ = {A_V} \times {\beta _{AC}}$
Complete step by step answer:
The power gain of an electrical network is defined as the ratio of an output power to an input power.
i.e. ${\text{Power gain}} = \dfrac{{{\text{o/p power}}}}{{{\text{i/p power}}}}$
Since it's a common emitter amplifier, then the input will be base and output will be a collector. So, now output power will be ${V_c} \times {I_c}$ and input power will be ${V_B} \times {I_B}$
Thus, AC Power gain can be written as,
Power gain $ = \dfrac{{{\text{Change in o/p power}}}}{{{\text{Change in i/p power}}}}$ $ = \dfrac{{\Delta {V_c} \times \Delta {I_c}}}{{\Delta {V_B} \times \Delta {I_B}}} = \left( {\dfrac{{\Delta {V_c}}}{{\Delta {V_B}}}} \right) \times \left( {\dfrac{{\Delta {I_c}}}{{\Delta {I_B}}}} \right) = {A_V} \times {\beta _{AC}}$
Where ${A_V}$ is voltage gain and ${\beta _{AC}}$ is AC current gain
Now, we know that
${A_V} = {\beta _{AC}} \times {\text{Resistance gain}}$
$ \Rightarrow {A_V} = {\beta _{AC}} \times \dfrac{{{R_o}}}{{{R_i}}}$
Here, it is given that:
Voltage gain $ = 50$;Input resistance, ${\text{ }}{{\text{R}}_i} = 100\Omega $ ; Output Resistance, ${{\text{R}}_o} = 200\Omega $
Thus,
$50 = {\beta _{AC}} \times \dfrac{{200\Omega }}{{100\Omega }}$
$ \Rightarrow {\beta _{AC}} = 25$
Now, AC power gain $ = {A_V} \times {\beta _{AC}} = 50 \times 25 = 1250$
So, the correct answer is “Option C”.
Additional Information: Current gain in a common emitter circuit is obtained from the base and the collector circuit currents. Because a very small change in base current can produce a large change in collector current, for a common-emitter circuit the current gain is always greater than unity.
Mathematically, current gain can be written as, Current gain $ = \dfrac{{{\text{voltage gain}}}}{{{\text{resistance gain}}}}$
Note: To solve questions like this we need to be clear with our concepts of impedance, voltage gain.
The ratio of the output voltage to the input voltage is known as voltage gain. It is a unitless quantity. Impedance is a measurement of a device’s resistance to the flow of electrical energy.
Formula used: ${\text{Power gain}} = \dfrac{{{\text{o/p power}}}}{{{\text{i/p power}}}}$
Power gain $ = {A_V} \times {\beta _{AC}}$
Complete step by step answer:
The power gain of an electrical network is defined as the ratio of an output power to an input power.
i.e. ${\text{Power gain}} = \dfrac{{{\text{o/p power}}}}{{{\text{i/p power}}}}$
Since it's a common emitter amplifier, then the input will be base and output will be a collector. So, now output power will be ${V_c} \times {I_c}$ and input power will be ${V_B} \times {I_B}$
Thus, AC Power gain can be written as,
Power gain $ = \dfrac{{{\text{Change in o/p power}}}}{{{\text{Change in i/p power}}}}$ $ = \dfrac{{\Delta {V_c} \times \Delta {I_c}}}{{\Delta {V_B} \times \Delta {I_B}}} = \left( {\dfrac{{\Delta {V_c}}}{{\Delta {V_B}}}} \right) \times \left( {\dfrac{{\Delta {I_c}}}{{\Delta {I_B}}}} \right) = {A_V} \times {\beta _{AC}}$
Where ${A_V}$ is voltage gain and ${\beta _{AC}}$ is AC current gain
Now, we know that
${A_V} = {\beta _{AC}} \times {\text{Resistance gain}}$
$ \Rightarrow {A_V} = {\beta _{AC}} \times \dfrac{{{R_o}}}{{{R_i}}}$
Here, it is given that:
Voltage gain $ = 50$;Input resistance, ${\text{ }}{{\text{R}}_i} = 100\Omega $ ; Output Resistance, ${{\text{R}}_o} = 200\Omega $
Thus,
$50 = {\beta _{AC}} \times \dfrac{{200\Omega }}{{100\Omega }}$
$ \Rightarrow {\beta _{AC}} = 25$
Now, AC power gain $ = {A_V} \times {\beta _{AC}} = 50 \times 25 = 1250$
So, the correct answer is “Option C”.
Additional Information: Current gain in a common emitter circuit is obtained from the base and the collector circuit currents. Because a very small change in base current can produce a large change in collector current, for a common-emitter circuit the current gain is always greater than unity.
Mathematically, current gain can be written as, Current gain $ = \dfrac{{{\text{voltage gain}}}}{{{\text{resistance gain}}}}$
Note: To solve questions like this we need to be clear with our concepts of impedance, voltage gain.
The ratio of the output voltage to the input voltage is known as voltage gain. It is a unitless quantity. Impedance is a measurement of a device’s resistance to the flow of electrical energy.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

Give 10 examples of unisexual and bisexual flowers

