Voltage Drop measurement using Multimeter in a 5A live line

[vinod]

Consider the below circuit with 28V supply and electronic load set at 5A Constant Current mode.

What voltage drop would I expect when I connect a digital multimeter between A & B ?
V(AB1)= V(R6)+V(R8)+V(R9)


What voltage drop would I expect when I connect a digital multimeter between C & D ?
V(CD1)= V(R7)+V(R8)+V(R10)

Consider the second circuit.

What voltage would I expect when I connect a multimeter between A & B ?
V(AB2)= V(R6)+V(R8)+V(R9)


What voltage would I expect when I connect a multimeter between C & D ?
V(CD2)= V(R8)

Am I right ? And V(AB2)>V(AB1)

[sghioto]

This doesn't make any sense. How do you expect to calculate the voltage if you don't know the values of the resistors.

[vinod]

This doesn't make any sense. How do you expect to calculate the voltage if you don't know the values of the resistors.

Sorry, I have modified my post, It was a mistake --I want to measure voltage drop.

[sghioto]

Sorry, I have modified my post, It was a mistake --I want to measure voltage drop.

I still don't see any resistor values.
What do you mean by a 5 amp load ? Is this a resistor also, or are you saying the 28 volt supply is delivering a constant 5 amps through the resistor network?

Steve G

[MrAl]

Consider the below circuit with 28V supply and electronic load set at 5A Constant Current mode.

What voltage drop would I expect when I connect a digital multimeter between A & B ?
V(AB1)= V(R6)+V(R8)+V(R9)


What voltage drop would I expect when I connect a digital multimeter between C & D ?
V(CD1)= V(R7)+V(R8)+V(R10)

Consider the second circuit.

What voltage would I expect when I connect a multimeter between A & B ?
V(AB2)= V(R6)+V(R8)+V(R9)


What voltage would I expect when I connect a multimeter between C & D ?
V(CD2)= V(R8)

Am I right ? And V(AB2)>V(AB1)

Hello there,


Let me give you an example for the circuit on the left. You want the voltage between points A and B, and the impedance of the measuring device is infinite (very high).

The idea is to use current division for the resistors R2, R6, and R3,R4,R7, to calculate the current through R6. Ditto for the other branch on the other side of the circuit, use current division there too. Doing this will lead to two currents, one through each resistor in question, and then you can calculate the voltage drop for each resistor and then knowing the full 5A has to flow through R8 you can sum all three voltages to get Vab.

The result is this:
vR6=(I1*R6*(R7+R4+R3))/(R7+R6+R4+R3+R2)
vR8=R8*I1
vR9=(I1*(R16+R13+R10)*R9)/(R9+R16+R13+R11+R10)

so the total voltage across A and B is:
Vab=(I1*R6*(R7+R4+R3))/(R7+R6+R4+R3+R2)+I1*R8+(I1*(R16+R13+R10)*R9)/(R9+R16+R13+R11+R10)

where I1 is the load current of 5 amps. And note that the equation is valid for Vab<=Vs where Vs is the source voltage which is 28 volts in your circuit.

You can do the other questions the same way.

Just a small note:
If the measuring device impedance is not very high but is comparable to the other resistor values, then we have to take that into consideration too and that complicates the math quite a bit because the meter then partially bridges the points A and B, so we might have to use something like nodal analysis.

What made this analysis so simple is the fact that the load current is already known. That means we know the current through R8 for example, as well as the dual branches so all we had to do was figure out how much of that 5 amps was going through each sub branch and then use Ohm's Rule to calculate the voltage drop across two resistors, and of course one more time for R8.

Just to note, current division works like this:
For two resistors in parallel, R1 and R2, with total current I flowing through them both, the currents are:
iR1=I*R2/(R1+R2), the current through R1, and
iR2=I*R1/(R1+R2), the current through R2.

and for this circuit we first have to lump R2 and R6, and then R3, R4, and R7 (and again ditto for the other half of the circuit). Since they are in series, we just add R2 and R6, then add the other three and we have our two parallel resistors we can then use current division (above) on.