How should a sense circuit be strapped

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jeepnjeff

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I need help on a practice exam question.
I cant find anything in any of my text books to help me figure this problem.
Please open image problem jpeg.
 

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Someone can explain why you place sense circuit in parallel with the load the sense current... if you need to deliver 10 amps to the load, shouldn't the sense circuit be in serie with the load!? I understand that none of the answer would comply but....
 
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hi,
The voltage 'sense' wires are across the load.

The 'current wires that connect the power supply to the load do have some resistance, so if the voltage is sensed at the power supply it would not compensate for the voltage drop in the 'current' wires.

example: if the load had a resistance of say 10R and the psu voltage is set to 10V and the 'leads' connecting the psu and load together had a resistance of 1R then there would be a voltage drop in the lead.

So the load would not have a 10V supply.

The 'sense' wires measure the actual voltage at the load and correct for any voltage drop in the 'current' leads by increasing the psu output voltage by the amount dropped in the current leads.

OK.?
 
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Hum... i thought that the power supply set to 10 amps not 10V. In that case, you couldn't measure voltage at load cause if the load changes, the voltage changes. It says that the PSU is set to deliver a constant 10 amps not 10V... that's why I would place sense circuit in series with the load (assuming that the sense circuit impedance is relay ***** low)... Maybe I'm wrong to... it's just i don't understand how you control a current flow with a voltage at an unknown load...
 

hi,

It says the psu is set to deliver 10A to the load, this means the psu VOLTAGE output would be set so that 10A, 'should' flow thru the load.

But has the load is some distance from the psu, setting the voltage at the psu will not guarantee 10A thru the load, because of the voltage drop in the long current carrying wires.

The 'voltage sense' wires from the psu to the load carry a very small current, so they measure [sense] the actual load voltage and correct the output of the psu voltage so that the load current will be 10A.
 
So, in that case, we assume that the load is constant right! That was probably the part I was missing!
 
So, in that case, we assume that the load is constant right! That was probably the part I was missing!

The example in the 'test' paper was set at 10A.

When a psu has Vsense function, there are 4 terminals, 2 are for the heavy current wires and the 'sense' 2 are for remote voltage sensing wires.

Often the 2 sense terminals are just linked on front of the psu to the current terminals.
 
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