More packet current lowers the equivalent resistance for the same voltage swing. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Equivalent resistance uses the same packet voltage and current

The inverse trend comes from current increasing while the voltage difference stays the same.

Req=ΔVIavgR_{\text{eq}}={\Delta V\over I_{\text{avg}}}
Switched-capacitor resistanceThe resistance row cites the packet-current row.dV 4 VIavg 4 AReq 1 ohm

More capacitance or more clock rate lowers equivalent resistance

The two scan groups show the inverse relation through the cited packet current, not by a standalone resistance label.

scanCfIavgReqC changes1 F12 Hz2 A2 ohmC changes2 F12 Hz4 A1 ohmC changes4 F12 Hz8 A12 ohmf changes2 F14 Hz2 A2 ohmf changes2 F12 Hz4 A1 ohmf changes2 F1 Hz8 A12 ohm\begin{array}{c|c|c|c|c}\text{scan}&C&f&I_{\text{avg}}&R_{\text{eq}}\\C\ \text{changes}&1\ \text{F}&\tfrac{1}{2}\ \text{Hz}&2\ \text{A}&2\ \text{ohm}\\C\ \text{changes}&2\ \text{F}&\tfrac{1}{2}\ \text{Hz}&4\ \text{A}&1\ \text{ohm}\\C\ \text{changes}&4\ \text{F}&\tfrac{1}{2}\ \text{Hz}&8\ \text{A}&\tfrac{1}{2}\ \text{ohm}\\f\ \text{changes}&2\ \text{F}&\tfrac{1}{4}\ \text{Hz}&2\ \text{A}&2\ \text{ohm}\\f\ \text{changes}&2\ \text{F}&\tfrac{1}{2}\ \text{Hz}&4\ \text{A}&1\ \text{ohm}\\f\ \text{changes}&2\ \text{F}&1\ \text{Hz}&8\ \text{A}&\tfrac{1}{2}\ \text{ohm}\\\end{array}

The final resistance row divides the cited packet voltage by current

The displayed resistance uses the same voltage difference as the cited packet row.

4 V/4 A=1 ohm4\ \text{V}/4\ \text{A}=1\ \text{ohm}
Switched-capacitor resistanceThe resistance row cites the packet-current row.dV 4 VIavg 4 AReq 1 ohm