Operational Amplifier (3)

From the previous problem, we know that the output voltage $V = 4$ . The current through $R_6$ is easily calculated.

$I_{R6} = \frac{4}{500} = \frac{8}{1000} = 8 mA$

The current through $R_6$ is $I = \dfrac {V_{\rm out}}{R_6} = \dfrac {V_{\rm out}}{500} = 2 V_{\rm out} \ \rm mA$ . To find $V_{\rm out}$ , we note that 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 consider that the current flowing in the non-inverting input (marked "+") is zero. Then we have:

$\begin{aligned} \frac {V_1 -V_{\rm in}}{R_1} + \frac {V_2 -V_{\rm in}}{R_2} + \frac {V_3 -V_{\rm in}}{R_3} & = 0 \\ \frac {2 -V_{\rm in}}{100} + \frac {-1.2 -V_{\rm in}}{100} + \frac {3.2 -V_{\rm in}}{100} & = 0 \\ 2-1.2+3.2-3V_{\rm in} & = 0 \\ \implies V_{\rm in} & = \frac 43 \\ \implies V_{\rm out} & 3 \times \frac 43 = 4 \ \rm V \end{aligned}$

Therefore $I = 2 V_{\rm out} = \boxed 8 \ \rm mA$ .