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 + I_load.

I₄ = V₄/R₄ = V₁/R₃.

V₁/R₃ = (V₁ - 5V)/R₃ + I_load

I_load = 5V / R₃ = 1.67 mA

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