Voltage Drop Help

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snoogins1236

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Location
Colorado
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Electrician
Hi,

I am studying for my next license using snapz. I keep getting VD answers wrong because sometimes the question wants me to use the formula Evd=i X r and other times is the Evd= (1.73 for 3p) 2 k i d.

How are you supposed to know which one to use in a scenario? It seems like branch circuits mostly get the I x R, but i cant figure it out. Super frustrating
 
Electromotive Force-noted E or ε or else- it is the voltage measured between terminals
of a circuit produced only by changes in magnetic flux -inner and outer this open circuit.
For instance, the voltage measured in a transformer secondary open circuit if the primary is energized.
If the system is a star-connected 3 phases, then we have VLN the supply voltage or ELN from the entrance terminal up to star center-neutral point- or VLL the supply voltage or ELL between two entrance terminals.
So, ELN=VLN+Iy*Zy and ELL=sqrt(3)*ELN=VLL+sqrt(3)*Iy*Zy
In star case, Iline=Iy the current from the source terminal up to load terminal is equal to the load current.
In a Delta connected system we have Iline and Idelta different. IfDelta to Star calculation.jpg we use the equivalent star of the delta we are in the same above case.
 
snoogins1236 said:
Hi,

I am studying for my next license using snapz. I keep getting VD answers wrong because sometimes the question wants me to use the formula Evd=i X r and other times is the Evd= (1.73 for 3p) 2 k i d.

How are you supposed to know which one to use in a scenario? It seems like branch circuits mostly get the I x R, but i cant figure it out. Super frustrating
Click to expand...
Vd = DIR for a balanced system with a neutral where D = length of the conductors, I = current, and R = resistance of the conductors
%Vd = (Vd/V)(100%) where V = line to neutral voltage
Works for three phase or split phase; either the sqrt(3) or the 2 in the numerator and denominator cancel and the current in the neutral is zero.
 
snoogins1236 said:
I keep getting VD answers wrong because sometimes the question wants me to use the formula Evd=i X r and other times is the Evd= (1.73 for 3p) 2 k i d.

How are you supposed to know which one to use in a scenario?
Click to expand...
V = I * R applies to any resistive load, such as the resistance from conductors. To use that formula, you must know the total resistance of the conductors in the circuit (including both legs for a 2-wire circuit). That resistance will depend on wire type, size, and length.

Vd = 2 * K * I * D is the same formula, just with the substitution that R = 2 * K * D, where D is the one way length, the 2 accounts for a 2-wire circuit, and K is the resistance per unit length of the given wire type, size, and length.

So if you already know the total resistance, you can use V = I * R. If you need to calculate the wire resistance, you can use R = 2 * K * D and then V = I * R, or use Vd = 2 * K * I * D to do the same thing in one step.

Cheers, Wayne
 
wwhitney said:
I think you mean R is the resistance of (all) the conductors per unit length.

Cheers, Wayne
Click to expand...
I mean the resistance of one phase conductor per unit length in whatever units D is measured in. Sorry if I wasn't clear.

My point is that if you use the phase to neutral voltage to calculate %Vd it doesn't matter whether it is single phase or three phase as long as the current is balanced.
 
It is hard to tell without any particular questions that you are getting wrong, but when doing voltage drop there are a couple of scenarios I can think of that might be causing you trouble.

Every wire has a resistance and reactance. That makes up their impedance. If it is just DC, then the resistance is the impedance. The angle of the voltage is zero and therefore no reactive part is applicable.

You can think of it like the current leaving the source, passes through the wire's resistance, through the load, and back to the source. It is a Kirchhoff's voltage law equation. How we get that resistance and reactance for each wire is where we use the tables. What voltage do we use in the calculation and how do we rectify polyphase vs single phase. All of that is fairly complicated math, so we try to simplify it into equations that are more digestible.

Wayne's equations work well but swap resistance with the word impedance so that you don't forget about reactance when the questions gives you a power factor, leading or lagging angles,.
 
Elect117 said:
It is hard to tell without any particular questions that you are getting wrong, but when doing voltage drop there are a couple of scenarios I can think of that might be causing you trouble.

Every wire has a resistance and reactance. That makes up their impedance. If it is just DC, then the resistance is the impedance. The angle of the voltage is zero and therefore no reactive part is applicable.

You can think of it like the current leaving the source, passes through the wire's resistance, through the load, and back to the source. It is a Kirchhoff's voltage law equation. How we get that resistance and reactance for each wire is where we use the tables. What voltage do we use in the calculation and how do we rectify polyphase vs single phase. All of that is fairly complicated math, so we try to simplify it into equations that are more digestible.

Wayne's equations work well but swap resistance with the word impedance so that you don't forget about reactance when the questions gives you a power factor, leading or lagging angles,.
Click to expand...
Depends on what kind of precision you are looking for as well. Conductor temperature is also a factor in it's resistance.

For licensing exams like OP is apparently concerned with they use a basic formula with K factor at a certain temperature (likely around 30C) and assume power factor of 1.0. This formula is generally "good enough" for in the field type calculations in most instances.
 
snoogins1236 said:
Hi,

I am studying for my next license using snapz. I keep getting VD answers wrong because sometimes the question wants me to use the formula Evd=i X r and other times is the Evd= (1.73 for 3p) 2 k i d.

