Solar Wiring from First Principles: Resistance, Voltage Drop and Cable Sizing

A solar string works at several hundred volts and more than ten amperes, through copper lying on a roof that can be 25 °C or more hotter than the air. The cable looks passive, yet it is a resistor in series with your array. It wastes energy every sunny hour, and if it is undersized or badly terminated it heats the insulation and the connectors.

This post builds cable sizing from the ground up: charge, current and Ohm's law, then resistance, voltage drop, annual energy loss, ampacity, connector heating, three-phase AC runs and finally Kelvin's law for the most economical size. Every symbol is defined before use, and every number was computed with Python before it was written.

Module and inverter values are typical datasheet figures chosen for illustration, so use your own datasheets. The work is for qualified, licensed electricians. For context, read what a solar inverter does first, and keep the string sizing mathematics for Voc, Vmp and MPPT windows nearby, since the voltage you size against comes from there.

Charge, current, voltage and power

Symbols for this section:

• $Q$: electric charge, in coulombs (C)
• $t$: time, in seconds (s)
• $I$: current, in amperes (A)
• $W$: energy, in joules (J)
• $V$: potential difference, or voltage, in volts (V)
• $P$: power, in watts (W)
• $R$: resistance, in ohms (Ω)

Current is the rate at which charge crosses a section of conductor. Potential difference is the energy given up per unit of charge moved between two points.

Moving a small charge $\mathrm{d}Q$ through $V$ transfers energy $\mathrm{d}W = V\,\mathrm{d}Q$. Power is energy per unit time, so dividing by $\mathrm{d}t$ gives the most used equation in electrical engineering.

In a metal at constant temperature, current is proportional to the voltage across it, and the constant is the resistance (Ohm's law). Substituting it into the power equation gives the heat produced in a resistance.

Sanity check: a string at 600 V and 13.2 A delivers 600 × 13.2 = 7,920 W. The form $I^{2}R$ is the one this post leans on. Doubling the current quadruples the heat in a wire.

Resistance from geometry and temperature

Symbols for this section:

• $\rho$: resistivity, in Ω·mm²/m, which lets $A$ be written in mm²
• $L$: conductor length, in metres (m)
• $A$: conducting cross-section, in mm²
• $\alpha$: temperature coefficient of resistivity, per kelvin (1/K)
• $T$: conductor temperature, in °C; $R_{20}$: resistance at 20 °C, in Ω

Assume a uniform conductor at a uniform temperature, and picture identical parallel filaments of metal. Doubling the length puts two lengths in series, so resistance doubles. Doubling the area adds a parallel path, so resistance halves.

For copper at 20 °C, $\rho_{20} = 0.0172\ \Omega\,\mathrm{mm^{2}/m}$, the same as $1.72\times10^{-8}\ \Omega\,\mathrm{m}$. One metre of 6 mm² copper therefore has $R = 0.0172 \times 1 / 6 = 2.87\ \mathrm{m\Omega}$. Aluminium has $\rho_{20} = 0.0282\ \Omega\,\mathrm{mm^{2}/m}$, about 1.64 times higher.

Why a hot roof matters

Resistivity rises as the lattice vibrates harder. Over a cable's working range the rise is close to linear, good to about one percent.

For copper, $\alpha = 0.00393\ \mathrm{K^{-1}}$. Datasheets usually quote resistance at 20 °C, but a conductor under hot modules can sit at 70 to 90 °C.

A designer who uses the 20 °C value understates both drop and loss by 20% to 28%. We use 70 °C for the string examples, which is slightly conservative for a cable that is cooler for much of the day.

The loop, voltage drop and power loss

Additional symbols:

• $L$: one-way route length from array to inverter, in metres (m)
• $\rho_T$: resistivity at conductor temperature $T$, in Ω·mm²/m
• $\Delta V$: voltage lost along the cable, in volts (V); $V_s$: voltage at the source end, in volts (V)
• $\varepsilon$: voltage drop as a percentage of $V_s$
• $P_{\mathrm{loss}}$: power dissipated as heat in the cable, in watts (W)

A DC circuit is a closed loop. Current leaves on the positive conductor and returns on the negative one, so it passes through two lengths of cable. Assume both conductors share material, area and temperature.

The heat produced is the power equation applied to the same resistance.

Now the identity that makes percentages so convenient. The source delivers $P_s = V_s I$ into the series circuit. Dividing the loss by it cancels the current.

The proof needs only that the loss is in series, so one current flows through cable and source. A 1% voltage drop is a 1% power loss at that instant. Dividing by the far-end voltage instead changes the figure by a factor of 1/(1 − ε), about 1% of its value for a 1% drop. The same relation shows why higher voltage helps. At fixed power $P$, $I = P/V$, so:

Raising string voltage from 600 V to 1,500 V in the same cable cuts the percentage loss to (600/1,500)² = 0.16 of its former value, one reason large plants moved to 1.5 kV DC.

