Voltage Drop Question: Calculate w/o I

[corvairbob]

i have a 24 volt buss in a machine and i need to connect an lvdt sensor to the plc i show a 1.5k resistor that is .5 watts on the lvdt. what need to find out is the voltage drop at the resistor. i formula asks for I but I but i don't have I so what might the votage drop be. answer or formula will help thanks

1. Homework Statement

Homework Equations

The Attempt at a Solution

 

[berkeman]

i have a 24 volt buss in a machine and i need to connect an lvdt sensor to the plc i show a 1.5k resistor that is .5 watts on the lvdt. what need to find out is the voltage drop at the resistor. i formula asks for I but I but i don't have I so what might the votage drop be. answer or formula will help thanks

1. Homework Statement

Homework Equations

The Attempt at a Solution

Can you post the schematic of the system and the datasheet for the sensor?

 

[corvairbob]

i don't have a diagram nor any data for the sensor at this time but if you draw the left side of the ladder at 24 volt + and then put the .5 watt resistor on a rung then attache the 24 + power input sensor wire to that. then the sensor has the ground and the wiper on that lvdt is the output. i gather the sensor is a Imax < 1ua that is what i see on it.

so what I'm trying to figure out is the voltage drop across that resistor i used the ohm's law but i asks for I and i don't have I i have 24 volt + and i have 1.5k ohm's and i have .5 watts for the resistor.

i'm thinking the sensor is changing current to the output to the plc and not the volts but I'm not sure. so that is why i asked the question all the numbers i plug into this show a voltage that doesn't make sense to me. maybe someone here can help figure this one out. thanks

 

[SammyS]

i don't have a diagram nor any data for the sensor at this time but if you draw the left side of the ladder at 24 volt + and then put the .5 watt resistor on a rung then attache the 24 + power input sensor wire to that. then the sensor has the ground and the wiper on that lvdt is the output. i gather the sensor is a Imax < 1ua that is what i see on it.

so what I'm trying to figure out is the voltage drop across that resistor i used the ohm's law but i asks for I and i don't have I i have 24 volt + and i have 1.5k ohm's and i have .5 watts for the resistor.

i'm thinking the sensor is changing current to the output to the plc and not the volts but I'm not sure. so that is why i asked the question all the numbers i plug into this show a voltage that doesn't make sense to me. maybe someone here can help figure this one out. thanks

If you want to know the voltage drop across the resistor, put a volt meter across it and measure the voltage drop directly.

 

[CWatters]

i don't have a diagram nor any data for the sensor at this time but if you draw the left side of the ladder at 24 volt + and then put the .5 watt resistor on a rung then attache the 24 + power input sensor wire to that. then the sensor has the ground and the wiper on that lvdt is the output. i gather the sensor is a Imax < 1ua that is what i see on it.

I cannot understand that description. Can I suggest you make a drawing and post it here.

Are you trying to use a 1.5K potentiometer as a potential divider to provide power for the sensor?