Inductance divides the same voltage-time push. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Inductance divides the same voltage-time push

The voltage-time area is fixed at 6 V for 1 s. Larger inductance spreads that push into a smaller current change.

ΔI1Lwhen VΔt is fixed\Delta I\propto {1\over L}\quad\text{when }V\Delta t\text{ is fixed}
Inductance denominator setupThe same voltage-time area is divided by L.L 4 HV 6 Vdt 1 sIi 0 AdI 3/2 AIf 3/2 A

Increasing inductance slows the current ramp

The rows hold voltage and time fixed while inductance grows.

LVΔtΔI1 H6 V1 s6 A2 H6 V1 s3 A4 H6 V1 s32 A\begin{array}{c|c|c|c}L&V&\Delta t&\Delta I\\1\ \text{H}&6\ \text{V}&1\ \text{s}&6\ \text{A}\\2\ \text{H}&6\ \text{V}&1\ \text{s}&3\ \text{A}\\4\ \text{H}&6\ \text{V}&1\ \text{s}&\tfrac{3}{2}\ \text{A}\\\end{array}

A larger inductance slows the same voltage-time ramp

Voltage and interval stay fixed. Only inductance changes, so the current change is halved.

ΔI=6 V1 s/4 H=32 A\Delta I=6\ \text{V}\cdot1\ \text{s}/4\ \text{H}=\tfrac{3}{2}\ \text{A}
Larger inductor rampThe same voltage-time area is divided by a larger L.L 4 HV 6 Vdt 1 sIi 0 AdI 3/2 AIf 3/2 A