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An amplifier without feedback has a voltage gain of 50, input resistance of 1 kΩ & Output resistance of 2.5 kΩ.The input resistance of the current shunt negative feedback amplifier using the above amplifier with a feedback factor of 0.2 is
Answer & Solution
Correct Answer:
Option
A
The correct answer is Option A: \( \frac{1}{11} \) kΩ
Explanation:
The input resistance of a current shunt negative feedback amplifier is given by the formula:
\[ R_{in(f)} = \frac{R_{in}}{1 + A\beta} \]
where,
\( R_{in} \) = Input resistance without feedback = 1 kΩ
\( A \) = Open-loop voltage gain = 50
\( \beta \) = Feedback factor = 0.2
Substituting the values:
\[ R_{in(f)} = \frac{1}{1 + (50 \times 0.2)} \]
\[ R_{in(f)} = \frac{1}{1 + 10} = \frac{1}{11} \text{ kΩ} \]
Thus, the input resistance of the current shunt negative feedback amplifier is \( \frac{1}{11} \) kΩ, which corresponds to Option A.
Explanation:
The input resistance of a current shunt negative feedback amplifier is given by the formula:
\[ R_{in(f)} = \frac{R_{in}}{1 + A\beta} \]
where,
\( R_{in} \) = Input resistance without feedback = 1 kΩ
\( A \) = Open-loop voltage gain = 50
\( \beta \) = Feedback factor = 0.2
Substituting the values:
\[ R_{in(f)} = \frac{1}{1 + (50 \times 0.2)} \]
\[ R_{in(f)} = \frac{1}{1 + 10} = \frac{1}{11} \text{ kΩ} \]
Thus, the input resistance of the current shunt negative feedback amplifier is \( \frac{1}{11} \) kΩ, which corresponds to Option A.
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LoginInput Resistance is Ri/D.
Output Resistance is Ro*D.
Here D is Desensivity which is equal to (1+BA), here B is feedback factor, A is open loop gain.
now calculate is
B=0.2, A=50 then D=11
We have to find Inputs resistance thai is Ri/D=
1/11.
This is answer of this solution.
1k ohm/ (1 + 50×0.2)