The sizing formula and a worked string

Set a maximum permitted drop $\Delta V_{\max}$ and rearrange the drop equation for area.

Here $I$ is the current that normally flows, the maximum-power current $I_{mp}$, not the short-circuit current. Choose the next standard size above $A_{\min}$. Solar cable commonly comes in 4, 6 and 10 mm², with 2.5 and 16 mm² on either side.

Worked example 1. A string has $I_{mp} = 13.2\ \mathrm{A}$ and $V_{mp} = 600\ \mathrm{V}$, with a 45 m one-way run. Take the conductor at 70 °C, so $\rho_T = 0.0172 \times 1.1965 = 0.02058\ \Omega\,\mathrm{mm^{2}/m}$, and allow 1% drop, so $\Delta V_{\max} = 6\ \mathrm{V}$.

That is just above 4 mm², so the next standard size is 6 mm². To see what 4 mm² would do, compute directly.

That is 1.02% of 600 V, marginally over target, with a loss of 13.2² × 0.463 = 80.7 W. With 6 mm² the loop resistance is 0.309 Ω, the drop 4.07 V (0.68%) and the loss 53.8 W. The 6 mm² drop would be 0.57% at 20 °C and 0.72% at 90 °C. Sanity check: 6 mm² is 1.5 times the area, and 1.02% divided by 1.5 is 0.68%.

The percentage you accept is a design choice. Many designers aim for about 1% to 2% on the DC side, but the project specification and the current edition of the applicable standard decide. The cable must also pass the thermal check described later.

Annual energy loss: the mean of the square

The percentage at peak sun is not the percentage over a year. New symbols:

• $I(t)$: string current over time, in amperes (A); $I_{pk}$: its peak, in amperes (A)
• $T_{\mathrm{day}}$: length of the generating day, in hours (h)
• $\langle \cdot \rangle$: the time average over the generating day
• $H$: generating hours in a year, in hours (h)
• $E$: energy delivered, and $E_{\mathrm{loss}}$: energy lost in the cable, in kilowatt-hours (kWh)

Assume the string voltage stays near $V_{mp}$, the resistance $R$ is constant, and the current over a day is a half sine wave.

Energy delivered follows the average current, $E = V\,\langle I \rangle\, T_{\mathrm{day}}$. Energy lost follows the average of the square, $E_{\mathrm{loss}} = R\,\langle I^{2} \rangle\, T_{\mathrm{day}}$, because instantaneous loss is $I(t)^{2}R$. Integrating the half sine wave:

Squaring the mean current therefore understates the true loss by about 19%. Any varying current gives a ratio of at least one, which is why you must average the square. Compare with the peak-hour percentage $f_{pk} = R\,I_{pk}/V$. The fraction lost over the day is:

Over a year the daily peak is itself well below the standard-test $I_{mp}$, because of irradiance, temperature and soiling. Use energy matching. Assume 1,500 kWh per kWp per year, an approximation within the usual range for this region, and a 12 h day, so $H = 4{,}380\ \mathrm{h}$. The 7.92 kWp string produces 11,880 kWh. Matching $E = V\,(2/\pi)\,I_{pk}\,H$ gives an equivalent peak current:

That is 0.54 of $I_{mp}$, so the annual loss fraction is $0.54 \times 0.785 = 0.42$ times the peak-hour percentage. Day-to-day variation raises the mean square, so treat 0.42 as central and expect about 0.4 to 0.6 in practice.

Worked example 1, continued. With $I_{pk,\mathrm{eff}}^{2}/2 = 25.2\ \mathrm{A^{2}}$, the 4 mm² loop of 0.463 Ω loses 0.463 × 25.2 × 4,380 / 1,000 = 51.1 kWh a year. The table gathers the results.

The annual percentage is 0.42 of the peak percentage in every row, as it must be, since both scale with $R$. Going from 4 to 6 mm² saves 17 kWh a year, about ₹100 at an illustrative ₹6 per kWh. That is small, but it never stops.

Ampacity, derating and design current

Voltage drop protects your yield. Ampacity protects the insulation. Symbols for the heat balance:

• $R'$: resistance per metre of one conductor, in Ω/m
• $h$: effective heat transfer coefficient from the cable surface to its surroundings, in W/(m²·K), lumping insulation, convection and radiation
• $p$: cable perimeter, in metres (m)
• $T_c$: conductor temperature, and $T_a$: ambient temperature, both in °C
• $I_{\max}$: the steady current that holds the conductor at $T_c$, in amperes (A)

In steady state, heat generated per metre equals heat lost per metre. Generation is $I^{2}R'$. Loss is proportional to surface area per metre, $p$, and the temperature difference.

