Reservoir ripple grows with load current because more charge leaves the capacitor during the same interval, reducing the valley margin. With 10 V peak, 3 s interval, and 3 F capacitance, load currents 1, 2, and 3 A make ripple 1, 2, and 3 V with margins 1, 0, and -1 V exactly for the ideal reservoir model; real reservoirs add capacitance tolerance, ESR, leakage, rectifier losses, load variation, ripple waveform shape, heating, and voltage/current/time measurement uncertainty.

Load current moves ripple and therefore the minimum valley margin. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

1 A ripple row

Source peak stays 10 V, interval stays 3 s, and capacitance stays 3 F. Load current 1 A makes ripple 1 V and margin 1 V.

ΔV=1 V,m=1 V\Delta V=1\ \text{V},\quad m=1\ \text{V}
Reservoir current boundaryCurrent is the numerator of the ripple ledger.Iload 1 Adt 3 sC 3 FdV 1 VVmin 8 Vgate pass

2 A ripple row

Source peak stays 10 V, interval stays 3 s, and capacitance stays 3 F. Load current 2 A makes ripple 2 V and margin 0 V.

ΔV=2 V,m=0 V\Delta V=2\ \text{V},\quad m=0\ \text{V}
Reservoir current boundaryCurrent is the numerator of the ripple ledger.Iload 2 Adt 3 sC 3 FdV 2 VVmin 8 Vgate pass

3 A ripple row

Source peak stays 10 V, interval stays 3 s, and capacitance stays 3 F. Load current 3 A makes ripple 3 V and margin negative 1 V.

ΔV=3 V,m=−1 V\Delta V=3\ \text{V},\quad m=-1\ \text{V}
Reservoir current boundaryCurrent is the numerator of the ripple ledger.Iload 3 Adt 3 sC 3 FdV 3 VVmin 8 Vgate fail