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

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

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

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Resistivity is the measure of how strongly a material opposes the flow of current through them. Check FAQs

ρ = P loss \cdot \frac{A}{2 \cdot {(I)}^{2} \cdot L}

ρ - Resistivity?P_loss - Line Losses?A - Area of Underground AC Wire?I - Current Underground AC?L - Length of Underground AC Wire?

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

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Here is how the Resistivity using Load Current (1-Phase 2-Wire US) equation looks like with Values.

Here is how the Resistivity using Load Current (1-Phase 2-Wire US) equation looks like with Units.

Here is how the Resistivity using Load Current (1-Phase 2-Wire US) equation looks like.

0.0009 = 2.67 \cdot \frac{1.28}{2 \cdot {(9)}^{2} \cdot 24}

0.0009Ω*m = 2.67W \cdot \frac{1.28m²}{2 \cdot {(9A)}^{2} \cdot 24m}

0.0009 = 2.67 \cdot \frac{1.28}{2 \cdot {(9)}^{2} \cdot 24}

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

Follow our step by step solution on how to calculate Resistivity using Load Current (1-Phase 2-Wire US)?

FIRST Step Consider the formula

ρ = P loss \cdot \frac{A}{2 \cdot {(I)}^{2} \cdot L}

Next Step Substitute values of Variables

∴

ρ = 2.67W \cdot \frac{1.28m²}{2 \cdot {(9A)}^{2} \cdot 24m}

Next Step Prepare to Evaluate

∴

ρ = 2.67 \cdot \frac{1.28}{2 \cdot {(9)}^{2} \cdot 24}

Next Step Evaluate

∴

ρ = 0.000879012345679012Ω*m

LAST Step Rounding Answer

∴

ρ = 0.0009Ω*m

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

Variables

Resistivity

Resistivity is the measure of how strongly a material opposes the flow of current through them.

Symbol: ρ

Measurement: Electric ResistivityUnit: Ω*m

Note: Value can be positive or negative.

Formulas to find Resistivity

Line Losses

Line Losses is defined as the total losses occurring in an Underground AC line when in use.

Symbol: P_loss

Measurement: PowerUnit: W

Note: Value can be positive or negative.

Formulas to find Line Losses

Area of Underground AC Wire

Area of Underground AC Wire is defined as the cross-sectional area of the wire of an AC supply system.

Symbol: A

Measurement: AreaUnit: m²

Note: Value can be positive or negative.

Formulas to find Area of Underground AC Wire

Current Underground AC

Current Underground AC is defined as the current flowing through the overhead ac supply wire.

Symbol: I

Measurement: Electric CurrentUnit: A

Note: Value can be positive or negative.

Formulas to find Current Underground AC

Length of Underground AC Wire

Length of Underground AC Wire is the total length of the wire from one end to other end.

Symbol: L

Measurement: LengthUnit: m

Note: Value should be greater than 0.

Formulas to find Length of Underground AC Wire

Other Formulas to find Resistivity

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

ρ = A \cdot ({V m}^{2}) \cdot P loss \cdot \frac{{(\cos (Φ))}^{2}}{(4) \cdot L \cdot (P^{2})}

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

ρ = V \cdot P loss \cdot \frac{{(V m \cdot (\cos (Φ)))}^{2}}{8 \cdot {(P \cdot L)}^{2}}

Go Resistivity using Line Losses (1-Phase 2-Wire US)

ρ = P loss \cdot A \cdot \frac{{(V m \cdot \cos (Φ))}^{2}}{4 \cdot {(P)}^{2} \cdot L}

Go Resistivity using Constant (1-Phase 2-Wire US)

ρ = (K \cdot P loss \cdot \frac{{V m}^{2}}{4 \cdot {(P \cdot L)}^{2}})

Other formulas in Resistance and Resistivity category

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

R = V \cdot \frac{ρ}{2 \cdot (A^{2})}

Go Resistance (1-Phase 2-Wire US)

R = ρ \cdot \frac{L}{A}

Go Resistance using Line Losses (1-Phase 2-Wire US)

R = \frac{P loss}{2 \cdot {(I)}^{2}}

Go Resistance using Constant (1-Phase 2-Wire US)

R = K \cdot P loss \cdot \frac{{V m}^{2}}{A \cdot L \cdot 4 \cdot {(P)}^{2}}

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

Resistivity using Load Current (1-Phase 2-Wire US) evaluator uses Resistivity = Line Losses*Area of Underground AC Wire/(2*(Current Underground AC)^2*Length of Underground AC Wire) to evaluate 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 evaluate Resistivity using Load Current (1-Phase 2-Wire US) using this online evaluator? To use this online evaluator 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.

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

What is the formula to find Resistivity using Load Current (1-Phase 2-Wire US)?

The formula of Resistivity using Load Current (1-Phase 2-Wire US) is expressed as Resistivity = Line Losses*Area of Underground AC Wire/(2*(Current Underground AC)^2*Length of Underground AC Wire). Here is an example- 0.000879 = 2.67*1.28/(2*(9)^2*24).

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

With Line Losses (P_loss), Area of Underground AC Wire (A), Current Underground AC (I) & Length of Underground AC Wire (L) we can find Resistivity using Load Current (1-Phase 2-Wire US) using the formula - Resistivity = Line Losses*Area of Underground AC Wire/(2*(Current Underground AC)^2*Length of Underground AC Wire).

What are the other ways to Calculate Resistivity?

Here are the different ways to Calculate Resistivity-

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)

Can the Resistivity using Load Current (1-Phase 2-Wire US) be negative?

Yes, the Resistivity using Load Current (1-Phase 2-Wire US), measured in Electric Resistivity can be negative.

Which unit is used to measure Resistivity using Load Current (1-Phase 2-Wire US)?

Resistivity using Load Current (1-Phase 2-Wire US) is usually measured using the Ohm Meter[Ω*m] for Electric Resistivity. Ohm Centimeter[Ω*m], Ohm Inch[Ω*m], Microhm Centimeter[Ω*m] are the few other units in which Resistivity using Load Current (1-Phase 2-Wire US) can be measured.

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

​GoResistivity using Area of X-Section (1-Phase 2-Wire US) Formula

​GoResistivity using Volume of Conductor Material (1-Phase 2-Wire US) Formula

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

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

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

Other Formulas to find Resistivity

Other formulas in Resistance and Resistivity category

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

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

Variable Name

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

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