Measuring Voltage Across a Resistor

By the way why don't you isolate measuring instrument?

Anyway if its not isolated then you ll have to take the differential of values observed at both end of resistor.Otherwise you can simply measure between resistor ends.

Hello world6:

Do not feel alone, I had to read the question several times before I realized what he is trying to say.

I'm assuming by now you're familiar with ohms law and Kirchoff's law. For simplicity's sake we will assume that it is a single resistance circuit, therefore the entire supply voltage will be dissipated in that resistor. We would read the potential of the circuit across the resistor.

In the event the meter is grounded we in essence induced a new current path (parallel resistances) therefore some of the potential of the circuit would flow through that current path to ground and you would no longer read the full potential of the circuit across the resistor.

I find the question to be somewhat misleading in that the circuit would have to be grounded to produce this effect. Unfortunately I have to meet someone this morning or I would provide a more detailed explanation. Hopefully if you're still having problems someone will provide that more detailed explanation, will check back later good luck.

And hopefully my explanation is correct, we'll see what others say.

Hello TPV45:

I see your point, however the way the question reads the circuit is not connected to ground, so there would be zero potential between ground in the isolated circuit.

You're definitely right if there is no resistance between a power supply that is grounded, and the ground there is going to be a nasty short.

I was originally thinking of the internal resistance of the meter itself, I don't know I'll stick to DMM's, that run on batteries. That way if I want to read voltage to ground I will intentionally ground one side of the meter. No chance of a dead short to ground.

After thoroughly rereading the question I'm going to vote for you, but I still think the question is misleading.

PS: I stand corrected.

I agree. I wouldn't own a line powered multimeter unless it was one of the fancy old Keithley jobs, in which case I'd spring for an isolation transformer with a yes/no ground switch so I could use ground when I wanted.

YWR,

"the way the question reads the circuit is not connected to ground, so there would be zero potential between ground in the isolated circuit."

Read it again. It says "even though neither side of the resistor is connected to ground."

Jon

So according to you if we connect grounded voltmeter's black lead to one end of resistor and red lead to another end...example to 2V point we connect red lead and to 3v point we connect black lead..the black lead will make potential of 3v point 0 by shorting the voltage. I got this point can u please tell what will be the potential in this case and what effect would it carry on the main circuit(by shorting voltage to 0)..as in normal case the potential will be (3v-2v-1v)

It's a little tricky to cover every possible case, but generally the potential will be 0.0 V once you've shorted the 3.0 V to ground. As for the mains, who knows? It depends on the actual circuit and whether or not the firemen turn off the power when they arrive. I actually set fire to a chassis once - flames and smoke - when I created a "sneak" ground by accident.

You wrote;

I read this in a text Book "Voltage can normally be measured across a resistor, even though neither side of the resistor is connected to ground.

Should this be " iether side" it would make the question more relevent?

Garth.

maybe "either side" instead of "iether side".

Garth,

"Neither" is correct.

Jon

First of all, you did not say whether the voltage is AC or DC. DC is much easier, since there is no phase difference at each measurement point. Anyway, the best way to measure eaither AC or DC across a floating R is with a differential, high impedance probe. These are available for any oscilloscope. The differential probe's input R needs to be at least 10 times larger than the floating R being measured.

If no differential probe is available, you can still use a scope to make this measurement. The scope's probe needs to be of the passive, high impedance type (these are typically 10 Mohms). For DC volts, measure the voltage on the high side to ground (even if ground is not one side of the R), then on the low side to the same ground point. The drop across the R will be the difference of these two V's.

If it is AC, the scope should be used to measure the RMS of the two voltages, and then take the difference.

This will work except in the rare case where the impedance to ground happens to be close to the probe impedance. In that case, only a differential probe can be used.

The original textbook question makes more sense if you think of the resistance as being a metering shunt. This is a common technique for measuring DC current. For a long time CTs have been used for doing AC measurements; hence my assumption the text is really referring to a DC circuit measurement. All too often the academics writing text books use rather arcane language that is so abstract.

World6,

To give you an idea of why you would take measurements that way. 16,000 Volt dc supply:

In this example the measurements would be taken with a battery operated Meter.

Which meter will burn? Meter A or Meter B

Meter A common lead is connected to Supply Ground. Meter A high lead is the broken lines. The intention is to measure one end of resistor R2 indicated by the dashed line, then the other end indicated by the dash-dot line. The circuit is more than 16,000Volts above ground. Meter A is only designed to measure a maximum of 1000Volts dc. Say goodbye to the meter and the circuit you were measuring.

Meter B common lead is connected to one end of R2 and the high lead to the other end. If the meter resistance is 100,000 ohms you would include that in the calcultions for the true Voltage and current of the measured resistor. R2 25,000 Ohms in parallel with the 100,000 Ohms meter resistance.

If R2 is an open resistor you may have to replaced the meter unless it has adequate overvoltage protection.

The round cornered box represents the user.

Note: 16,000 Volts @ 20mA can kill you.

Jon

Under normal circumstances, when you measure a voltage, this is basically measuring a current through the instrument. The voltage (drop) measured along the resistor equals the voltage drop along the wires + instrument (the 2 ends (resistor & wire) are connected together, so their potential with respect to the ground (which is assumed zero (0)) is the same. The voltage equals the difference between the 2 potentials.

Voltmeters have to be isolated from ground because if not, the potential of that part will be brought to the potential of the ground (0) and the determination is altered.

Assume for instance a circuit with a source of 20V ( one terminal (terminal 1) at 0 and the other (terminal 2) at +20V). This source is connected in a circuit involving wire between terminal 1 and one end (end 3) of a resistor, the resistor itself ( between end 3 and end 4) and the second wire between resistor end 4 and source terminal 2. Assume the resistor has 14 ohm, with the wire resistances of 3 ohm + 3 ohm, a total of 20 ohm, with the corresponding current of 1A, constant along the circuit.

The voltage distribution along that circuit will be:

terminal 1: 0V

resistor end 3: +3V

resistor end 4: +17V

terminal 2: + 20V

The voltage between end 3 and 4 of the resistor will be 14V.

If the potential (with respect to ground) of terminal 1 is not forced to 0 (for instance by grounding) by applying an instrument with one lead grounded (assume to resistor end 3), the potential of that point will be forced to 0, while the potential of the other points will be:

terminal 1: -3V

resistor end 3: 0V

resistor end 4: +14V

terminal 2: +17V

The voltage along the resistor remained at 14V, even if the potentials of different points floated freely.

If the potential of that point (terminal 1) is grounded, the resistance of portion between terminal 1 and end 3 is 0 (they are at the same potential), so the total resistance of the circuit would be 17 ohm, with a current of 20/17 = 1.176A, which leads to a potential at the same points considered in the previous situation as:

terminal 1: 0V

resistor end 3: 0V

resistor end 4: 14 x 1.176 = 16.46 V

terminal 2: + 20V

with a voltage along the resistor of 16.46V. This is an operation alteration.

It reads like a badly translated indian to english text!!!!!