Load Current using Line Losses (1-Phase 2-Wire US) Calculator

✖Line Losses is defined as the total losses occurring in an Underground AC line when in use.ⓘ Line Losses [P_loss]			+10% -10%
✖Area of Underground AC Wire is defined as the cross-sectional area of the wire of an AC supply system.ⓘ Area of Underground AC Wire [A]			+10% -10%
✖Resistivity is the measure of how strongly a material opposes the flow of current through them.ⓘ Resistivity [ρ]			+10% -10%
✖Length of Underground AC Wire is the total length of the wire from one end to other end.ⓘ Length of Underground AC Wire [L]			+10% -10%

✖Current Underground AC is defined as the current flowing through the overhead ac supply wire.ⓘ Load Current using Line Losses (1-Phase 2-Wire US) [I]

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Formula

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Load Current using Line Losses (1-Phase 2-Wire US)

Formula

`"I" = sqrt("P"_{"loss"}*"A"/(2*"ρ"*"L"))`

Example

`"64.71658A"=sqrt("2.67W"*"1.28m²"/(2*"0.000017Ω*m"*"24m"))`

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Load Current using Line Losses (1-Phase 2-Wire US) Solution

STEP 0: Pre-Calculation Summary

Formula Used

Current Underground AC = sqrt(Line Losses*Area of Underground AC Wire/(2*Resistivity*Length of Underground AC Wire))
I = sqrt(P_loss*A/(2*ρ*L))
This formula uses 1 Functions, 5 Variables

Functions Used

sqrt - A square root function is a function that takes a non-negative number as an input and returns the square root of the given input number., sqrt(Number)

Variables Used

Current Underground AC - (Measured in Ampere) - Current Underground AC is defined as the current flowing through the overhead ac supply wire.
Line Losses - (Measured in Watt) - Line Losses is defined as the total losses occurring in an Underground AC line when in use.
Area of Underground AC Wire - (Measured in Square Meter) - Area of Underground AC Wire is defined as the cross-sectional area of the wire of an AC supply system.
Resistivity - (Measured in Ohm Meter) - Resistivity is the measure of how strongly a material opposes the flow of current through them.
Length of Underground AC Wire - (Measured in Meter) - Length of Underground AC Wire is the total length of the wire from one end to other end.

STEP 1: Convert Input(s) to Base Unit

Line Losses: 2.67 Watt --> 2.67 Watt No Conversion Required
Area of Underground AC Wire: 1.28 Square Meter --> 1.28 Square Meter No Conversion Required
Resistivity: 1.7E-05 Ohm Meter --> 1.7E-05 Ohm Meter No Conversion Required
Length of Underground AC Wire: 24 Meter --> 24 Meter No Conversion Required

STEP 2: Evaluate Formula

Substituting Input Values in Formula

I = sqrt(P_loss*A/(2*ρ*L)) --> sqrt(2.67*1.28/(2*1.7E-05*24))

Evaluating ... ...

I = 64.7165766563533

STEP 3: Convert Result to Output's Unit

64.7165766563533 Ampere --> No Conversion Required

FINAL ANSWER

64.7165766563533 ≈ 64.71658 Ampere <-- Current Underground AC

(Calculation completed in 00.004 seconds)

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< 17 Current & Voltage Calculators

Maximum Voltage using Area of X-Section (1-Phase 2-Wire US)

Go Maximum Voltage Underground AC = sqrt((4*Length of Underground AC Wire*Resistivity*(Power Transmitted^2))/(Area of Underground AC Wire*Line Losses*(cos(Phase Difference))^2))

RMS Voltage using Area of X-Section (1-Phase 2-Wire US)

Go Root Mean Square Voltage = sqrt((2*Length of Underground AC Wire*Resistivity*(Power Transmitted^2))/(Area of Underground AC Wire*Line Losses*((cos(Phase Difference))^2)))

Maximum Voltage using Volume of Conductor Material (1-Phase 2-Wire US)

Go Maximum Voltage Underground AC = sqrt(8*Resistivity*(Power Transmitted*Length of Underground AC Wire)^2/(Line Losses*Volume Of Conductor*(cos(Phase Difference))^2))

Maximum Voltage using Line Losses (1-Phase 2-Wire US)

Go Maximum Voltage Underground AC = 2*Power Transmitted*sqrt(Resistivity*Length of Underground AC Wire/(Area of Underground AC Wire*Line Losses))/cos(Phase Difference)

RMS Voltage using Line Losses (1-Phase 2-Wire US)

Go Root Mean Square Voltage = 2*Power Transmitted*sqrt(2*Resistivity*Length of Underground AC Wire/(Area of Underground AC Wire*Line Losses))/cos(Phase Difference)

RMS Voltage using Volume of Conductor Material (1-Phase 2-Wire US)

Go Root Mean Square Voltage = sqrt(4*Resistivity*(Power Transmitted*Length of Underground AC Wire)^2/(Line Losses*(cos(Phase Difference))^2*Volume Of Conductor))

Load Current using Constant (1-Phase 2-Wire US)

Go Current Underground AC = sqrt(Constant Underground AC*Line Losses/(2*Resistivity*(Length of Underground AC Wire*cos(Phase Difference))^2))

Maximum Voltage using Constant (1-Phase 2-Wire US)

Go Maximum Voltage Underground AC = sqrt(4*Resistivity*(Power Transmitted*Length of Underground AC Wire)^2/(Constant Underground AC*Line Losses))

