Instrument Power Cable Sizing Part 1: The Basics

Note: due to blogger limitation, the symbol ohm and delta couldn’t shown here, as a replacement it is shown as OHM and Delta.

What is an instrument power cable sizing? Instrument power cable sizing is a calculation of voltage drop across some distance from instrument marshalling cabinet to field devices. It is needed to make sure that the supply voltage at the field devices is sufficient (within device operating voltage or within specified requirement). For example, we have a solenoid valve put at 250 meter from marshalling cabinet. This solenoid valve requires a supply voltage 24 VDC. Due to a long distance separated, the voltage received at solenoid valve will not be 24 VDC. It may be 23.5 VDC, 22 VDC, or even 15 VDC. It depends on the size of cables that used. So, the purpose of the instrument power cable sizing is to select the appropriate cable size so that the voltage drop received at field devices is within an acceptable range or within specified requirements. So, how to perform such calculation?

First of all, let’s take a look at below circuit.

This schematic is an illustration of equipment that has 1 K OHM resistance and separated far away from the power supply by 100 meter. Let’s assume that the cable has an inherent resistance 1 OHM/meter so that in this case, the resistance from power supply to equipment will be 100 OHM. We can replace the above schematic with below simplified schematic to calculate the voltage drop received at equipment.


V = 24 Volt

R1 = 100 OHM

R2 = 1K OHM

R3 = 100 OHM

Now we could look for the equivalent replacement resistor of this 3 resistor.

RTOT = R1 + R2 + R3

RTOT = (100 + 1000 + 100)  OHM

RTOT = 1200  OHM

After we calculate the equivalent replacement resistor (RTOT) now we could get the current that flow through this circuit.

I = V / RTOT

I = 24 / 1200

I = 0.02 A

Then we shall go back to above circuit to find the voltage received by the R2 (which is our equipment).

VR2 = I * R2

VR2 = 0.02 * 1000

VR2 = 20 Volt

From above calculation we could determine that the voltage drop is 4 volt. Each drop is contributed by R1 and R2 as follows:

VR1 = VR3 = I * 100 = 0.02 * 100 = 2 Volt.

Each resistor (R1 & R3) contribute 2 volt voltage drop.


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