# What is Voltage Drop? Advanced Voltage Drop Calculator with Solved Examples

**Advanced Voltage Drop Calculator with Solved Examples & Formulas in IEC and IEC**

### What is Voltage Drop?

**Voltage drop** refers to the reduction in voltage as electric current flows through a conductor, such as a wire, resistor, or other components in a circuit. This happens due to the inherent resistance or impedance of the conductor, which causes some of the electrical energy to be lost as heat. The greater the distance and the smaller the conductor size, the higher the voltage drop will be.

In practical terms, voltage drop can affect the performance of electrical systems. If too much voltage is lost along the way, devices may not function correctly because they aren’t receiving the required voltage.

**Key Factors Affecting Voltage Drop:**

**Conductor length**– Longer conductors lead to higher voltage drops.**Conductor size (gauge)**– Smaller diameter wires have higher resistance and cause more voltage drop.**Current flow (amperage)**– Higher current increases the voltage drop.**Conductor material**– Materials like copper have lower resistance, while others like aluminum have higher resistance.**Temperature**– Higher temperatures can increase resistance in a conductor, leading to a larger voltage drop.

**How to Minimize Voltage Drop in Cables?**

Managing voltage drop is essential in electrical installations to ensure that electrical devices receive adequate power and avoid efficiency losses.

**Use a larger gauge cable**to reduce resistance. If the distance is larger than 50ft, the NEC suggests to add 20% more ampacity for each 100ft of distance to compensate the voltage drop.**Shorten the cable length**where possible.**Use higher-conductivity materials**such as copper instead of aluminum.**Reduce the load**or distribute the load among multiple circuits to lower the current.

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**What is the Allowable Voltage Drop?**

**Voltage Drop Value in NEC**

According to the **NEC** (National Electrical Code) **[210.19(A)(1)**, FPN No. 4] and **[215.2(A)(3)**, FPN No. 2], the **allowable voltage drop for feeders is** **3%**, and the **acceptable voltage drop for final sub-circuits and branch circuits is 5%** to ensure proper and efficient operation.

For example, if the Supply voltage is **120V**, then the value of allowable voltage drop should be;

Allowable Voltage Drop = 120V x (3/100) =** 3.6V**.

For long-distance runs over 50 feet (15.25 meters), consider upgrading to a larger gauge wire to compensate for voltage drop. According to **NEC Table 310-16**, for every 100 feet (30.50 meters) of wire length, add 20% ampacity to counter the voltage drop.

**Voltage Drop Value in IET and IEC**

According to **BS 7671 – TABLE 4Ab** and **IEC60364-5-52**, **article 525**, **table G.52.1**, the maximum **allowable voltage drop for lighting circuits is 3%**. The limit of maximum voltage drop for other heating and power appliances supplied by public low voltage distribution system is **5%**.

For private Low Voltage (LV) supply systems, the permissible voltage drop for general lighting circuits is 6% and 8% for other HVAC appliances.

**Voltage Drop Value in IEEE**

The **IEEE B-23** rule specifies that **voltage drop should not exceed 2.5% of the supply voltage**.

**Voltage Drop Value in AS/NZS 3008**

According to **AS3008**, the standard allowable voltage drop from the supply to any point in the circuit should not exceed **5%**. However, an exception is made when a low-voltage substation is installed on the premises and used as a dedicated circuit. In this case, the allowable voltage drop can increase to **7%**, as per AS3000.

- The NEC recommends a maximum voltage drop of
**3%**for branch circuits. - The NEC suggests the overall voltage drop from the service panel to the final load should not exceed
**5%**to ensure efficient operation. - In IET and IEC, the maximum allowable voltage drop is
**3%**and**5%**for lighting and inductive loads respectively. - In IEEE, the value of permissible voltage drop is
**2.5%**. - AS/NZS 3008 allow a voltage drop up to
**5%**and should no more than 7% in dedicated installation networks.

We have already discussed the selection of proper size of cable for electrical wiring installation in SI and British system with examples. In the above article, we have also explained the voltage drop calculation and voltage drop formula as well as online cable size calculator.

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**Voltage Drop Formulas**

#### Voltage Drop in DC Circuits:

The voltage drop across a conductor in single phase AC circuits and DC circuits can be calculated using Ohm’s law:

*Voltage Drop (V _{d}) = I × R*

Where:

*V*= is the voltage drop in volts._{d}*I*= is the current (in amperes),*R =*is the resistance of the conductor (in ohms).

