Lithium Battery Mathematics for Solar: Capacity, C-Rate, Efficiency and Cycle Life

A battery quotation usually reads something like "5 kWh, 6,000 cycles, 95% efficiency". Each number hides a calculation, and the conditions behind it decide whether it holds on your roof. This guide builds the mathematics of a lithium iron phosphate (LFP) battery from first principles, so that you can read a datasheet critically and size a backup system without guessing.

We move from electrochemistry and Faraday's law to amp-hours, C-rate, internal resistance, Peukert's law, state of charge and ageing, and end with a backup-sizing example for an Indian home. Symbols are defined with SI units before each idea. Every equipment value is an illustrative "typical datasheet" figure, so use the datasheet of the battery you are buying. Whether a battery is worth buying is a financial question, covered in Solar Plus Battery Storage in 2026, and system types are compared in On-Grid vs Off-Grid vs Hybrid Solar.

Chemistry and Faraday's law: what a cell can store

A battery stores energy as chemical potential. Lithium ions sit at a higher energy in one electrode than in the other. Charging pushes ions through the electrolyte from the positive electrode into the negative one, while the electrons travel through the external circuit. Discharging lets the ions fall back spontaneously, and the electrons do electrical work on the way round.

In an LFP cell the positive electrode (cathode, during discharge) is lithium iron phosphate and the negative electrode (anode) is graphite. Reading left to right as charging, the half-reactions are:

The first line is lithium leaving the iron phosphate as iron changes between Fe²⁺ and Fe³⁺. The second is lithium entering the layers of graphite (intercalation, meaning inserted between layers without destroying them). The iron couple sits at roughly 3.4 V against lithium metal and graphite at roughly 0.1 V, so the cell is about 3.3 V at rest. Datasheets print the nominal value, 3.2 V, closer to the average under load. Because the cathode changes between two distinct phases, LiFePO₄ and FePO₄, the voltage barely moves during charge or discharge. This flat plateau delivers steady power, but it will cause trouble when we estimate state of charge.

Faraday's law and the theoretical capacity

How much charge can a given mass of cathode hold? Each formula unit of LiFePO₄ releases one lithium ion and one electron, so counting electrons counts charge. Faraday's law says the charge moved is proportional to the amount of substance reacted. The symbols are:

• $Q$: charge passed, in coulombs (C)
• $z$: electrons transferred per formula unit, here $z = 1$ (dimensionless)
• $F$: the Faraday constant, the charge of one mole of electrons, $F = N_A q = 96{,}485.33\,\mathrm{C/mol}$
• $n$: amount of substance, in moles (mol)
• $m$: mass of active material, in grams (g)
• $M$: molar mass, in grams per mole (g/mol)

Dividing by the mass gives $F/M$ in C/g when $z = 1$. Capacity is quoted in milliamp-hours, and $1\,\mathrm{mAh} = 10^{-3}\,\mathrm{A} \times 3600\,\mathrm{s} = 3.6\,\mathrm{C}$. That is the origin of the factor 3.6: dividing coulombs by 3.6 gives mAh. The molar mass of LiFePO₄ is one Li, one Fe, one P and four O atoms:

Sanity check: the often-quoted theoretical capacity of LFP is about 170 mAh/g, so the derivation agrees with the literature. At 3.2 V this is $170 \times 3.2 \approx 544\,\mathrm{Wh/kg}$ of cathode material alone. A finished cell stores far less per kilogram, because its mass also includes graphite, electrolyte, current collectors and casing, and not every lithium ion is used. A 100 Ah cell moves $100 \times 3600 / 96{,}485 = 3.73\,\mathrm{mol}$ of lithium, needing at least 589 g of LiFePO₄ at theoretical capacity. The 170 mAh/g figure is a ceiling, not a datasheet value.

From amp-hours to watt-hours: cells, strings and packs

Amp-hours count charge, while loads and bills count energy. The bridge is the voltage at which the charge is delivered. The symbols are:

• $V$: terminal voltage, in volts (V), which changes as the cell discharges
• $q$: charge discharged so far, in amp-hours (Ah)
• $E$: energy, in watt-hours (Wh)
• $V_{\text{nom}}$: nominal voltage, a single representative value (3.2 V per LFP cell)
• $Q_{\text{nom}}$: rated capacity, in Ah

Energy is power integrated over time, and since current is the rate of flow of charge, the integral becomes one over charge:

The approximation replaces the voltage curve by its average. To see the error, take an illustrative discharge curve for a 100 Ah LFP cell and add up the area under it segment by segment (trapezoidal rule).

The integral gives 321.95 Wh against $3.2\,\mathrm{V} \times 100\,\mathrm{Ah} = 320\,\mathrm{Wh}$, a difference of 0.6%. Because LFP is flat, the shortcut is accurate to about a percent, which is why it is used everywhere.

