Resistivity using Load Current (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%
✖Current Underground AC is defined as the current flowing through the overhead ac supply wire.ⓘ Current Underground AC [I]			+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%

✖Resistivity is the measure of how strongly a material opposes the flow of current through them.ⓘ Resistivity using Load Current (1-Phase 2-Wire US) [ρ]

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

STEP 0: Pre-Calculation Summary

Formula Used

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

Variables Used

Resistivity - (Measured in Ohm Meter) - Resistivity is the measure of how strongly a material opposes the flow of current through them.
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.
Current Underground AC - (Measured in Ampere) - Current Underground AC is defined as the current flowing through the overhead ac supply wire.
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
Current Underground AC: 9 Ampere --> 9 Ampere No Conversion Required
Length of Underground AC Wire: 24 Meter --> 24 Meter No Conversion Required

STEP 2: Evaluate Formula

Substituting Input Values in Formula

ρ = P_loss*A/(2*(I)^2*L) --> 2.67*1.28/(2*(9)^2*24)

Evaluating ... ...

ρ = 0.000879012345679012

STEP 3: Convert Result to Output's Unit

0.000879012345679012 Ohm Meter --> No Conversion Required

FINAL ANSWER

0.000879012345679012 ≈ 0.000879 Ohm Meter <-- Resistivity

(Calculation completed in 00.004 seconds)

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< Resistance and Resistivity Calculators

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

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

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

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

Resistance (1-Phase 2-Wire US)

LaTeX Go Resistance Underground AC = Resistivity*Length of Underground AC Wire/Area of Underground AC Wire

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

LaTeX Go Resistance Underground AC = Volume Of Conductor*Resistivity/(2*(Area of Underground AC Wire^2))

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

LaTeX Go

Resistivity = Line Losses*Area of Underground AC Wire/(2*(Current Underground AC)^2*Length of Underground AC Wire)
ρ = P_loss*A/(2*(I)^2*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 Resistivity using Load Current (1-Phase 2-Wire US)?

Resistivity using Load Current (1-Phase 2-Wire US) calculator uses Resistivity = Line Losses*Area of Underground AC Wire/(2*(Current Underground AC)^2*Length of Underground AC Wire) to calculate the Resistivity, The Resistivity using Load Current (1-Phase 2-Wire US) formula is defined as a characteristic property of each material, resistivity is useful in comparing various materials on the basis of their ability to conduct electric currents. High resistivity designates poor conductors. Resistivity is denoted by ρ symbol.

How to calculate Resistivity using Load Current (1-Phase 2-Wire US) using this online calculator? To use this online calculator for Resistivity using Load Current (1-Phase 2-Wire US), enter Line Losses (P_loss), Area of Underground AC Wire (A), Current Underground AC (I) & Length of Underground AC Wire (L) and hit the calculate button. Here is how the Resistivity using Load Current (1-Phase 2-Wire US) calculation can be explained with given input values -> 0.000879 = 2.67*1.28/(2*(9)^2*24).

FAQ

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

The Resistivity using Load Current (1-Phase 2-Wire US) formula is defined as a characteristic property of each material, resistivity is useful in comparing various materials on the basis of their ability to conduct electric currents. High resistivity designates poor conductors and is represented as ρ = P_loss*A/(2*(I)^2*L) or Resistivity = Line Losses*Area of Underground AC Wire/(2*(Current Underground AC)^2*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, Current Underground AC is defined as the current flowing through the overhead ac supply wire & Length of Underground AC Wire is the total length of the wire from one end to other end.

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

The Resistivity using Load Current (1-Phase 2-Wire US) formula is defined as a characteristic property of each material, resistivity is useful in comparing various materials on the basis of their ability to conduct electric currents. High resistivity designates poor conductors is calculated using Resistivity = Line Losses*Area of Underground AC Wire/(2*(Current Underground AC)^2*Length of Underground AC Wire). To calculate Resistivity using Load Current (1-Phase 2-Wire US), you need Line Losses (P_loss), Area of Underground AC Wire (A), Current Underground AC (I) & Length of Underground AC Wire (L). With our tool, you need to enter the respective value for Line Losses, Area of Underground AC Wire, Current Underground AC & 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 Resistivity?

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

Resistivity = Area of Underground AC Wire*(Maximum Voltage Underground AC^2)*Line Losses*((cos(Phase Difference))^2)/((4)*Length of Underground AC Wire*(Power Transmitted^2))
Resistivity = Volume Of Conductor*Line Losses*(Maximum Voltage Underground AC*(cos(Phase Difference)))^2/(8*(Power Transmitted*Length of Underground AC Wire)^2)
Resistivity = Line Losses*Area of Underground AC Wire*(Maximum Voltage Underground AC*cos(Phase Difference))^2/(4*(Power Transmitted)^2*Length of Underground AC Wire)

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