In more complex AC circuits, the impedance “Z” (including resistance “R” and reactance “X” (either inductive reactance “X_{L}” or capacitive reactance “X_{C}“) needs to be considered.

**Voltage Drop in Single-Phase AC System**

The voltage drop for a single-phase AC system can be calculated using the following formula:

*Voltage Drop (V _{d}) = (2 × I × L × Z) ÷ 1000*

Where:

*V*= is the voltage drop in volts._{d}*I*= Current in amperes,*L*= Length of the cable in meters,*Z*= Impedance of the cable (ohms per kilometer).

Voltage drop in a single-phase system when power factor is involved

*V _{d} = IL [2 (R_{C} Cosθ + X_{C} Sinθ)]*

**Good to know:**

- Z is the overall resistance of AC circuit.
- In pure resistive circuit, impedance “Z” = resistance “R”.
- In capacitive circuit, Z = √(R
^{2}+ X_{C}^{2}) while in inductive circuit, Z = √(R^{2}+ X_{L}^{2}) - Z = √(R
^{2}+ X_{C}^{2}) where R is reactance and X_{C}and X_{L}are capacitive reactance and inductive reactance respectably. - There is no impedance in DC circuits due to the absence of frequency and phase angle difference.
- When wire length is in meters, put R or Z = wire resistance in [Ω/km].
- When Wire length is in feet, put R or Z = wire resistance in [Ω/kft].

**Voltage Drop in Three-Phase AC System**

*Voltage Drop (V _{d}) = (√3 × I × L × Z) ÷ 1000*

*V _{d }= *(

*1.732 × I × L× R*)

*/ 1000*

The voltage drop calculation formulas is applicable for both 3-Phasem 3-Wire system i.e. delta connection) and 3-Phase. 4-Wire systems i.e. Star Connection

Where;

*V*= is the voltage drop in volts._{d}- $√3 = 1.732$is the square root of 3 for three-phase power.
*I, L,*and*R/Z*remain the same as in the single-phase equation.

Voltage drop in a balanced three-phase system when power factor is involved

*V _{d} = IL [ √3 (R_{C} Cosθ + X_{C} Sinθ)]*

Voltage drop in an unbalanced three-phase system

*Vd = IA LA Z _{C} A + IN LN Z_{C} N*

**Voltage Drop Formula for Steel Conduit**

This is the approximate voltage drop formula at unity power factor, cabal temperature 75˚C, and cable conductors in steel conduit.

*V _{D} = (2 × k × Q × I × D) / CMA ….. for 1-Phase.*

*V _{D} = (1.732 × k × Q × I × D) / CMA ….. for 3-Phase.*

Where;

*CMA*= The cross sectional area of the conductor in circular mils.*D*= The one way distance in feet.*I*= The circuit current in amperes.*Q*= The ratio between AC resistance and DC resistance (R_{AC}/R/_{DC}) for conductor larger than 2/0 for skin effect.*k*= Specific resistivity = 21.2 for aluminum and 12.9 for copper.

**Voltage Drop at the End of the Cable**

V_{End} = V – V_{d}

Where;

- V
_{End}= Supply voltage at the end of the cable. - V = Supply voltage.
- V
_{d}= Voltage Drop in the cable conductors.

**Voltage Drop Formula for Circular Mils**

V_{d} = *ρ* P L I / A

Where;

- V
_{d}= Voltage drop in volts**.** *ρ*= rho = specific resistivity in (*Ω – circular mils/foot*).- P = Phase Constant = 2 (for single phase and DC system) and = √3 = 1.732 (for three phase system)
- L = wire length in feet.
- A = wire area in circular mils.

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**Voltage Drop Calculations**

Today, we are introducing the most **Advanced Online Voltage Drop Calculator** along with detailed voltage drop formulas and solved examples for both NEC and IEC voltage systems. Calculating and measuring voltage drop is important especially when sizing the wire and cable and sizing a circuit breaker for wiring installation. This practice makes sure the circuit is protected and working efficiently for long run.

**How to Calculate Voltage Drop for 1-Phase and 3-Phase? – Solved Examples**

**Examples Based on NEC**

**Example 1: 120V – Single Phase – NEC**

Calculate the approximate voltage drop in a #14 AWG copper conductor having length of 33ft (10 m) from the 120V panel to the final 12A load point.

**Solution:**

We will use the following formula to determine the voltage drop in 120V, single phase AC circuit.

V_{d} = (2 *× *K *×* I *×* L) ÷ CMA

Where:

- V
_{d}= Voltage drop in volts. - 2 = There are two conductors in a single phase supply and voltage drop occurs in both.
- K = Approximate resistance in Ω per circular-mil foot at 75°C.
- For uncoated copper wire, use 12Ω for K.
- For aluminum wire, use 20Ω for K.

