Assume that the op-amp in the circuit shown is ideal.

The value of $\frac{V_x}{I_x}$ (in kΩ) is _______

In the range: -4 - -4



By applying KCL at inverting terminal,

$\frac{V^{-}}{1}+\frac{V^{-}-V_0}{2}=0$

$\Rightarrow V^{-}=\frac{V_0}{3}$

By using the concept of virtual ground, Vₓ = V⁻

$V_x=\frac{V_0}{3}$

⇒ V₀ = 3Vₓ

By applying KCL at the non-inverting terminal,

$I_x=\frac{V_x-V_0}{8}$

$\Rightarrow I_x=\frac{V_x-3V_x}{8}=\frac{-V_x}{4}$

$\Rightarrow \frac{V_x}{I_x}=-4\mathrm{k\Omega }$