According to my textbook, it is said that, for an ammeter:

It is essential that the resistance $R_A$ of the ammeter be very much smaller than other resistances in the circuit. Otherwise, the very presence of the meter will change the current to be measured.

Which makes definite intuitive sense to me. One would hope its resistance is very low, because otherwise it would be like trying to measure the speed of a car by putting spikes under its tires just before.

However, what it says for voltmeters is not intuitive for me:

It is essential that the resistance $R_V$ of a voltmeter be very much larger than the resistance of any circuit element across which the voltmeter is connected. Otherwise, the meter alters the potential difference that is to be measured.

Why would it having a small resistance cause a hamper on anything? Why does restricting the current of charge carriers a necessity in measuring potential difference in the circuit? This is coming from someone who is very unfamiliar with circuits in general, but trying to learn incrementally.

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    Did you know there is an Electrical Engineering Stackexchange? If you search there you should find this question has already been asked many times. – The Photon Feb 18 at 17:07
  • Consider the extreme: What if the resistance of the voltmeter is really low... like nearly a short. Can you imagine how putting a short at some arbitrary position in your circuit might have undesirable side effects? – Cort Ammon Feb 20 at 23:40

Consider how a voltmeter and an ammeter are connected to the circuit and how the resistance of the meter will alter the total resistance of the circuit.

A voltmeter should have a much larger resistance compared to any circuit element across which it is connected because a low internal resistance voltmeter would draw a current from the circuit which changes the very voltage across the circuit element you are trying to determine. A very high internal resistance and thus very small current through the voltmeter ensures that there is a negligible disturbance of the currents in the circuit and thus of the voltage to measured.

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    Oh, so you're saying it's analogous to someone trying to count the number of puppies running around on a given road from your house on the road where you can measure it, but, given their spryness, the puppies will try to run into your home to greet you, which would cause an inaccurate number of puppies on the road to be counted if some have retreated into your house? So your door must be kept closed (high resistance) rather than open (low resistance) to make sure you can get the right number? That was a bizarre analogy, but what came to mind first. – sangstar Feb 18 at 17:28
  • Is my understanding of why an ammeter needs a low resistance correct, though? – sangstar Feb 18 at 17:29
  • @sangstar- Yes your ammeter inserted into the current path of a circuit hs to have an internal resistance that is a slow as possible in order not to significantly change the current in the measured loop. – freecharly Feb 18 at 18:21
  • And is my analogy the correct idea for voltmeter? – sangstar Feb 18 at 18:49

A simple model for a voltmeter is a known resistance, say $R_m$ in series with an ammeter. The current reading can then be recalibrated in voltage by Ohm's Law. The term potential difference gives a clue how a voltmeter is used to measure the voltage across some resistor, say $R$: you place the voltmeter across the resistance $R$. Measuring the current through $R_m$ gives the voltage across both $R_m$ and $R$. You want the voltmeter to influence the current flowing through the original circuit so that you need the value of $R_m$ in parallel with $R$ to be as close to $R$ as possible, which means $R_m$ is much larger than any resistors in the circuit. There is, of course, a trade-off, making $R_m$ infinite means no current flows through the ammeter.

Lets say we have an ideal 2V supply and connect two identical resistors of resistance $r\Omega$ in series across that supply. We should have 1V across each resistor.

Now we take our bog standard DMM which has an input resistance of $10\mathrm{M\Omega}$ and try to use it to measure the voltage across one of the resistors. What is our measured voltage?

Well the meter is now in parallell with our resistor, so we need to take that into account when we run our voltage divider calculation, the value of our resistor and meter in paralell is $\frac{r \times10^7}{r +10^7}$, so our voltage divider equation becomes

$$v = 2\frac{\frac{r \times10^7}{r +10^7}}{\frac{r \times10^7}{r +10^7} + r} = 2\frac{r \times10^7}{r \times10^7 + r({r +10^7})} = 2\frac{10^7}{10^7 + r +10^7} = \frac{1}{1+\frac{r}{2*10^7}}$$

Now lets try plugging in some values for r.

$$ r = 10^0 \rightarrow v \approx 0.99999995$$ $$ r = 10^3 \rightarrow v \approx 0.99995000$$ $$ r = 10^6 \rightarrow v \approx 0.95238095$$ $$ r = 10^7 \rightarrow v \approx 0.66666667$$ $$ r = 10^9 \rightarrow v \approx 0.01960784$$

We see that when r is much smaller than the input resistance of our meter our meter only has a tiny effect on the circuit, probablly too small to measure, but as the resistances in our circuit get larger our meter has a progressively larger affect on the circuit until it is completely changing the behaviour.

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