- I = Load current in amperes flowing through the conductor.
- L = Length in feet from beginning of circuit to the final load point.
- CMA = cross-sectional area of the conductors in circular mils. Use the table below based on Table 8, Chapter 9, NEC.

Substituting the values in the formula:

V_{d} = (2 *× *12 *×* 12A *×* 33ft) ÷ 4110

V_{d} = 2.31V

This indicates the 2.31V of voltage drop is less than the 3% of allowable voltage drop which is 120V x 3% = 3.6V. Hence, the wiring system is efficient and according to the code.

**Example 2: 240V – Single Phase – NEC**

Calculate the voltage drop for 6 AWG Type TW copper or 8 AWG Type THHN copper which is used for the a 240-volt, air-conditioner circuit having Minimum Circuit Ampacity 40 Amperes. There is no-neutral wire and the AC unit is located 50ft (15.24 m) away from the panel.

**Solution:**

We will be using the same formula used in the above example 1 for single phase AC circuits

V_{d} = (2 *× *K *×* I *×* L) ÷ CMA

We will use #8AWG Type THHN copper for this circuit based on NEC Table 310.15(B)(16).

Putting the values

V_{d} = (2 *× *12 *×* 40A *×* 50ft) ÷ 16,510

V_{d} = 2.9V

The measured voltage drop is well below than the recommended value of voltage drop which is 240V x 3% = 7.2V. Hence, the wiring is acceptable and according to the codes.

**Voltage Drop Calculation based on Table and Factors**

The voltage drop in copper conductors can be easily calculated using the simple formula and voltage drop factor provided below, along with the table for reference.

V_{d}* = F × *I *×* *l*

Where:

*F*= Factor from the table below.- I = Current in amperes.
*l*= Conductor length in 100 feet).

(Refer to the solved example below the table for a clearer understanding.)

**Example – 3: 1/3 – Phase, NEC**:

Suppose the voltage is 120V (single phase), the current is 15A, the conductor length is 45, and the wire gauge is #12 AWG. Calculate the voltage drop.

**Solution**:

The voltage drop can be calculated using the following formula:

V_{d}* = F × *I *×* *l*

From the table, the factor for a #12 AWG copper conductor is 0.313. Now, substituting the values into the formula:

$Vd=0.313× 15×()$

$Vd=2.11V$

The calculated voltage drop is less than the allowable voltage drop which is 120V x 3% = 3.6V. Hence, the wiring system is acceptable and in accordance with codes.

**Example 4: 480V – Three Phase – NEC**

Measure the voltage drop for #12AWG, 10 stranded, uncoated copper wire in a steel conduit used for 480V branch circuits. The distance from the load center to the load point is 200ft (60.96 m).

Solution:

We will use the same formula as for single-phase circuits, but for three-phase circuits, we will use √3 (1.732) instead of 2. The rest of the equation remains the same, as shown below:

V_{d} = (√3 *× *K *×* I *×* L) ÷ CMA

V_{d} = (1.732 *× *12 *×* 20A *×* 200ft) ÷ 6530

V_{d} = 12.73V

This value of calculated voltage drop is acceptable for our 480V circuit where the allowable voltage drop is = 480V x 3% = 14.4V. Hence, the wiring system is acceptable and comply with the codes.

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**Examples Based on IET & IEC**

**Example 1: – 230V – Single Phase**

Determine the voltage drop for 15A, 230V – single phase AC circuit fed by 1.5mm^{2} cable laying 20 meters far away from the consumer unit. Take resistance of copper conductor = 9.5 Ω/km

*Voltage Drop (V _{d}) = (2 × I × L × R) ÷ 1000*

Putting the values

*V _{d }= *(

*2 × 15A × 20m× 9.5Ω*)

*/ 1000*

*V _{d }= 5.7V*

The calculated 5.7V of voltage drop is less than the allowable voltage drop which is 230V × 3% = 6.9V. Hence, the wiring system is perfect.

**Example 2: 230V – 1-Phase**

What is the voltage drop at the end of a 15-meter long lighting circuit if the current flowing through a 2.5mm^{2} CSA copper conductor is 20A?