Series and parallel

Let $N_s$ be the number of cells in series and $N_p$ the number of parallel groups (both counts). Cells in series carry the same current and their voltages add. Cells in parallel share the current and their capacities add:

Sanity check: sixteen cells of 320 Wh each is 5,120 Wh, the same answer. A 16s2p pack gives 200 Ah and 10.24 kWh at the same 51.2 V. With typical cell limits of 3.65 V full and 2.5 V empty (check your datasheet), the pack swings between 58.4 V and 40 V, which is why this class is called a "48 V" system.

At 51.2 V a 1 kW load draws $1000/51.2 = 19.5\,\mathrm{A}$ and a 5 kW load draws 97.7 A, so cable size, fuse rating and connection quality matter. Conductor sizing is worked through in Solar Wiring from First Principles. In a series string the weakest cell limits the whole pack, so the battery management system (BMS) monitors and balances every cell.

C-rate, discharge time and depth of discharge

A 100 Ah and a 200 Ah battery behave differently at the same 50 A. C-rate expresses current relative to capacity. The symbols are:

• $C$: C-rate, in units of 1/h, usually written as "0.5C" or "1C"
• $I$: current, in amperes (A)
• $Q_{\text{nom}}$: rated capacity, in Ah
• $t_{\text{dis}}$: time to discharge from full to empty, in hours (h)

The time formula is the ideal case, assuming the full rated charge is available at that current. Peukert's law and internal resistance qualify it shortly. For our 16s1p, 100 Ah, 51.2 V pack:

Depth of discharge and usable energy

Depth of discharge (DoD, a fraction from 0 to 1) is the share of rated capacity you draw before recharging. Limiting it protects the battery and leaves a reserve:

At 90% DoD our pack gives $5.12 \times 0.9 = 4.61\,\mathrm{kWh}$, and at 80% DoD 4.10 kWh. Check whether a quoted kWh figure is nominal or already usable. Power matters too: a 3 kW inverter at full load on this pack draws $3000/5120 = 0.59C$, so the pack must be rated for that continuous rate.

Internal resistance and round-trip efficiency

Real cells waste some energy as heat. The simplest useful model is an ideal voltage source in series with a resistance. We assume constant open-circuit voltage, constant resistance, steady current, and every coulomb that goes in comes back out. The symbols are:

• $V_{\text{oc}}$: open-circuit voltage, in volts (V), here the mean pack voltage 51.2 V
• $R_{\text{int}}$: total DC internal resistance of the pack, in ohms ($\Omega$), including cells, busbars, fuse and the BMS switch
• $I$: magnitude of the current, in amperes (A)
• $x = I R_{\text{int}} / V_{\text{oc}}$: the fractional voltage drop (dimensionless)

By Ohm's law the terminal voltage sags when current is drawn and rises when current is forced in:

On discharge, the fraction of chemical power reaching the terminals is $V_{\text{dis}}/V_{\text{oc}}$. On charge, the fraction stored is $V_{\text{oc}}/V_{\text{chg}}$. The rest is $I^2 R_{\text{int}}$ heat:

For the round trip, push charge $Q$ in and take the same charge out. Energy in is $(V_{\text{oc}} + I R_{\text{int}})\,Q$ and energy out is $(V_{\text{oc}} - I R_{\text{int}})\,Q$, so:

Each direction loses about $x$, so the round trip loses about twice that. Since $x$ is proportional to current, efficiency falls as C-rate rises, and the heat rises with the square of current. Take an illustrative pack resistance of 15 m$\Omega$ (sixteen cells of roughly 0.5 m$\Omega$, plus about 7 m$\Omega$ for busbars, fuse and BMS switch), a charger efficiency of 97% and an inverter efficiency of 95%.

Moving from 0.2C to 1C multiplies the current by five and the heat by twenty-five, from 6 W to 150 W. The shortcut $1 - 2x$ gives 94.14% at 1C against the exact 94.31%. Real cells also show polarisation and a resistance that rises in the cold and at low state of charge, so datasheet round-trip figures depend on the stated test conditions.

Stages multiply

Energy passing through several stages loses a fraction at each, so system efficiency is the product of stage efficiencies. For solar energy stored and later used by an AC load:

That is the 0.5C row of the table: about one-tenth of the stored solar energy is lost before it reaches your fan. Inverter efficiency varies with load because switching and conduction losses do not scale with output power, as explained in How Inverters Make AC. There is also a fixed cost. An inverter and BMS consuming an illustrative 30 W around the clock use $30 \times 24 = 720\,\mathrm{Wh}$ a day, more than the roughly 0.47 kWh lost in conversion when 4 kWh a day is delivered at 89.5%. For a lightly cycled battery, standby consumption can dominate.

