To calculate \( i_o \) in the given op-amp circuit, we will analyze each stage step by step. ### Step 1: Analyze the First Op-Amp This op-amp is configured as a non-inverting amplifier. The voltage at the non-inverting terminal is: \[ V_+ = V_S \times \frac{3 \, \text{k}\Omega}{1 \, \text{k}\Omega + 3 \, \text{k}\Omega} = 1.2 \, \text{V} \times \frac{3}{4} = .9 \, \text{V} \] The voltage gain \( A \) for a non-inverting amplifier is given by: \[ A = 1 + \frac{5 \, \text{k}\Omega}{2 \, \text{k}\Omega} = 1 + 2.5 = 3.5 \] Thus, the output voltage \( V_1 \) of the first op-amp is: \[ V_1 = A \times V_+ = 3.5 \times .9 \, \text{V} = 3.15 \, \text{V} \] ### Step 2: Analyze the Second Op-Amp The second op-amp is configured as an inverting amplifier. The input to this op-amp is \( V_1 = 3.15 \, \text{V} \). The voltage gain for an inverting amplifier is given by: \[ A = -\frac{R_f}{R_{in}} = -\frac{10 \, \text{k}\Omega}{2 \, \text{k}\Omega} = -5 \] Therefore, the output voltage \( V_o \) is: \[ V_o = A \times V_1 = -5 \times 3.15 \, \text{V} = -15.75 \, \text{V} \] ### Step 3: Calculate Output Current \( i_o \) Using Ohm's Law, \( i_o \) is calculated as follows: \[ i_o = \frac{V_o}{4 \, \text{k}\Omega} = \frac{-15.75 \, \text{V}}{4 \, \text{k}\Omega} = -3.9375 \, \text{mA} = -3937.5 \, \mu\text{A} \] Therefore, the value of \( i_o \) is approximately \(-3937.5 \, \mu\text{A}\).