# B Why you can't measure voltage and current at the same time?

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1. Oct 12, 2017

### Karagoz

In my physics textbook it says that ammeter in circuit C shows higher current than ammeter in circuit A and D.

And the voltmeter shows higher voltage in D than in B, and C.

It's explained in the book that way: some electric current is passing through the voltmeter in C, thus the ammeter will show higher and current. And in D there's some voltage on ammeter too, so the voltmeter show higher voltage.

But why "some current passing through the voltmeter" would cause more current passing through ammeter?

And why some voltage on the ammeter would cause increased voltage on voltmeter?

Last edited by a moderator: Oct 12, 2017
2. Oct 12, 2017

Staff Emeritus
First, in physics we speak of "current", not "amperage".

Second, replace lightbulbs and voltmeters by resistors. Now try and answer the questions.

3. Oct 12, 2017

### Karagoz

Thank you for the correction.

But wouldn't higher resistance on the circuit cause less current since I = V/R (higher the R, less the I is)?

And so adding a voltmeter in the circuit would cause increased resistance, then decreased amps?

And why would more current pass through ammeter in C than in D?

4. Oct 12, 2017

### f95toli

Is the voltmeter in series of parallel to the lightbulb?
What happens if you replace the lightbulb and voltmeter with resistors?
What happens to the total resistance if you have two resistors in parallel?

5. Oct 12, 2017

### NFuller

If the resistors are in parallel, then the resistance will decrease and the current will increase. This is the case in C.
In D, some of the current is bypassing the ammeter through the voltmeter.

As an aside, you should keep in mind that the change in these values shown on the meter will be very very small. Most voltmeters have internal resistances of many megaohms and ammeters have resistances of less than an ohm.

6. Oct 12, 2017

### Karagoz

Yes.

Nothing. But the total resistance will be less than the resistance of the individual resistors.

The total resistance will be less than the individual resistors. Increasing number of resistors connected in parallel will decrease the total resistance.

E.g. if the resistance of lamb is 40, and resistance of Voltmeter is 200. Then in circuit A the sum resistance will be 40 ohm. In circuit C it'll be ca 33, also less. Decreased resistance will cause increased current when the voltage is same.

But why is voltage higher in D, than in C and E?

Last edited by a moderator: Oct 12, 2017
7. Oct 12, 2017

### cnh1995

Assume supply voltage=10V, resistance of the voltmeter=100 ohm, resistance of the ammeter= 1 ohm.

Apply Ohm's law to B, C and E and see what you get. I believe that would answer your "why" questions.

8. Oct 12, 2017

### sophiecentaur

It's a general principle of Physics (of all life, in fact): measuring something will disturb the system and spoil it for the next person.

9. Oct 12, 2017

### CWatters

This exercise is all about the difference between ideal and real meters. I recommend you have a think about what properties you would like your meters to have and then you might understand the consequences and effects of them not being perfect.

10. Oct 12, 2017

### CWatters

Here's a question for you to think about. If you have a multimeter that can measure volts, amps, resistance etc it's a good idea to leave it set on a high voltage range when not in use. Why?

11. Oct 12, 2017

### Karagoz

Isn't voltage the same in parallel circuits? How does the ammeter in D and C affect the voltage on voltmeter?

What about C and E? Ammeter in C or E will show higher values?

And voltmeter in C or E will show higher values?

12. Oct 12, 2017

### cnh1995

Take this question.

In circuit C, what is the voltage across the voltmeter? What is the current supplied by the power supply? Can you find the voltages across and currents through all the three components using Ohm's law and basic series-parallel rules?

It will be easier than explaining it with words. While working it out, you will understand how meters affect the actual circuit parameters and which meter should have very high or very low resistance.

13. Oct 12, 2017

### XZ923

If the ammeter is in fact creating a voltage drop*, then it has a resistance value and the light bulb is no longer the only load in the system. In D, your voltmeter will be equal to the battery voltage since it's connected across both loads. In C, it will read battery voltage minus voltage drop of created by the ammeter's resistance. As far as the voltmeter, its internal resistance* will be connected in parallel with the light bulb, again creating an additional load which will thus increase the system current read by the ammeter.

*-I don't know if it's really germane to this discussion (if not I apologize and hopefully a moderator will remove the comment), but what tools does your book assume you're using? Every voltmeter I've ever used had an impedance in the mega-ohm range to prevent the very thing the text seems to be describing (influencing the system by connecting a parallel load). Likewise, ammeters are usually shunts connected in series with a load, then with a high-impedance meter connected in parallel to the shunt to avoid a voltage drop.

14. Oct 12, 2017

### Karagoz

I assume the bulb has resistance of 2, ammeter 1, voltmeter 10 (I know these are far from realistic values of voltmeter and ammeter, it's not how voltmeter and ammeter works. But it's just to make calculation easier). Voltage of the circuits are 10 V.

