Aiding winding sources are checked against an output-voltage ceiling. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Aiding-polarity row 1

First source stays 6 V. Aiding second source 2 V gives output 8 V against a 10 V ceiling, so margin is 2 V.

m=2 Vpassm=2\ \text{V}\quad\text{pass}
Aiding-polarity output boundaryThe output label cites two winding sources and the aiding sign.Np 1Ns 1dPhi 6 Wbdt 1 sVp 6 VVs 6 Vmode aidingNp 1Ns 1dPhi 2 Wbdt 1 sVp 2 VVs 2 Vmode aidingV1 6 VV2 2 Vmode aidingVout 8 V

Aiding-polarity row 2

First source stays 6 V. Aiding second source 4 V gives output 10 V against a 10 V ceiling, so margin is 0 V.

m=0 Vpassm=0\ \text{V}\quad\text{pass}
Aiding-polarity output boundaryThe output label cites two winding sources and the aiding sign.Np 1Ns 1dPhi 6 Wbdt 1 sVp 6 VVs 6 Vmode aidingNp 1Ns 1dPhi 4 Wbdt 1 sVp 4 VVs 4 Vmode aidingV1 6 VV2 4 Vmode aidingVout 10 V

Aiding-polarity row 3

First source stays 6 V. Aiding second source 6 V gives output 12 V against a 10 V ceiling, so margin is negative 2 V.

m=2 Vfailm=-2\ \text{V}\quad\text{fail}
Aiding-polarity output boundaryThe output label cites two winding sources and the aiding sign.Np 1Ns 1dPhi 6 Wbdt 1 sVp 6 VVs 6 Vmode aidingNp 1Ns 1dPhi 6 Wbdt 1 sVp 6 VVs 6 Vmode aidingV1 6 VV2 6 Vmode aidingVout 12 V