I²R: why a short high current costs more than the same energy delivered slowly
Move a given quantity of charge out of a battery quickly and you waste more of it as heat than if you move the same charge slowly. Not a little more — proportionally more, in direct ratio to the current. This is the whole argument for buffering a transient, and it is four lines of algebra rather than an opinion.
The four lines
A cell has an internal resistance. Current through a resistance dissipates power as heat at a rate given by Joule’s law:
P = I²R
Over a period of time t, the energy lost as heat is I²Rt. Now hold fixed the thing the application actually cares about, which is not time but the quantity of charge delivered:
Q = I t, so t = Q / I
Substitute that into the loss:
Eloss = I²R · (Q / I) = I R Q
The squared term collapses. For a fixed amount of charge moved through a fixed resistance, the energy wasted as heat is directly proportional to the current used to move it. Draw it twice as fast and you pay twice the heat. Ten times as fast, ten times the heat.
A worked example, so the number is not abstract
Take a pack with an internal resistance of 20 milliohms, and a job that requires 1,000 ampere-seconds of charge to be delivered. Two ways to do it:
| Slow | Fast | |
|---|---|---|
| Current | 10 A | 100 A |
| Duration | 100 s | 10 s |
| Charge delivered | 1,000 A·s | 1,000 A·s |
| Heat, I²Rt | 10² × 0.02 × 100 = 200 J | 100² × 0.02 × 10 = 2,000 J |
Identical charge. Ten times the waste heat, deposited in ten times less time, which also means it arrives faster than the pack can carry it away. The peak temperature reached is worse than the ten-times figure alone suggests.
Three consequences, in the order they bite
- You get less out of the pack. The terminal voltage under load is the open-circuit voltage minus the drop across the internal resistance. Pull harder and the terminal voltage sags further, so the pack reaches its cut-off voltage sooner and delivers less usable energy. This is why a cell’s capacity is always quoted at a stated discharge rate, and why the same cell reports a smaller number at a higher one.
- The heat lands in the material that ages. The loss is dissipated inside the cell, not in a resistor bolted to a heatsink. Degradation rates in lithium-ion cells rise with temperature, so heat generated by a transient accelerates the ageing of the very component the transient is drawn from.
- It gets worse as the pack gets worse. Internal resistance rises with age and rises at low temperature. Since the loss is proportional to R as well as to I, a cold or aged pack pays more for the same transient than a warm new one — which is a feedback loop, not a one-off penalty.
Why this makes transients disproportionately expensive
Consider a duty cycle where brief high-current events account for a small share of the total energy delivered but a large share of the peak current. Because loss scales with current and not with energy share, those brief events can account for a share of the total heat generated far larger than their share of the work done. A transient that is a footnote in the energy budget can be the main entry in the heat budget.
That asymmetry is the entire reason to route the transient somewhere else. It is also why the right specification to look at is not the average current but the shape of the current over time — the point of the note on peak versus RMS current.
What this argument does and does not license
It establishes that moving a transient off the cells reduces the heat generated in them, and that the reduction is proportional rather than marginal. That is arithmetic, and it does not depend on any product.
It does not establish how much longer any particular pack will last. Cell ageing depends on temperature, state of charge, depth of discharge and chemistry together, and the relationship between a reduction in ohmic heating and an extension of service life is empirical, not algebraic. Anyone quoting a life-extension multiple from I²R alone is over-reaching, and this site does not do it. The distinction between the two dials that actually govern ageing is set out in the technical note on state of charge versus depth of discharge.
It also does not establish that buffering prevents thermal runaway. Ohmic self-heating is one contributor among several — internal short circuits, mechanical damage, overcharge and manufacturing defects are others — and reducing one contributor is not the same as eliminating a failure mode. Any claim that this architecture eliminates thermal-runaway risk is a claim this site does not make and does not support.
The counter-argument worth taking seriously
Routing the transient elsewhere does not delete it. It moves it into a capacitor bank, a converter and the conductors between them, each of which has resistance of its own and dissipates a share of what was saved. Whether the net is favourable depends on how much lower the combined resistance of that path is, how efficient the converter is at the currents involved, and what the mass and cost of the additional hardware buy in return. That is the calculation that decides the architecture, and it is application-specific — stated as an open question in the list of what I cannot yet answer rather than assumed in the architecture’s favour.