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

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

ρ - Resistivity?K - Constant Underground AC?P_loss - Line Losses?V_m - Maximum Voltage Underground AC?P - Power Transmitted?L - Length of Underground AC Wire?

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

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

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

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

0.0006 = (0.87 \cdot 2.67 \cdot \frac{230^{2}}{4 \cdot {(300 \cdot 24)}^{2}})

0.0006Ω*m = (0.87 \cdot 2.67W \cdot \frac{{230V}^{2}}{4 \cdot {(300W \cdot 24m)}^{2}})

0.0006 = (0.87 \cdot 2.67 \cdot \frac{230^{2}}{4 \cdot {(300 \cdot 24)}^{2}})

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

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

FIRST Step Consider the formula

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

Next Step Substitute values of Variables

∴

ρ = (0.87 \cdot 2.67W \cdot \frac{{230V}^{2}}{4 \cdot {(300W \cdot 24m)}^{2}})

Next Step Prepare to Evaluate

∴

ρ = (0.87 \cdot 2.67 \cdot \frac{230^{2}}{4 \cdot {(300 \cdot 24)}^{2}})

Next Step Evaluate

∴

ρ = 0.000592599392361111Ω*m

LAST Step Rounding Answer

∴

ρ = 0.0006Ω*m

Resistivity using Constant (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

Constant Underground AC

Constant Underground AC is defined as the constant of line of a Overhead supply system.

Symbol: K

Measurement: NAUnit: Unitless

Note: Value can be positive or negative.

Formulas to find Constant Underground AC

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

Maximum Voltage Underground AC

Maximum Voltage Underground AC is defined as the peak amplitude of the AC voltage supplied to the line or wire.

Symbol: V_m

Measurement: Electric PotentialUnit: V

Note: Value can be positive or negative.

Formulas to find Maximum Voltage Underground AC

Power Transmitted

Power Transmitted is the amount of power that is transferred from its place of generation to a location where it is applied to perform useful work.

Symbol: P

Measurement: PowerUnit: W

Note: Value can be positive or negative.

Formulas to find Power Transmitted

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 Load Current (1-Phase 2-Wire US)

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

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}

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 Constant (1-Phase 2-Wire US)?

Resistivity using Constant (1-Phase 2-Wire US) evaluator uses Resistivity = (Constant Underground AC*Line Losses*(Maximum Voltage Underground AC^2)/(4*(Power Transmitted*Length of Underground AC Wire)^2)) to evaluate the Resistivity, The Resistivity using Constant (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 Constant (1-Phase 2-Wire US) using this online evaluator? To use this online evaluator for Resistivity using Constant (1-Phase 2-Wire US), enter Constant Underground AC (K), Line Losses (P_loss), Maximum Voltage Underground AC (V_m), Power Transmitted (P) & Length of Underground AC Wire (L) and hit the calculate button.

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

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

The formula of Resistivity using Constant (1-Phase 2-Wire US) is expressed as Resistivity = (Constant Underground AC*Line Losses*(Maximum Voltage Underground AC^2)/(4*(Power Transmitted*Length of Underground AC Wire)^2)). Here is an example- 0.000593 = (0.87*2.67*(230^2)/(4*(300*24)^2)).

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

With Constant Underground AC (K), Line Losses (P_loss), Maximum Voltage Underground AC (V_m), Power Transmitted (P) & Length of Underground AC Wire (L) we can find Resistivity using Constant (1-Phase 2-Wire US) using the formula - Resistivity = (Constant Underground AC*Line Losses*(Maximum Voltage Underground AC^2)/(4*(Power Transmitted*Length of Underground AC Wire)^2)).

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/(2*(Current Underground AC)^2*Length of Underground AC Wire)

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

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

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

Resistivity using Constant (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 Constant (1-Phase 2-Wire US) can be measured.

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

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

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

Other Formulas to find Resistivity

Other formulas in Resistance and Resistivity category

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

FAQs on Resistivity using Constant (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