A balanced difference amplifier subtracts two inputs and applies an exact resistor ratio. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

The resistor ratios are matched

Both resistor ratios are 2, so the difference network is balanced.

RfRin=RbottomRtop=2\frac{R_f}{R_{\text{in}}} = \frac{R_{\text{bottom}}}{R_{\text{top}}} = 2
Matched ratiosThe helper rejects mismatched difference-amplifier ratios.+-op amprails -12 V to 12 VVout 6 VV+ 4 VV- 4 VI+ 0 AI- 0 AV1 3 VV2 6 VRin 1 ohmRf 2 ohmtop 1 ohmbottom 2 ohmV+ = V-difference

Subtract first, then multiply

The inputs differ by 3 volts, and ratio 2 gives 6 volts.

Vout=2(6 V(3 V))=6 VV_{\text{out}} = 2\cdot(6\ \text{V}-\left(3\ \text{V}\right)) = 6\ \text{V}
Difference amplifierThe output is checked from matched resistor ratios.+-op amprails -12 V to 12 VVout 6 VV+ 4 VV- 4 VI+ 0 AI- 0 AV1 3 VV2 6 VRin 1 ohmRf 2 ohmtop 1 ohmbottom 2 ohmV+ = V-difference