Question

Consider the circuit shown in the figure below.

$\text{The circuit is a current amplifier circuit with amplification factor}{A}_I=\frac{I_L}{I_i}=21\text{. If the op-amp is ideal then the value of resistance R is equal to}$

• A. $1.5\text{k}\Omega$
• B. $1\text{k}\Omega$
• C. $500\Omega$
• D. $2\text{k}\Omega$

Answer 1

The correct option is C $500\Omega$


now, from the figure it is clear that,
$I_1\left(10\text{k}\Omega \right)=I_{2}R\text{[where}{I}_1=I_i]$
$\text{also,}{I}_1+I_2=I_L$
$\therefore I_1\times 10\text{k}\Omega =(I_L-I_1).R$
$10\text{k}\Omega =[\frac{I_L}{I_I}-1]$R
$R=500\Omega$