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
A. $$\frac{1}{{11}}{\text{k}}\Omega $$
B. $$\frac{1}{5}{\text{k}}\Omega $$
C. 5 kΩ
D. 11 kΩ
Answer: Option A
Solution (By Examveda Team)
The correct answer is Option A: \( \frac{1}{11} \) kΩExplanation:
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.
This Question Belongs to Electronics And Communications Engineering >> Analog Electronics
In current Shunt Feedback, Input resistance decreases and Out resistance increases.
Input 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.
1/5
R input/ (1+beta. gain)
Ri/(1 + beta.volt gain)
1k ohm/ (1 + 50×0.2)
Could you please explain?