**Solution**:

Voltage drop in a cable is then given by:

V_{d} = K *×* I_{FLA} *×* L

- V
_{d }= Voltage drop in the cable - K = is given by the table (see below)
- I
_{FLA}= is the full-load current in amps - L = is the length of cable in km

Since the value of K for 2.5mm^{2} is 19 for lighting circuit in the following voltage drop table;

Substituting the values

V_{d} = 19 *×* 20A *×* (15/1000)

V_{d} = 5.7V

This value of calculated voltage drop is satisfactory, being less than the maximum permitted voltage drop of 3% which is 3% × 230V = 6.9V.

**Example 3: 415V – Three Phase**

Calculate the voltage drop in a 35mm^{2} copper conductor carrying 100A in a three-phase, 415V supply system. The cable length is 50 meters from the distribution board to the load. Assume a resistance of 0.55 ohms per kilometer for the 35mm^{2} cable.

Voltage drop in a cable for three phase system is given by:

*Voltage Drop (V _{d}) = (√3 × I × L × Z) ÷ 1000*

Substituting the values

*V _{d }= *(

*1.732 × 100A × 50m× 0.55*)

*/ 1000*

*V _{d }= 4.76V*

As the calculated value of voltage drop is 4.76V is less than the permissible voltage drop which is 415V × 3% = 12.45. Therefore, the wiring system is perfect and comply the codes.

**Example 4: 400V – Three Phase**

What is the approximate voltage drop in a 3-phase, 4-wire copper line with a cross-sectional area (CSA) of 50mm^{2}, carrying 130A of current, over a distance of 60 meters from the 400V main distribution board to the final circuit?

**Solution**:

Voltage drop in a cable is then given by same formula used in example 2 (based on table):

V_{d} = K *×* I_{FLA} *×* L

In the voltage drop table given above for example 2, the value of K for 50mm^{2} is 0.86 for lighting circuit in a a Three-phase balanced circuit.

Putting the values:

V_{d} = 0.86 *×* 130A *×* (60/1000)

V_{d} = 6.7V … Phase to Phase

This value of calculated voltage drop (6.7V) is satisfactory and less than the allowable voltage drop which is 400V × 3% = 12V.

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**Advanced Voltage Drop Calculator**

Note: This Calculator is also available in our Free Electrical Technology Android App

**Enter value and click on calculate. Result will be displayed**

*More Related Electrical & Electronics Engineering Calculators*

**P.S:**The voltage drop calculator above provides approximate values, and we cannot guarantee 100% accuracy. Results may vary depending on real-world factors such as different cables, conductors, wire materials, resistivity, the number of strands, temperature, weather conditions, conduit, and PVC. If unsure, use the manual methods or consult a licensed electrician for precise calculations.

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Really great detailed post, thanks for that!

thank you very muh

Update any new formula please informed me on my e – mail Id

thank you very much

The voltage drop formula must include the reactance impedance as well. Too many design software programs omit voltage drop and other safety code requirements, resulting in design deficiencies and fiascos like the Apollo 13 disaster. Contractors make multiple profits over eir original bid with change orders.

In your great calculator program, do I enter the cable length (the actual distance) or the total wire length (there and back)?

Can you please provide a reference from where are the formulas developed ??

bus duct system manufacturer providing the following characteristics in th catalogue

rating(amp) conductor(mm) impedence

2500 6×230 R 1.97 x 0.74 z 2.10

units in 10^-3

calculate the voltage drop of 2500A busbar used to supply 8floors.Each floor having 135kw load and total length of bus bar is 325 feet………….plz solve this question

It appears that the TEMPERATURE option is not taking effect on the RESISTANCE output value.

For example, setting input as 20C, Copper, 16 AWG, 150 ft, DC, 12V, 3.5A calculates to 4.2062 V, 35.0520 %, 1.2018 ohms, 7.7938 Ve. However, changing the temperature setting to 65C and pressing Calculate now shows 4.9634 V, 41.3614 %, 1.2018 ohms, 7.0366 Ve. Shouldn’t the resistance have changed as well as the other values?

Can anyone confirm? Can the calculator be fixed if so? (Or am I wrong?) Thank you!

Thanks a lot. This was very beneficial as I was going to use wrong wire gauge in my landscape project. I have not never had the though that 12v circuit will run 10 times AMP current that would be run in 110v circuit.

Great Post.

I would like to know what is the reason behind different voltage drop for 3phase 3 wire system and 3 phase 4 wire system ?

ENTES DIN rail mount timer Operating voltage: 230 V AC DTR-20 2 change-overs 16 A 250 V AC Astronomical

DTR-20 astronomic timer. This astronomic timer relay have it’s LED screen program setting not displayed in English language. Kindly assist with programming manual for timer relay

Super sir