
The voltage gain of an amplifier with $9\% $ negative feedback is 10. The voltage gain without feedback will be.
A. 90
B. 100
C. 10
D. 1.25
Answer
568.2k+ views
Hint: When a part of output energy gets converted into the input energy is then this process is known as feedback. If the feedback energy is not in the phase with input energy then, there is opposition between input signal and feedback t and this is known as negative feedback.
Formula Used: The voltage gain with feedback is given by, ${{\text{A}}_{{\text{vf}}}} = \dfrac{{{{\text{A}}_{\text{b}}}}}{{{{1 + \beta }}{{\text{A}}_{\text{v}}}}}$ here ${{\text{A}}_{{\text{vf}}}}$ is the voltage gain with feedback, ${{\text{A}}_{\text{b}}}$ is voltage gain without feedback and ${{\beta }}$ is the negative feedback.
Complete step by step answer:
The formula for voltage gain with feedback is given as${{\text{A}}_{{\text{vf}}}} = \dfrac{{{{\text{A}}_v}}}{{{{1 + \beta }}{{\text{A}}_{\text{v}}}}}$, where ${{\text{A}}_{{\text{vf}}}}$ is the voltage gain with feedback, ${{\text{A}}_{\text{b}}}$ is voltage gain without feedback and ${{\beta }}$ is the negative feedback.
Step 2.
The given values are voltage gain with feedback${A_{vf}} = 10$, the negative feedback$\beta = 0.09$.
Step 3.
Let us replace the given values in the equation${{\text{A}}_{{\text{vf}}}} = \dfrac{{{{\text{A}}_v}}}{{{{1 + \beta }}{{\text{A}}_{\text{v}}}}}$,
${{\text{A}}_{{\text{vf}}}} = \dfrac{{{{\text{A}}_v}}}{{{{1 + \beta }}{{\text{A}}_{\text{v}}}}}$
Replace ${A_{vf}} = 10$ and $\beta = 0.09$ in above equation.
${{\text{A}}_{{\text{vf}}}} = \dfrac{{{{\text{A}}_v}}}{{{{1 + \beta }}{{\text{A}}_{\text{v}}}}}$
$10 = \dfrac{{{{\text{A}}_v}}}{{1 + \left( {0.09} \right){{\text{A}}_v}}}$
$\Rightarrow {{\text{A}}_v} = 10\{ 1 + \left( {0.09} \right){{\text{A}}_v}\} $
$\Rightarrow {{\text{A}}_v} = 10 + 0.9{{\text{A}}_v}$
$\Rightarrow {{\text{A}}_v} - 0.9{{\text{A}}_v} = 10$
$\Rightarrow 0.1{{\text{A}}_v} = 10$
$\Rightarrow {{\text{A}}_v} = \dfrac{{10}}{{0.1}}$
$\Rightarrow {{\text{A}}_v} = 100$
So, the value of voltage gain without feedback is ${{\text{A}}_v} = 100$. The correct answer for this problem will be option B.
Note: While solving this type of questions students should not take negative feedback value to be negative as this is opposition between the feedback energy and input signal. Feedback energy can be either voltage or current.
Formula Used: The voltage gain with feedback is given by, ${{\text{A}}_{{\text{vf}}}} = \dfrac{{{{\text{A}}_{\text{b}}}}}{{{{1 + \beta }}{{\text{A}}_{\text{v}}}}}$ here ${{\text{A}}_{{\text{vf}}}}$ is the voltage gain with feedback, ${{\text{A}}_{\text{b}}}$ is voltage gain without feedback and ${{\beta }}$ is the negative feedback.
Complete step by step answer:
The formula for voltage gain with feedback is given as${{\text{A}}_{{\text{vf}}}} = \dfrac{{{{\text{A}}_v}}}{{{{1 + \beta }}{{\text{A}}_{\text{v}}}}}$, where ${{\text{A}}_{{\text{vf}}}}$ is the voltage gain with feedback, ${{\text{A}}_{\text{b}}}$ is voltage gain without feedback and ${{\beta }}$ is the negative feedback.
Step 2.
The given values are voltage gain with feedback${A_{vf}} = 10$, the negative feedback$\beta = 0.09$.
Step 3.
Let us replace the given values in the equation${{\text{A}}_{{\text{vf}}}} = \dfrac{{{{\text{A}}_v}}}{{{{1 + \beta }}{{\text{A}}_{\text{v}}}}}$,
${{\text{A}}_{{\text{vf}}}} = \dfrac{{{{\text{A}}_v}}}{{{{1 + \beta }}{{\text{A}}_{\text{v}}}}}$
Replace ${A_{vf}} = 10$ and $\beta = 0.09$ in above equation.
${{\text{A}}_{{\text{vf}}}} = \dfrac{{{{\text{A}}_v}}}{{{{1 + \beta }}{{\text{A}}_{\text{v}}}}}$
$10 = \dfrac{{{{\text{A}}_v}}}{{1 + \left( {0.09} \right){{\text{A}}_v}}}$
$\Rightarrow {{\text{A}}_v} = 10\{ 1 + \left( {0.09} \right){{\text{A}}_v}\} $
$\Rightarrow {{\text{A}}_v} = 10 + 0.9{{\text{A}}_v}$
$\Rightarrow {{\text{A}}_v} - 0.9{{\text{A}}_v} = 10$
$\Rightarrow 0.1{{\text{A}}_v} = 10$
$\Rightarrow {{\text{A}}_v} = \dfrac{{10}}{{0.1}}$
$\Rightarrow {{\text{A}}_v} = 100$
So, the value of voltage gain without feedback is ${{\text{A}}_v} = 100$. The correct answer for this problem will be option B.
Note: While solving this type of questions students should not take negative feedback value to be negative as this is opposition between the feedback energy and input signal. Feedback energy can be either voltage or current.
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