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Load Current Using Area Of X-section(2-phase 4-wire OS) Solution

STEP 0: Pre-Calculation Summary
Formula Used
load_current = sqrt(Line Losses*Area Of X-Section/((32)*Resistivity*Length))
Il = sqrt(W*a/((32)*ρ*l))
This formula uses 1 Functions, 4 Variables
Functions Used
sqrt - Squre root function, sqrt(Number)
Variables Used
Line Losses - Line Losses is defined as the losses that are produced in the line. (Measured in Watt)
Area Of X-Section - Area Of X-Section is defined as the cross-sectional area simply as the square of the wire's diameter in mils and calls that our area in units of “circular mils.” (Measured in Square Meter)
Resistivity - Resistivity is the measure of how strongly a material opposes the flow of current through them. (Measured in Ohm Meter)
Length - Length is the measurement or extent of something from end to end. (Measured in Meter)
STEP 1: Convert Input(s) to Base Unit
Line Losses: 0.6 Watt --> 0.6 Watt No Conversion Required
Area Of X-Section: 5 Square Meter --> 5 Square Meter No Conversion Required
Resistivity: 1.7E-05 Ohm Meter --> 1.7E-05 Ohm Meter No Conversion Required
Length: 3 Meter --> 3 Meter No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
Il = sqrt(W*a/((32)*ρ*l)) --> sqrt(0.6*5/((32)*1.7E-05*3))
Evaluating ... ...
Il = 42.8746462856272
STEP 3: Convert Result to Output's Unit
42.8746462856272 Ohm --> No Conversion Required
FINAL ANSWER
42.8746462856272 Ohm <-- Load current
(Calculation completed in 00.018 seconds)

9 Area Of X-Section Calculators

Power Transmitted Using Area Of X-Section(2-phase 4-wire OS)
power_transmitted = sqrt((2*Area Of X-Section*(Maximum Voltage^2)*Line Losses*((cos(Theta))^2))/(Resistivity*Length)) Go
Maximum Voltage Using Area Of X-section(2-phase 4-wire OS)
maximum_voltage = sqrt((Length*Resistivity*(Power Transmitted^2))/(2*Area Of X-Section*Line Losses*((cos(Theta))^2))) Go
RMS Voltage Using Area Of X-Section(2-phase 4-wire OS)
rms_voltage = sqrt((Length*Resistivity*(Power Transmitted^2))/(Area Of X-Section*Line Losses*((cos(Theta))^2))) Go
Power Factor Using Area Of X-section(2-phase 4-wire OS)
power_factor = sqrt((Power Transmitted^2)*Resistivity*Length/(2*Area Of X-Section*Line Losses*(Maximum Voltage^2))) Go
Line Losses Using Area Of X-Section(2-phase 4-wire OS)
line_losses = (Length*Resistivity*(Power Transmitted^2))/(2*Area Of X-Section*(Maximum Voltage^2)*((cos(Theta))^2)) Go
Length Of Wire Using Area Of X-section(2-phase 4-wire OS)
length = 2*Area Of X-Section*(Maximum Voltage^2)*Line Losses*((cos(Theta))^2)/(Resistivity*(Power Transmitted^2)) Go
Resistivity Using Area Of X-Section(2-phase 4-wire OS)
resistivity = 2*Area Of X-Section*(Maximum Voltage^2)*Line Losses*((cos(Theta))^2)/(Length*(Power Transmitted^2)) Go
Load Current Using Area Of X-section(2-phase 4-wire OS)
load_current = sqrt(Line Losses*Area Of X-Section/((32)*Resistivity*Length)) Go
Volume Of Conductor Material Using Area Of X-Section(2-phase 4-wire OS)
volume_of_conductor_material = (4)*Area Of X-Section*Length Go

Load Current Using Area Of X-section(2-phase 4-wire OS) Formula

load_current = sqrt(Line Losses*Area Of X-Section/((32)*Resistivity*Length))
Il = sqrt(W*a/((32)*ρ*l))

What is the value of maximum voltage and volume of conductor material in 3-phase 4-wire system?