For a round conductor of area $A$, the diameter is $d = \sqrt{4A/\pi}$, so $p = \pi d = 2\sqrt{\pi A}$, and $R' = \rho/A$. For this derivation use SI units, with $A$ in m² and $\rho$ in Ω·m. Substituting:

Current capacity grows more slowly than area, because heat is made throughout the volume (proportional to $A$) but escapes through the surface (proportional to $\sqrt{A}$). If a 4 mm² cable has some rating in an installation, an otherwise identical 6 mm² is rated near (6/4)^0.75 = 1.36 times that, not 1.5 times, and 10 mm² near 1.99 times.

Why derating factors exist

Each derating factor changes $h$ or $T_c - T_a$. With the insulation limit $T_{\max}$, a reference ambient $T_{ref}$ and an actual ambient $T_a$, the heat balance gives:

Illustration: with $T_{\max} = 90\ ^{\circ}\mathrm{C}$ and a 30 °C reference, a 60 °C ambient gives $k_T = \sqrt{30/60} = 0.71$. Bundling cables or running them in conduit lowers $h$, because neighbours heat each other and trapped air resists heat flow. If $h$ halves, the rating falls to $\sqrt{0.5} = 0.71$. This is first-order physics. The factors in standards are measured and tabulated, and differ in detail.

Design current: why not just Imp?

For drop and loss we used $I_{mp}$, the current that normally flows. For ampacity we cover the worst credible current. The short-circuit current $I_{sc}$ is the most a healthy module delivers. It rises with temperature and with irradiance, and brief irradiance above 1,000 W/m² can occur. Common practice sizes string conductors on a multiple of $I_{sc}$, most often 1.25 times, with further margins for irradiance and temperature where the code asks.

With a typical $I_{sc} = 14.0\ \mathrm{A}$, $I_{\mathrm{design}} = 17.5\ \mathrm{A}$. The cable's rating, after derating for ambient temperature and grouping, must be at least this. Some practices apply a second 1.25, giving 21.9 A, so check the current edition of the standard for your project. Protection around the cable is covered in earthing and surge protection for solar plants.

Connectors and contact resistance

Each MC4-type connector adds a small resistance where two contacts press together. With $R_c$ the contact resistance in ohms and $P_c$ the heat there in watts, the rule is the same as for the cable.

Illustrative example. At 13.2 A, a healthy 0.5 mΩ crimp dissipates $13.2^{2} \times 0.0005 = 0.087\ \mathrm{W}$. A poor 5 mΩ crimp dissipates $13.2^{2} \times 0.005 = 0.87\ \mathrm{W}$, ten times as much. Beside the 80 W lost along a 4 mm² cable that looks small, but the cable spreads its heat over 90 m of conductor while the connector concentrates it in a few cubic centimetres of plastic. With an illustrative thermal resistance of 40 K/W to air, the healthy contact warms by about 3.5 K and the bad one by about 35 K, on a roof that is already hot.

The danger is feedback. A hot, poor contact oxidises, its resistance rises and so does the heat. At 50 mΩ the same current would dissipate 8.7 W. A DC arc also has no natural zero crossing, so once struck it does not stop by itself.

Why you do not mix connector brands

A male and female pair is designed as a system. Contact spring force, plating, barrel dimensions and seal tolerances are matched, and the maker's current rating, ingress protection and contact resistance apply only to that pair. Connectors of different makes may click together yet mate loosely, with higher $R_c$ and a weaker seal. The crimp tool must also match the contact and conductor size. Use one maker's connectors and its specified tool throughout.

The AC side: three-phase drop, copper and aluminium

After the inverter, reactance joins resistance. Symbols:

• $P$: real power, in watts (W); $V_{LL}$: line-to-line voltage, 415 V; $V_{ph} = V_{LL}/\sqrt{3}$: phase voltage
• $I$: line current, in amperes (A)
• $\cos\varphi$: power factor, with $\varphi$ the angle between voltage and current
• $R'$, $X'$: resistance and reactance per metre of one conductor, in Ω/m
• $L$: one-way cable length, in metres (m)

For a balanced three-phase load, the line current is:

Take one phase, with the receiving-end voltage as the reference direction. The current lags by $\varphi$, so $\mathbf{I} = I(\cos\varphi - j\sin\varphi)$. The cable impedance is $R + jX$, with $R = R'L$ and $X = X'L$. The phase drop is:

The first term lies along the voltage and changes its size. The second is at right angles and mainly turns the phasor. A perpendicular part $\delta_{\perp}$ changes the magnitude by only about $\delta_{\perp}^{2}/2V_{ph}$, which is negligible, so the phase drop is approximately $I L (R'\cos\varphi + X'\sin\varphi)$.

