This circuit shows a current source. The current through the load is the same regardless of the position of the switch.

The op-amp attempts to keep the voltages at both terminals the same, so V– = V+. Call V1 the voltage across R1, and I1 the current across R1. Then V2 = V4. Since I1 = I2 and R1 = R2, V1 = V2 = V4.

V– = V+ = V1 - 5V

I3 = V+/R3 = (V1 - 5V)/R3.

I4 = I3 + Iload.

I4 = V4/R4 = V1/R3.

V1/R3 = (V1 - 5V)/R3 + Iload

Iload = 5V / R3 = 1.67 mA

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