Peukert's law: why lead-acid shrinks and lithium barely does

Lead-acid sheets state capacity at a slow rate, for example 100 Ah over 20 hours. Draw current faster and fewer amp-hours come out. Peukert's law is an empirical fit for this. The symbols are:

• $H$: the rated discharge time, in hours (h), for example 20 h
• $C_p$: the capacity at that rating, in Ah
• $I_p = C_p/H$: the rated current, in A
• $I$: the actual discharge current, in A
• $k$: the Peukert exponent (dimensionless), 1 for an ideal battery
• $t$: the actual runtime, in hours (h)

Multiplying by $I$ gives the delivered capacity, $C_{\text{eff}} = I t$. Using $I_p = C_p/H$:

At $I = I_p$, $C_{\text{eff}} = C_p$, as it must. For larger $I$ and $k \gt 1$, the bracket is below one and capacity shrinks. Typical exponents are about 1.1 to 1.3 for lead-acid and about 1.0 to 1.05 for lithium-ion. These are typical ranges, not guarantees. Compare two 100 Ah batteries both rated at 5 A (20 hours), with $k = 1.2$ for lead-acid and $k = 1.02$ for lithium:

At 50 A the lead-acid battery delivers only 63% of rated capacity, so a 2-hour ideal runtime becomes 1.26 hours, while the lithium battery delivers over 95%. Peukert's law is a fit over a limited current range, not a law of nature. For lithium, the remaining shortfall mostly comes from the resistive drop described earlier and the low-voltage cut-off.

State of charge by coulomb counting

State of charge (SOC) is the fuel gauge: the fraction of present capacity still stored. Coulomb counting integrates the current measured by a sensor. The symbols are:

• $\text{SOC}$: state of charge, between 0 and 1 (dimensionless)
• $\text{SOC}_0$: the state of charge at the start of counting
• $Q$: present full capacity, in Ah
• $i(\tau)$: measured current, in A, positive on discharge and negative on charge
• $\eta_c$: charge (coulombic) efficiency, slightly below 1, applied to charging current only

With $Q$ in Ah, time is in hours. The method is smooth, but integrals accumulate error. Suppose the sensor reads $i + \varepsilon$, where $\varepsilon$ is a constant offset in amperes. The SOC error then grows linearly with time:

An illustrative offset of 0.1 A, plausible for a Hall-effect sensor, produces 2.4 Ah, or 2.4% of a 100 Ah pack per day and 24% after ten days. A shunt with a 10 mA offset gives 0.24% per day. Temperature drift and the approximate value of $\eta_c$ add further error.

Recalibration and the flat LFP curve

A BMS therefore resets the count whenever it gets a trustworthy signal. When cells reach the charge-limit voltage and the current tapers below a small threshold, the pack is physically full. The BMS sets SOC to 100% and may update its estimate of $Q$ from the charge counted since the last empty state. This is why the displayed percentage can jump at full charge.

Why not read the voltage instead? For LFP, open-circuit voltage maps poorly to SOC because of the flat plateau. If the plateau rises by an illustrative 0.08 V between 20% and 80%, the slope is $0.08\,\mathrm{V} / 60 = 1.3\,\mathrm{mV}$ per percentage point. A 10 mV measurement error then means a 7.5% SOC error. A 0.5 m$\Omega$ cell carrying 50 A drops 25 mV, which looks like a 19% error unless the BMS compensates. So an LFP BMS combines coulomb counting in the middle of the range with voltage corrections at the steep ends.

Cycle life, throughput and the Arrhenius law

Batteries age through cycling and through calendar time and temperature. We treat the first as a throughput calculation and the second with the Arrhenius law. The symbols for throughput are:

• $N$: number of cycles to end of life (a count)
• $E_{\text{usable}}$: usable energy per cycle when new, in kWh
• $f(n)$: capacity fraction at cycle $n$, 1 when new (dimensionless)
• $f_{\text{end}}$: the capacity fraction defining end of life, commonly 0.8
• $\bar{f}$: the average capacity fraction over life

Each cycle delivers energy proportional to the remaining capacity, so lifetime energy is a sum over cycles. If capacity fades linearly from 1 to $f_{\text{end}}$, the average is the midpoint:

Suppose, purely as an illustration and not a promise for any product, a datasheet shows 6,000 cycles at 80% DoD to 80% capacity. Our 5.12 kWh pack at 80% DoD has $E_{\text{usable}} = 4.096\,\mathrm{kWh}$, so $E_{\text{life}} \approx 6000 \times 4.096 \times 0.9 = 22{,}118\,\mathrm{kWh}$, about 22 MWh. Without the 0.9 factor the figure would be overstated by 11%. Real fade is rarely linear, and cycle counts depend on the test temperature, C-rate and DoD. Check against time as well: one cycle a day for 10 years is only 3,650 cycles, so a lightly cycled home battery is more likely limited by calendar ageing than by cycle count.