The $\sqrt{3}$ comes from the geometry. A line-to-line voltage is the difference of two phase voltages whose phasors are 120° apart: $2V_{ph}\sin 60^{\circ} = \sqrt{3}\,V_{ph}$. In a balanced system every phase has the same drop relative to its own voltage, so the whole triangle scales together and the line-to-line drop is $\sqrt{3}$ times the phase drop.

In DC or single-phase the go and return paths give a factor of 2. Balanced three-phase has no neutral current, so the factor is $\sqrt{3} = 1.73$.

Worked example 2. A 50 kW inverter output at 415 V and $\cos\varphi = 0.95$ feeds a panel 60 m away. Many inverters run near unity, where the $X'$ term vanishes, but 0.95 shows both terms. The current is $I = 50{,}000 / (\sqrt{3} \times 415 \times 0.95) = 73.2\ \mathrm{A}$. Take a conductor at 90 °C, which is conservative, and an illustrative multicore $X' = 0.08\ \mathrm{m\Omega/m}$. For 25 mm² copper, $R' = 0.0172 \times 1.2751 / 25 = 0.877\ \mathrm{m\Omega/m}$.

That is 1.57% of 415 V. The loss is $3I^{2}R'L = 3 \times 73.2^{2} \times 0.877\times10^{-3} \times 60 = 847\ \mathrm{W}$, or 1.69% of 50 kW. At unity power factor the current is 69.6 A, and the same cable gives 6.34 V (1.53%) and 764 W (1.53%), so the earlier identity holds exactly when only resistance acts.

The table assumes $X' = 0.08\ \mathrm{m\Omega/m}$ for every row and an aluminium coefficient of about 0.0040 per kelvin. Each option still needs its ampacity checked against the datasheet.

Copper versus aluminium

At 90 °C aluminium resistance is about 1.65 times that of copper of equal area, so 25 mm² of copper corresponds to about 41 mm² of aluminium, and the next standard size is 50 mm². Even so, the aluminium conductor weighs about half as much, since the densities are 2.70 and 8.96 g/cm³, which is why it is common on larger AC feeders. It needs compatible lugs, joint compound and torque checks, because it creeps under pressure. Rooftop DC string cable is normally copper, so follow the cable standard you work to.

Economic optimisation: Kelvin's law

A thicker cable costs more upfront but wastes less, so there is an optimum. Symbols:

• $c$: conductor cost per mm² per metre, in ₹/(mm²·m); $a = 2Lc$: capital cost per mm² for the whole run, in ₹/mm²
• $b$: present value of lost energy multiplied by area, in ₹·mm²
• $C(A)$: total cost over the cable's life, in ₹
• $A^{*}$: the cross-section that minimises $C$, in mm²

Assume capital cost is proportional to area, and the value of lost energy is proportional to $R$, hence to $1/A$.

Setting the slope to zero finds the minimum. The second derivative, $2b/A^{3}$, is positive, so it is a minimum.

At the optimum the two cost terms are equal. Fixed costs per metre, such as the jacket, add a constant to $C$ and do not move the minimum. Both $a$ and $b$ are proportional to $L$, so the length cancels: the optimum depends on current, energy value and cable price, not distance.

Worked example (illustrative assumptions, not quotes). Take the 45 m string, copper at $c = ₹12$ per mm² per conductor-metre, energy worth ₹6 per kWh, a 25-year life at an 8% discount rate. The present value of a yearly cost is the annuity factor:

One metre of 1 mm² conductor at 70 °C would lose $0.02058 \times 25.2 \times 4{,}380 / 1{,}000 = 2.27\ \mathrm{kWh}$ a year. A loop of length $2L$ and area $A$ loses that figure multiplied by $2L/A$, so $b = 90 \times 2.27 \times 6 \times 10.67 \approx 13{,}100\ ₹\cdot\mathrm{mm^{2}}$, and $a = 90 \times 12 = ₹1{,}080$ per mm².

At $A^{*}$ the total is ₹7,523, with both terms at ₹3,761. The curve is flat, and 4 mm² is within about 1% of the minimum. The optimum scales with the square root of the energy price: at ₹12 per kWh it moves to 4.9 mm², and about ₹18 per kWh would be needed to favour 6 mm². Substitute your own cable quotes, tariff and life.

Kelvin's law ignores the drop limit. In Example 1 the 1% limit pushed the answer to 6 mm² although the economic minimum is near 4 mm². The constraint sets the floor, and the economics tells you whether going higher pays.

Ray2Volt can run this analysis for your own site, using your layout, cable lengths and datasheets. Start with a free power audit, or see our approach to commercial and industrial solar. The I-V curve behind $I_{mp}$ and $V_{mp}$ is explained in how a solar module works.

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