Operational Amplifier (2)

We know that the voltage at the non-inverting input is $V_+ = \frac{4}{3}$ . Therefore, the voltage at the inverting input is $V_- = \frac{4}{3}$ . We also know that the current into the inverting input must be zero.

$\frac{V - 4/3}{200} + \frac{0 - 4/3}{100} = 0$

Solving results in $V = 4$

The voltage at the node circled is the output voltage of the OpAmp $V_{\rm out}$ . The given OpAmp circuit is a non-inverting OpAmp configuration . The closed loop voltage gain of the non-inverting OpAmp is given by:

$\begin{aligned} A_V & = \frac {R_5+R_4}{R_4} \\ \frac {V_{\rm out}}{V_{\rm in}} & = \frac {200+100}{100} \\ V_{\rm out} & = 3 V_{\rm in} \end{aligned}$

To find $V_{\rm in}$ , we note that the current flowing in the non-inverting input (marked "+") is zero. Then we have:

$\begin{aligned} \frac {V_1 -V_{\rm in}}{R_1} + \frac {V_2 -V_{\rm in}}{R_2} + \frac {V_3 -V_{\rm in}}{R_3} & = 0 \\ \frac {2 -V_{\rm in}}{100} + \frac {-1.2 -V_{\rm in}}{100} + \frac {3.2 -V_{\rm in}}{100} & = 0 \\ 2-1.2+3.2-3V_{\rm in} & = 0 \\ \implies V_{\rm in} & = \frac 43 \ \rm V \end{aligned}$

Therefore, $V_{\rm out} = 3 V_{\rm in} = 3 \times \dfrac 43 = \boxed 4 \ \rm V$ .