In C, the ammeter will drop the voltage by -1 since its resistance is 1. So the voltage on the voltmeter will be 9.

But what about the voltmeter in E, D and B?

Will the voltmeter be higher in D than in B? If so why?

What about E: yhe ammeter in E cause voltage drop on voltmeter?

15. Oct 12, 2017

### Drakkith

Staff Emeritus
Not quite. If the lightbulb has a resistance of 2 ohms and the ammeter 1 ohm, then the total resistance of the circuit (before connecting the voltmeter) is 3 ohms. Applying Ohm's law, the current through the circuit is $I = V/R = 10V/3Ω = 10/3 = 3.333 A$. Applying Ohm's law to the ammeter to find the voltage across it yields $V = IR = (10/3)(1) = 10/3 = 3.333 V$ Applying Ohm's law to the lightbulb yields $V = (10/3)(2) = 20/3 = 6.667 V$.

Assuming the resistance of the voltmeter is extremely high, much higher than 10 ohms, attaching it to the circuit will have a negligible impact and it will measure 6.667 V across the lightbulb.

Now, if we have a poor-quality voltmeter, such that the resistance is not quite so high, then it may impact the circuit in a non-negligible way. If we use 10 ohms as the resistance of the voltmeter, then attaching it across the lightbulb in C will put the two in parallel with each other. The equivalent resistance of these two components will be $\frac{1}{R}=\frac{1}{R1}+\frac{1}{R2} = \frac{1}{2}+\frac{1}{10} = \frac{5}{10}+\frac{1}{10}=\frac{6}{10}$. Dividing both sides by their denominators yields $R = \frac{10}{6} = \frac{5}{3} = 1.667 Ω$.

Total resistance of the circuit is now $1.667 + 1 = 2.667 Ω$. The current is now $I = V/R = \frac{10}{2.667} = 3.75 A$.

So attaching our (poor quality) voltmeter has increased the current from 3.333 Amps to 3.75 Amps.

E is identical to C. The order of the components does not matter in this case. As for what's happening with the voltmeter in B and D, I can't answer that. I can't see why the voltmeter would read a higher voltage.

Hmmm. The last part of this doesn't make much sense to me. The voltmeter is in parallel with the entire other branch (lightbulb in B and lightbulb + ammeter in D) and should read the same voltage as far as I know. Inserting the ammeter into circuit D merely increases the resistance of that branch, but has no effect on the current through the voltmeter, so the voltmeter's reading shouldn't change. The voltage drop across the ammeter + lightbulb branch is split among both devices, but the total voltage drop for that branch remains the same as before. I don't know if the book is wrong or if I'm wrong (I'm far from an expert. I'm in the middle of ECE 220, "Basic Circuits").

16. Oct 12, 2017

### NFuller

Something is definitely up with the explanation in the text. If the power supply is a voltage source, then the volt meter in both B and D will read the same voltage; they will read the voltage of the voltage source. If the power supply is a current source, then the voltages will be different since the total resistance of the circuit will change.
I believe the text is trying to explain the difference between only C and D here. Since the ammeter has some internal resistance, there will be a voltage drop across the ammeter. Thus the voltage measured in D will be the voltage of the power source and the voltage in C will be the voltage drop across the light bulb.

17. Oct 12, 2017

### Karagoz

Would the ammeter cause voltage drop in E like it does in C?

18. Oct 12, 2017

### davenn

yes, why not ? ... tell me why you think it wouldn't ?

19. Oct 12, 2017

### Karagoz

I don't know how the "voltage drop" works. I thought it like this: electrons like some cars in some on road that move from (-) to (+). So in C, the cars slow down when moving through ammeter (like cars slow down in speed bumps), so when they reach voltmeter they have "lower speed".

But in E, since the cars reach the voltmeter first before any speed bump, there's no drop on speed.

Similarly electrons lose some voltage when moving through the ammeter before reaching the voltmeter.

And this is what's written in the text book, I'm translating it more precisely (from Norwegian):

(the source is a voltage source, DC).

"If we want to measure both the voltage on, and the current through the light bulb at the same time, we can connect the measuring devices in two different ways (the picture above). Do you think these two connections will give same results? If that should be the case, the current through the voltmeter and the voltage on the ammeter must be equal to zero. It's same as saying that the resistance in the voltmeter must be infinitely big, and the resistance in the ammeter must be equal to zero. In reality some current goes through the voltmeter, so that the ammeter in 11a show a bit too high current. Furthermore there are some voltage on the ammeter too, so the voltmeter in 11b shows a bit too high voltage."