Temperature and the Arrhenius law

Many ageing reactions, such as growth of the solid-electrolyte layer on graphite, speed up with temperature. The Arrhenius law describes a thermally activated rate. The symbols are:

• $k$: reaction rate (1/s)
• $A$: pre-exponential factor, same units as $k$
• $E_a$: activation energy, in joules per mole (J/mol)
• $R = 8.314\,\mathrm{J/(mol\,K)}$: the gas constant, $k_B N_A$
• $T$: absolute temperature, in kelvin (K)

Dividing the rates cancels $A$. For an illustrative $E_a = 50\,\mathrm{kJ/mol}$, from 25 °C (298.15 K) to 35 °C (308.15 K):

So the rate nearly doubles. A small-step expansion gives $\exp[E_a \Delta T/(R T^2)]$, about 1.97 here. But the ratio depends strongly on $E_a$, so "doubling per 10 °C" is only a rough rule.

Exactly doubling from 25 to 35 °C needs $E_a \approx 53\,\mathrm{kJ/mol}$, and the ratio shrinks slightly at higher temperature. Real activation energies are empirical fits that mix several mechanisms, so treat these as illustrations. The practical message holds: a cabinet at 45 °C instead of 30 °C ages about 2.5 times faster with $E_a = 50\,\mathrm{kJ/mol}$, since $\exp[6014 \times (1/303.15 - 1/318.15)] = 2.55$. Batteries therefore belong in a shaded, ventilated place, away from direct sun and inverter exhaust. Cold has limits too: charging below freezing can damage cells, so a BMS may restrict it.

Sizing a backup battery for a four-hour outage

Now we combine the pieces to find the battery energy needed to carry a home's critical loads through an outage. The symbols are:

• $P_i$: power drawn by load $i$, in watts (W)
• $t_i$: hours that load $i$ runs, in hours (h)
• $\text{DoD}$: the depth of discharge allowed (fraction)
• $\eta_{\text{inv}}$: inverter efficiency (fraction)
• $E_{\text{req}}$: required nominal battery energy, in Wh

The numerator is the energy the loads consume, and the denominator corrects for inverter loss and the share of the battery we agree to use. Below is an illustrative Indian home in a four-hour evening outage with only critical circuits on. Wattages are typical values, so measure your own appliances.

The loads total 2,595 Wh, or 2.6 kWh, an average of 649 W over four hours. With an illustrative DoD of 0.9 and inverter efficiency of 0.90 (small loads on a larger inverter run at modest efficiency, and the small battery discharge loss at these low rates is folded into the same factor):

Now allow for ageing. If the battery must still do the job at 80% of its original capacity, divide again: $3.20/0.8 = 4.0\,\mathrm{kWh}$. Our 5.12 kWh pack delivers $5.12 \times 0.9 \times 0.9 = 4.15\,\mathrm{kWh}$ to AC loads when new, enough for the average 649 W for about 6.4 hours. Sanity check: 3.20 kWh exceeds the 2.6 kWh consumed, as it should once reserve and inverter loss are included, and the pack has margin above the 4.0 kWh end-of-life figure.

The power check

The system must also supply the peak power. With everything on, the running peak is $220 + 60 + 150 + 15 + 100 + 60 + 750 = 1355\,\mathrm{W}$, counting the fridge at its running power. Induction motors draw several times their running current at start. An illustrative factor of 4 on the 750 W pump gives 3,000 W briefly. If the pump starts while the other loads (605 W) run, the peak is:

The inverter needs a surge rating covering about 3.6 kW for a few seconds, and the battery and BMS must permit about 78 A for the same period. A battery that is adequate on energy can fail on power. A soft-starter, or keeping the pump off the backup circuit, reduces the demand. The 78 A figure also drives fuse and cable sizing, as in the cable sizing article.

Recharging from solar

The battery must be refilled by solar, and every stage loses some energy. With an illustrative 0.97 for the charger, 0.97 for the battery round trip and 0.90 for the inverter:

As a rule of thumb, rooftop solar in this part of Andhra Pradesh generates roughly 4 to 4.5 units (kWh) per kW per day on annual average. Recharging one such outage then needs $3.06/4.5$ to $3.06/4.0$, roughly 0.7 to 0.8 kW of panels. These are approximations that a site audit refines. This is on top of your daytime consumption, and monsoon months generate less, so leave margin.

Ray2Volt can run this analysis for your own site, from your measured loads, outage pattern and the battery datasheet. The lithium-ion batteries we supply carry a 10-year product warranty. See our hybrid systems with battery page and book a free power audit.

Frequently asked questions

Ray2Volt Solar Private Limited

We design, install, and service rooftop solar for homes and businesses across Tirupati district and Andhra Pradesh, from PM Surya Ghar residential systems to commercial and industrial plants. Every enquiry starts with a free power audit.