A negative inductor voltage makes a negative current change. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Negative voltage means negative current change

The initial current is 3 A. A negative voltage removes current instead of adding it.

If=Ii+VΔtLI_f=I_i+{V\Delta t\over L}
Reverse ramp setupThe final current is checked from initial current plus signed change.L 2 HV -2 Vdt 3 sIi 3 AdI -3 AIf 0 A

Longer reverse time removes more current

The voltage and inductance stay fixed. Each longer interval makes the negative current change larger in magnitude.

IiVΔtΔIIf3 A2 V1 s1 A2 A3 A2 V2 s2 A1 A3 A2 V3 s3 A0 A\begin{array}{c|c|c|c|c}I_i&V&\Delta t&\Delta I&I_f\\3\ \text{A}&-2\ \text{V}&1\ \text{s}&-1\ \text{A}&2\ \text{A}\\3\ \text{A}&-2\ \text{V}&2\ \text{s}&-2\ \text{A}&1\ \text{A}\\3\ \text{A}&-2\ \text{V}&3\ \text{s}&-3\ \text{A}&0\ \text{A}\\\end{array}

Reverse voltage ramps the current down

The sign of the applied inductor voltage sets the sign of the current change; the final current is checked from the initial current.

If=Ii+ΔI=3 A+3 A=0 AI_f=I_i+\Delta I=3\ \text{A}+-3\ \text{A}=0\ \text{A}
Reverse-voltage rampA fixed negative voltage removes the stored current.L 2 HV -2 Vdt 3 sIi 3 AdI -3 AIf 0 A