20. Oct 12, 2017

### davenn

Hi Karagoz

no, it doesn't really work that way
A hint, in a series circuit, the measured current is the same everywhere

electrons don't loose voltage as they don't have voltage. An electron is a single negative charge

In 11a, you are measuring the voltage drop across the lamp ONLY. This will be a portion of the total voltage drop
In 11b, you are measuring the voltage drop across the lamp AND the ammeter and the measured value will be the value of the battery voltage

you could replace the lamp and ammeter with 2 resistors of different values ( I think some one mentioned this in an earlier post)

lets make the battery/voltage source 10V, the lamp a 100 Ohm resistor and the ammeter a 1 Ohm resistor

Knowing the voltage source = 10V and the values of the 2 resistors, we can work out the current in the circuit

tell me ......
1) what is the total resistance ? _________

2) knowing that, what is the current ? ___________

do you know how to do that using Ohm's Law ?

now knowing the current and the individual resistances you can work out the voltage drop across each resistor
Again use Ohm's Law

Vdrop across 100 Ohm = __________

Vdrop across 1 Ohm = ____________

lets see how you go with those problems

Dave

21. Oct 13, 2017

### BvU

The analogy is wrong: a pile-up of charge does not occur in that amount. For practical purposes it is more realistic to consider the charge carrier stream as incompressible, so:

In that respect the analogy with water flowing through wider or narrower pipes is a little better: you have to push harder to get it through the narrower pipe so the pressure difference over a narrow section is bigger than over a wider section.

22. Oct 13, 2017

### CWatters

Everything is relative. Try using your volt meter in circuits that have high value resistors and you may see exactly the effect described. Some op-amp circuits for example.

Likewise an ammeter measuring high currents can cause a significant voltage drop in some circuits. For example some batteries have an internal impedance in the milli Ohm range. The voltage drop due to the meter can be more than that due to the internal resistance of the battery.

23. Oct 13, 2017

### Drakkith

Staff Emeritus
Voltage is a measure of the electric potential energy per unit of charge between two points, much like how we can measure the gravitational energy between two heights. The voltage drop of a component is the amount of this electric potential energy lost per unit of charge passing through the component. Note that a unit of charge is a coulomb in this case, not a single electron. Volts, the SI unit of voltage, is measured in joules per coulomb. So in your problem above, if the circuit consists of just the voltage source of 10 volts and a resistor, then the resistor has a voltage drop of 10 volts, or 10 joules per coulomb. So for every coulomb that passes through the resistor, 10 joules of electric potential energy is converted into heat. How quickly this amount of charge passes through depends on the resistance of the resistor. A higher resistance means less current is flowing and so it takes longer for one coulomb to pass through. Hence the power absorbed by the resistor is less (power being energy/work per unit of time).

A silly analogy: let's say I were to get in a hot air balloon with a bunch of bowling balls and drop them on top of various objects on the ground. One of my bowling balls comes crashing down through your roof, passes through your kitchen table where you were just having a nice omelette, and comes to rest on the floor. If you were able to measure the energy lost as the bowling ball passed through the roof and then through your table, you could say that an identical roof to your own has a "potential drop" of X, and a table identical to your kitchen table has a "potential drop" of Y, where X and Y are the amount of energy spent per bowling ball (or per 10 bowling balls, or per 100 if we wanted to define it that way) to pass through the roof and the table respectively.

Similarly, we can measure the amount of energy spent per unit of charge while passing through a resistor or another component. A crucial difference here is that electrons are not like bowling balls. An electron moving through a resistor is not quite like a bowling ball smashing through a roof. The resistor is not damaged whatsoever because electrons are passing through. The electrons, instead of spending their energy by smashing the resistor, instead undergo various other interactions with nuclei and other electrons that cause them to lose energy. I won't go into detail on those interactions, as they are quite beyond my expertise and well outside the scope of this thread.

24. Oct 13, 2017

### Karagoz

1) Total resistance is: 101 ohm.
2) Current is: 10/101 = 0.099

Vdrop across lamp: 0.099 * 100 = 9.9 V. This is voltage on the lamp.
Vdrop across ammeter: 0.099 * 1 = 0.099 V. This is voltage on the ammeter.

In circuit C and E it measures voltage of the lamp only, but in D it measures voltage of both the lamp and the ammeter. That means in D, more current will pass through the voltmeter, since there are

In circuit C and D, the current will be divided in two in the first branch point, then get together in the second branch point.

In circuit D, because there are more resistors (the lamp and the ammeter) than in C (only the lamp) after the branch point, then when the current is divided, there'll be more current in I2 in D than in I2 in C. And more current causes more voltage (when the resistance is constant).

25. Oct 13, 2017

### Drakkith

Staff Emeritus
The current in I2 is determined solely by the resistance of I2 and the voltage across it. In C, the voltage drop across I2 is slightly less than the voltage source because the ammeter is in series with I1 and I2 and will have non-zero resistance and voltage drop. So some of the voltage is spent in the ammeter instead of in I1 and I2. In D, the ammeter is not in series with I1 and I2, it is part of I1. Since I2 is in parallel with I1, which now includes the ammeter, and has no other components in series with it, the voltage across it is equal to the voltage source. Similarly, the voltage drop across I1 in D is also equal to the voltage source, though it is divided between the ammeter and the resistor.