How are you supposed to know which one to use in a scenario? It seems like branch circuits mostly get the I x R, but i cant figure it out. Super frustrating
Click to expand...
Consider the effective round trip length of the wiring. For single phase and DC, it's twice the 1-way length. For 3-phase relative to the interphase voltage, it's sqrt(3) times 1-way length.

What's really happening in 3-phase, is that each phase-to-neutral source sends current from source to load, and has a phase-conductor voltage drop for the 1-way length. The other two phases carry their own outbound current that adds up to the negative of this current at any given instant. So 3-phase relative to the phase-to-neutral voltage, the effective round trip length is just the 1-way length. Multiply by 1, i.e. sqrt(3)/sqrt(3) to make it relative to the interphase voltage, and the sqrt(3) in the numerator sticks around as the sqrt(3) factor in the 3-phase voltage drop calculation.
 
Why not simply say line voltdrop is square root 3 time phase drop as line volt square root 3 time phase volt in balanced 3 phase circuit
 
Carultch said:
Consider the effective round trip length of the wiring. For single phase and DC, it's twice the 1-way length. For 3-phase relative to the interphase voltage, it's sqrt(3) times 1-way length.

What's really happening in 3-phase, is that each phase-to-neutral source sends current from source to load, and has a phase-conductor voltage drop for the 1-way length. The other two phases carry their own outbound current that adds up to the negative of this current at any given instant. So 3-phase relative to the phase-to-neutral voltage, the effective round trip length is just the 1-way length. Multiply by 1, i.e. sqrt(3)/sqrt(3) to make it relative to the interphase voltage, and the sqrt(3) in the numerator sticks around as the sqrt(3) factor in the 3-phase voltage drop calculation.
Click to expand...
In the balanced case where there is a neutral with no current in it, the voltage drop is simply the drop in a phase / line conductor calculated by the application of Ohm's Law. No sqrt(3) or 2 is needed and it makes no difference whether it is single or three phase. For %Vd, compare Vd to the line to neutral voltage.
 
But
ggunn said:
In the balanced case where there is a neutral with no current in it, the voltage drop is simply the drop in a phase / line conductor calculated by the application of Ohm's Law. No sqrt(3) or 2 is needed and it makes no difference whether it is single or three phase. For %Vd, compare Vd to the line to neutral voltage.
Click to expand...
But for line to line voltage drop multiply square root 3.
 
In
ggunn said:
In the balanced case where there is a neutral with no current in it, the voltage drop is simply the drop in a phase / line conductor calculated by the application of Ohm's Law. No sqrt(3) or 2 is needed and it makes no difference whether it is single or three phase. For %Vd, compare Vd to the line to neutral voltage.
Click to expand...
In single phase voltage drop not require 2, you mean balanced mwbc
 
Deltaforce said:
In
In single phase voltage drop not require 2, you mean balanced mwbc
Click to expand...
No, I don't. If you use 240V you use 2D but if you use 120V you just use D. If you use line to neutral voltages, single phase and three phase calculations are the same.
 
What I understood by Zy [or Z∆] is a complex number of R and X for a.c. circuits. The complex number is a symbolic group of two elements, one real -the resistance- and second an imaginary X- the reactance.The resistance depends on type of metal -usually copper or aluminum-cross-section area of conductor and the length. The system frequency does not intervene directly but through skin and proximity effect, but temperature is a factor that could raise the resistance a lot.
So, you have to know the specific conductor [ metal] resistance, that means a resistance of one unit length per unit cross-section area at a standard temperature [usually 20oC] = ρ.
Then R=ρ*length/Area*(234.5+Tc)/(234.5+20)*(1+yskin+yprox+yconduit) for copper conductor, for instance [ρ=1/58 if length is m and Area mm^2].
X=2*π*f*L where f it is frequency and L is the inductance [different assembly of conductors different method of calculations] for 2 parallel conductors L=µo/2/π*ln(dist/dc) ln=natural logarithm; dist-center-to-center distance; dc conductor diameter. µo= permeability of free space= 4π x 10-7.
The complex number then it is Z=R+jX where j it is the imaginary j[or i]=sqrt(-1).
 
Julius Right said:
What I understood by Zy [or Z∆] is a complex number of R and X for a.c. circuits. The complex number is a symbolic group of two elements, one real -the resistance- and second an imaginary X- the reactance.The resistance depends on type of metal -usually copper or aluminum-cross-section area of conductor and the length. The system frequency does not intervene directly but through skin and proximity effect, but temperature is a factor that could raise the resistance a lot.
So, you have to know the specific conductor [ metal] resistance, that means a resistance of one unit length per unit cross-section area at a standard temperature [usually 20oC] = ρ.
Then R=ρ*length/Area*(234.5+Tc)/(234.5+20)*(1+yskin+yprox+yconduit) for copper conductor, for instance [ρ=1/58 if length is m and Area mm^2].
X=2*π*f*L where f it is frequency and L is the inductance [different assembly of conductors different method of calculations] for 2 parallel conductors L=µo/2/π*ln(dist/dc) ln=natural logarithm; dist-center-to-center distance; dc conductor diameter. µo= permeability of free space= 4π x 10-7.
The complex number then it is Z=R+jX where j it is the imaginary j[or i]=sqrt(-1).
Click to expand...
Upto particular conductir size X may ignored
 
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