Answer

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**Hint:**In the above question, we are given that the amplifier gain is $50$ , which is denoted by $A$. We are also given that the gain after feedback falls to $25$ , which is denoted by ${A_f}$ . We also know that Voltage gain after feedback is ${A_f} = \dfrac{A}{{1 + A\beta }}$ . Where, $\beta $ is the feedback ratio. Now, By substituting the values of amplifier gain and voltage gain after feedback falls, we will get our answer.

**Formula used:**

Voltage gain after feedback falls, ${A_f} = \dfrac{A}{{1 + A\beta }}$.

Where, $\beta $ is the feedback ratio and $A$ is amplifier gain.

**Complete step by step solution:**

From the above question, we know that

Here, amplifier gain is $A = 50$ and Voltage gain after feedback falls is ${A_f} = 25$

Now, we know that

Formula used for voltage gain after feedback falls is ${A_f} = \dfrac{A}{{1 + A\beta }}$

Here, we have the values of amplifier gain and voltage gain after feedback falls,

Now, substituting the values of amplifier gain and voltage gain after feedback falls, which is $A = 50$ and ${A_f} = 25$ respectively in the above formula,

We get,

\[

\Rightarrow {A_f} = \dfrac{A}{{1 + A\beta }} \\

\Rightarrow 25 = \dfrac{{50}}{{1 + 50\beta }} \\

\]

Now, simplifying the above equation,

\[

\Rightarrow 25\left( {1 + 50\beta } \right) = 50 \\

\Rightarrow 1 + 50\beta = 2 \\

\Rightarrow \beta = \dfrac{1}{{50}} \\

\Rightarrow \beta = 0.02 \\

\]

Hence, The feedback ratio is \[\beta = 0.02\].

**Hence, the answer is 0.02.**

**Note:**We know that amplifier gain is the amplified difference between the input and output. Now, in the problem, we have given the negative feedback which means that the feedback decreases after the gain and this results in reducing distortion and noise. Hence, it improves the bandwidth and signal. Now, by using the above formula carefully, we can get our answer correct.

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