The volume of conductor material required in this system is 1/2cos2θ times that of 2-wire d.c.system with the one conductor earthed. The maximum voltage between conductors is 2vm so that r.m.s. value of voltage between them is √2vm.

How to Calculate Load Current Using Area Of X-section(2-phase 4-wire OS)?

Load Current Using Area Of X-section(2-phase 4-wire OS) calculator uses load_current = sqrt(Line Losses*Area Of X-Section/((32)*Resistivity*Length)) to calculate the Load current, The Load Current Using Area Of X-section(2-phase 4-wire OS) formula is defined as the current that flows into the load of the two-phase four-wire overhead system. Load current and is denoted by Il symbol.

How to calculate Load Current Using Area Of X-section(2-phase 4-wire OS) using this online calculator? To use this online calculator for Load Current Using Area Of X-section(2-phase 4-wire OS), enter Line Losses (W), Area Of X-Section (a), Resistivity (ρ) and Length (l) and hit the calculate button. Here is how the Load Current Using Area Of X-section(2-phase 4-wire OS) calculation can be explained with given input values -> 42.87465 = sqrt(0.6*5/((32)*1.7E-05*3)).

FAQ

What is Load Current Using Area Of X-section(2-phase 4-wire OS)?
The Load Current Using Area Of X-section(2-phase 4-wire OS) formula is defined as the current that flows into the load of the two-phase four-wire overhead system and is represented as Il = sqrt(W*a/((32)*ρ*l)) or load_current = sqrt(Line Losses*Area Of X-Section/((32)*Resistivity*Length)). Line Losses is defined as the losses that are produced in the line, Area Of X-Section is defined as the cross-sectional area simply as the square of the wire's diameter in mils and calls that our area in units of “circular mils.”, Resistivity is the measure of how strongly a material opposes the flow of current through them and Length is the measurement or extent of something from end to end.
How to calculate Load Current Using Area Of X-section(2-phase 4-wire OS)?
The Load Current Using Area Of X-section(2-phase 4-wire OS) formula is defined as the current that flows into the load of the two-phase four-wire overhead system is calculated using load_current = sqrt(Line Losses*Area Of X-Section/((32)*Resistivity*Length)). To calculate Load Current Using Area Of X-section(2-phase 4-wire OS), you need Line Losses (W), Area Of X-Section (a), Resistivity (ρ) and Length (l). With our tool, you need to enter the respective value for Line Losses, Area Of X-Section, Resistivity and Length 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 Load current?
In this formula, Load current uses Line Losses, Area Of X-Section, Resistivity and Length. We can use 9 other way(s) to calculate the same, which is/are as follows -
  • line_losses = (Length*Resistivity*(Power Transmitted^2))/(2*Area Of X-Section*(Maximum Voltage^2)*((cos(Theta))^2))
  • maximum_voltage = sqrt((Length*Resistivity*(Power Transmitted^2))/(2*Area Of X-Section*Line Losses*((cos(Theta))^2)))
  • power_factor = sqrt((Power Transmitted^2)*Resistivity*Length/(2*Area Of X-Section*Line Losses*(Maximum Voltage^2)))
  • power_transmitted = sqrt((2*Area Of X-Section*(Maximum Voltage^2)*Line Losses*((cos(Theta))^2))/(Resistivity*Length))
  • resistivity = 2*Area Of X-Section*(Maximum Voltage^2)*Line Losses*((cos(Theta))^2)/(Length*(Power Transmitted^2))
  • length = 2*Area Of X-Section*(Maximum Voltage^2)*Line Losses*((cos(Theta))^2)/(Resistivity*(Power Transmitted^2))
  • volume_of_conductor_material = (4)*Area Of X-Section*Length
  • load_current = sqrt(Line Losses*Area Of X-Section/((32)*Resistivity*Length))
  • rms_voltage = sqrt((Length*Resistivity*(Power Transmitted^2))/(Area Of X-Section*Line Losses*((cos(Theta))^2)))
Where is the Load Current Using Area Of X-section(2-phase 4-wire OS) calculator used?
Among many, Load Current Using Area Of X-section(2-phase 4-wire OS) calculator is widely used in real life applications like {FormulaUses}. Here are few more real life examples -
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