Why Real Batteries Deliver Less

Introduction

Theoretical electron stoichiometry and equilibrium voltage set an upper bound, not a guaranteed battery output. Under current, electrode reactions require activation driving force, ions move through resistive pathways, and concentration gradients develop. The terminal voltage falls on discharge and rises on charge, reducing energy returned per cycle.

Core explanation

For a discharging cell at a fixed state of charge, write a simplified relation Vdis ≈ Erev − ηactivation,total − ηconcentration,total − IR, where the loss terms are positive magnitudes. The reversible voltage Erev itself changes as activities and phases change during discharge. A measured polarization curve plots terminal voltage against current at a specified state and temperature; it should not be confused with a voltage-versus-capacity discharge curve.

Internal resistance includes electrolyte and separator ionic resistance, electronic resistance in active layers and current collectors, and contact resistance. A larger current creates a larger approximately ohmic voltage drop. Electrode kinetics contribute nonlinear activation polarization; reactant transport and product accumulation contribute concentration polarization. In a porous electrode, these losses can be distributed spatially, so a single lumped R is only an approximation.

During charging, applied voltage must exceed the reversible voltage to overcome analogous losses in the reverse direction. The charge and discharge voltage curves are therefore separated by hysteresis. Even if nearly all charge is recovered, the difference in voltage means discharge energy is below charge energy. Heat generation from irreversible processes can raise temperature, which in turn changes kinetics, transport and aging.

Accessible capacity falls at high rate if local reactant or lithium concentration becomes limiting or a cut-off voltage is reached before all active material is used. The remaining capacity is not necessarily chemically destroyed; some may be recovered at lower current after gradients relax. Long-term capacity fade, by contrast, can result from loss of cyclable ions, active material or electrical contact and may not be recovered by a short rest.

Cell-level specific energy also includes mass of inactive components, and manufacturing choices trade energy against power and safety. Thicker electrodes may store more active material per unit collector mass but have longer transport paths, increasing high-rate polarization. A good battery design balances these demands for the intended application.

Step-by-step reasoning

Begin with a reversible voltage and theoretical charge at a stated state of charge. For an operating current, subtract activation, concentration and resistive loss magnitudes from discharge voltage. Integrate the resulting voltage over delivered charge for energy, stopping at the specified cut-off. Compare with charge input and distinguish temporary rate-related capacity loss from permanent aging.

Visual explanation

Draw open-circuit voltage as a high reference line. Plot discharge voltage bending downward as current grows and charge voltage bending upward. Shade the gap as lost energy at the operating cycle. A second plot of voltage versus discharged capacity shows a high-rate curve reaching its cut-off earlier than a low-rate curve.

Real-world analogy

A loaded delivery truck can carry a large theoretical cargo, but steep roads and traffic reduce how much it can deliver by a deadline. Resistance and transport constraints similarly reduce usable battery output at a given rate. The analogy should not be read as a literal loss of all undelivered material; some capacity returns when operating conditions improve.

Real-world example

A phone battery may show voltage sag during a high-power task, then partially recover after the task ends. The quick change can reflect iR loss and concentration relaxation. If the cell has aged, its internal resistance may be higher and its usable capacity lower, making the same current cause a larger voltage drop.

Why?

Reversible Gibbs energy sets the ideal electrical work, but finite-rate processes dissipate energy. Current through resistance heats components, charge-transfer barriers require overpotential, and local composition changes shift interfacial equilibrium. These losses reduce voltage and can force an early cut-off before theoretical charge is extracted.

Common misconception

A battery delivering less at high current has not necessarily lost that amount of active material permanently. Rate-dependent polarization can hide capacity. Conversely, a high open-circuit voltage does not prove a cell will maintain voltage under load or retain capacity after repeated cycles.

Worked example

Question: At one state of charge, a cell has Erev = 3.80 V and carries 2.0 A. Its combined electrode polarization magnitude is 0.15 V and its lumped internal resistance is 0.050 Ω. Estimate discharge terminal voltage.

Reasoning: Ohmic drop is IR = 2.0 × 0.050 = 0.10 V. Total loss is 0.15 + 0.10 = 0.25 V. Subtract from 3.80 V to get 3.55 V. This is an operating-point estimate; losses can change as the cell discharges.

Answer: Approximately 3.55 V.

Quick check

1. If charge and discharge return equal Ah but charge voltage is higher, is energy efficiency necessarily 100%? Answer: No. Higher input voltage makes input Wh exceed output Wh.

Exam focus

Use loaded voltage rather than open-circuit voltage for delivered energy. Separate immediate resistive sag from kinetic and concentration effects, and state current, temperature and cut-off when comparing capacities. Distinguish reversible thermodynamic voltage from practical polarization curves.

Advanced insight

Equivalent-circuit parameters can depend on state of charge, temperature and current history. A single “internal resistance” value may be useful for rough estimates but can hide charge-transfer, diffusion and contact contributions with different timescales and aging behavior.

Summary

Real batteries deliver less than theoretical limits because activation, transport and resistive losses lower discharge voltage and can truncate accessible capacity. Charging needs extra voltage, creating energy hysteresis. Inactive mass and long-term degradation further separate active-material theory from cell performance.

Practice questions

1. What is the ohmic drop at 3 A through 0.10 Ω? Answer: 0.30 V. 2. Why can capacity appear lower at high rate? Answer: Polarization can reach cut-off voltage before all active material is utilized. 3. Does equal charge input and output guarantee equal energy input and output? Answer: No. Different charge and discharge voltage profiles create energy loss. 4. What additional mass lowers cell-specific energy relative to active-material theory? Answer: Electrolyte, separator, current collectors, binder, packaging and other components.