Instrument Power Cable Sizing Part 2: The Simplified Equations

At actual application we rarely know the resistance of our equipment. What usually specified by manufacturer is only a power consumption at nominal power supply (can be 24 VDC or 48 VDC). So, how to get the equation to calculate the voltage drop across some distance with using the power consumption data?

From previous illustration in the part 1, we could express the voltage drop (Delta V) as below:

Delta V = V – VR2

Delta V = V – I*R2

Because I=V/RTOT then,

Delta V = V – ( (V/RTOT) * R2)

We take out the same V so that we could get the following equation,

Delta V = V* (1 – (R2/RTOT) )

Since the RTOT = R1+R2+R3 then,

Delta V = V* (1 – (R2 / (R1+R2+R3) ) )

Since R1 = R2 = Length of cable (L) * Resistance of cable (Rcable) then,

Delta V = V* (1 – (R2 / ((L*Rcable) + R2 + (L*Rcable)) )

Delta V = V* (1 – (R2 / ((2*L*Rcable) + R2)

Unfortunately we don’t know the R2 of the equipment (but we only know the power consumption), then we should replace this in term of power consumption (P).

R2 = P/I2

While I = P/V, then

R2 = P / (P/V)2

R2 = V2 / P

Now we substitute this R2 value to voltage drop equation above.

Delta V = V* (1 – ( (V2 / P) / ((2*L*Rcable) + (V2 / P) )

Delta V = V* ( ((2*L*Rcable) + (V2/P) – (V2/P)) / (2*L*Rcable) + (V2/P) )

Delta V = V* ((2*L*Rcable) / (2*L*Rcable) + (V2/P))

While (2*L*Rcable) + (V2/P) = (2*L*Rcable) + R2 = RTOT

Then we could simplify the above equation to be:

Delta V = V/RTOT * (2*L*Rcable)

Delta V = I * (2*L*Rcable)


Delta V = Voltage drop across some distance (Volt)

I = electric current (Ampere) = P/V

L = Length of cable (km or m)

Rcable =  OHM/km or  OHM/m

Usually Rcable represents in OHM/km rather than in OHM/m. And length of cable represents in m. In this case the voltage drop equation will be:

Delta V = (I * 2*L*Rcable) / 1000

Let’s take some example.

We have a solenoid valve put at distance 250 meter from the system cabinet (power supply putted). This solenoid valve has a power consumption 5 watt@24 VDC. We need to check what cable size that can be used with the requirement of voltage drop must not exceed 5% at solenoid valve terminal. The following example cable data will be used:

Cable Cross Sectional Area
1.0 mmsq
20 OHM/km
1.5 mmsq
15 OHM/km
2.5 mmsq
10 OHM/km

 The calculation:

Calculate the electric current:

I = P/V = 5 watt / 24 V = 0.2083 A

Calculate the voltage drop by using formula:

Delta V = (I * 2*L*Rcable) / 1000 with the preliminary Rcable using the 1.0 mmsq properties.

Delta V = (0.2083*2*250*20)/1000

Delta V = 2.083 Volt

Delta V = 2.083 Volt / 24 Volt *100%

Delta V = 8.67%

It exceeds our requirement which is 5%. So using 1.0 mmsq cable isn’t sufficient for our application. Let’s try with 1.5 mmsq cable.

Delta V = (0.2083*2*250*15)/1000

Delta V = 1.56 Volt

Delta V = 1.56Volt / 24 Volt *100%

Delta V = 6.5%

By using 1.5 mmsq cables it still exceed the requirement. Let’s try again with 2.5 mmsq cable, looks like this cable will suit for this application.

Delta V = (0.2083*2*250*10)/1000

Delta V = 1.04 Volt

Delta V = 1.04Volt / 24 Volt *100%

Delta V = 4.34% which is less than 5% requirement.

Yes, this 2.5 mmsq cable is suitable for our applications. Thus this cable size will be selected for this application.


  1. salam wisdu,

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