For the op-amp circuit shown below, find the value of vO, where R1 = 19 Ω, R2 = 14 Ω, R3 = 18 Ω, R4 = 14 Ω, Rf = 11 Ω, VS1 = 15 V, and VS2 = 3 V. **Question 3: Operational Amplifier Circuit Analysis**For the op-amp circuit shown below, find the value of \( v_o \). The component values are:- \( R_1 = 19 \, \Omega \)- \( R_2 = 14 \, \Omega \)- \( R_3 = 18 \, \Omega \)- \( R_4 = 14 \, \Omega \)- \( R_f = 11 \, \Omega \)- \( V_{S1} = 15 \, V \)- \( V_{S2} = 3 \, V \)**Circuit Description:**The circuit diagram shows an operational amplifier (op-amp) configuration featuring resistors \( R_1, R_2, R_3, R_4, \) and \( R_f \). There are two voltage sources labeled as \( V_{S1} \) and \( V_{S2} \).The circuit setup involves:- Two input voltage sources, \( V_{S1} \) and \( V_{S2} \), connected to the non-inverting and inverting inputs, respectively.- Resistors \( R_1, R_2, \) and \( R_3 \) are connected in such a way as to set the input voltages.- The feedback resistor \( R_f \) is connected from the output back to the inverting input.- \( R_4 \) is connected from the inverting input to ground.The goal is to determine the output voltage \( v_o \) based on the given resistor values and input voltages.

for the ideal op-amp voltage at inverting and non-inverting terminal are same Vn = VpApply the KCL at VpVp -Vs2R3+VpR4=0Vp -318+Vp14=0Vp 118+114=318Vp = 1.3125Now apply KCL at VnVn-Vs1R1+Vn-Vs2R2+Vn-VoRf=0Vn-1519+Vn-314+Vn-Vo11=0154Vn -2310 +209Vn-627+266Vn-266Vo = 0629Vn - 266Vo = 2937629(1.3125) - 266Vo = 2937 (Vp = Vn)Vo = -7.93 V