RMS Voltage using Constant (1-Phase 2-Wire US)

Go Root Mean Square Voltage = 2*Power Transmitted*Length of Underground AC Wire*sqrt(2*Resistivity/(Line Losses*Constant Underground AC))

RMS Voltage using Resistance (1-Phase 2-Wire US)

Go Maximum Voltage Underground AC = 2*Power Transmitted*sqrt(2*Resistance Underground AC/Line Losses)/cos(Phase Difference)

Maximum Voltage using Resistance (1-Phase 2-Wire US)

Go Maximum Voltage Underground AC = 2*Power Transmitted*sqrt(Resistance Underground AC/Line Losses)/cos(Phase Difference)

Load Current using Line Losses (1-Phase 2-Wire US)

Go Current Underground AC = sqrt(Line Losses*Area of Underground AC Wire/(2*Resistivity*Length of Underground AC Wire))

Maximum Voltage using Load Current (1-Phase 2-Wire US)

Go Maximum Voltage Underground AC = (sqrt(2))*Power Transmitted/(Current Underground AC*(cos(Phase Difference)))

Load Current (1-Phase 2-Wire US)

Go Current Underground AC = Power Transmitted*sqrt(2)/(Maximum Voltage Underground AC*cos(Phase Difference))

RMS Voltage using Load Current (1-Phase 2-Wire US)

Go Root Mean Square Voltage = Power Transmitted/(Current Underground AC*cos(Phase Difference))

Load Current using Resistance (1-Phase 2-Wire US)

Go Current Underground AC = sqrt(Line Losses/(2*Resistance Underground AC))

RMS Voltage(1-Phase 2-Wire US)

Go Root Mean Square Voltage = Maximum Voltage Underground AC/sqrt(2)

Load Current using Line Losses (1-Phase 2-Wire US) Formula

Current Underground AC = sqrt(Line Losses*Area of Underground AC Wire/(2*Resistivity*Length of Underground AC Wire))
I = sqrt(P_loss*A/(2*ρ*L))

What is the value of maximum voltage and volume of conductor material in 1-phase 2-wire system?

The volume of conductor material required in this system is 2/cos²θ times that of 2-wire d.c.system with the one conductor earthed. The maximum voltage between conductors is v_m so that r.m.s. value of voltage between them is v_m/√2.

How to Calculate Load Current using Line Losses (1-Phase 2-Wire US)?

Load Current using Line Losses (1-Phase 2-Wire US) calculator uses Current Underground AC = sqrt(Line Losses*Area of Underground AC Wire/(2*Resistivity*Length of Underground AC Wire)) to calculate the Current Underground AC, The Load Current using Line Losses (1-Phase 2-Wire US) formula is defined as the current that flows into the load of the one-phase two-wire underground system. Current Underground AC is denoted by I symbol.

How to calculate Load Current using Line Losses (1-Phase 2-Wire US) using this online calculator? To use this online calculator for Load Current using Line Losses (1-Phase 2-Wire US), enter Line Losses (P_loss), Area of Underground AC Wire (A), Resistivity (ρ) & Length of Underground AC Wire (L) and hit the calculate button. Here is how the Load Current using Line Losses (1-Phase 2-Wire US) calculation can be explained with given input values -> 64.71658 = sqrt(2.67*1.28/(2*1.7E-05*24)).

FAQ

What is Load Current using Line Losses (1-Phase 2-Wire US)?

The Load Current using Line Losses (1-Phase 2-Wire US) formula is defined as the current that flows into the load of the one-phase two-wire underground system and is represented as I = sqrt(P_loss*A/(2*ρ*L)) or Current Underground AC = sqrt(Line Losses*Area of Underground AC Wire/(2*Resistivity*Length of Underground AC Wire)). Line Losses is defined as the total losses occurring in an Underground AC line when in use, Area of Underground AC Wire is defined as the cross-sectional area of the wire of an AC supply system, Resistivity is the measure of how strongly a material opposes the flow of current through them & Length of Underground AC Wire is the total length of the wire from one end to other end.

How to calculate Load Current using Line Losses (1-Phase 2-Wire US)?

The Load Current using Line Losses (1-Phase 2-Wire US) formula is defined as the current that flows into the load of the one-phase two-wire underground system is calculated using Current Underground AC = sqrt(Line Losses*Area of Underground AC Wire/(2*Resistivity*Length of Underground AC Wire)). To calculate Load Current using Line Losses (1-Phase 2-Wire US), you need Line Losses (P_loss), Area of Underground AC Wire (A), Resistivity (ρ) & Length of Underground AC Wire (L). With our tool, you need to enter the respective value for Line Losses, Area of Underground AC Wire, Resistivity & Length of Underground AC Wire and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.

How many ways are there to calculate Current Underground AC?

In this formula, Current Underground AC uses Line Losses, Area of Underground AC Wire, Resistivity & Length of Underground AC Wire. We can use 3 other way(s) to calculate the same, which is/are as follows -

Current Underground AC = Power Transmitted*sqrt(2)/(Maximum Voltage Underground AC*cos(Phase Difference))
Current Underground AC = sqrt(Line Losses/(2*Resistance Underground AC))
Current Underground AC = sqrt(Constant Underground AC*Line Losses/(2*Resistivity*(Length of Underground AC Wire*cos(Phase Difference))^2))

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