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Consider the picture below. An ammeter is shown to the left, a voltmeter to the right.

enter image description here

  • They consist of a d'Arsonval galvanometer, which is a coil (with a pointer attached) that can rotate in a magnetic field but is held back by a small spring. When current runs through the coild, it will be deflected proportionally to that current.

  • The extra resistance $R_{sh}$ or $R_s$ is added to extend the range to be measured to larger values; that is, to extend the scale.

Now, an ammeter ideally has negligible resistance and is added in series with other components. The voltmeter ideally has infinite resistance and is added in parallel over a component.

When looking at the setups in the picture there are two finite resistances in each. So, which is supposed to be the the negligible and the infinite resistance, respectively?

In addition

How will it ever work if e.g. $R_s$ is so large that no current flows through it? The voltmeter is based on $R_s$ being finite; else there will be no current and no deflection of the coil.

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  • $\begingroup$ And that is not how one does high precision measurements. This is how a high precision voltage measurement is done with instruments with finite internal resistance: en.wikipedia.org/wiki/Potentiometer_%28measuring_instrument%29. The method is still being used in metrology and inside very high precision digital voltmeters. $\endgroup$ – CuriousOne Jan 6 '15 at 8:41
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To really understand the answer to your question we should make sure we understand how the ammeter works. Once we understand that we will see that neither of the resistors in the meter is "negligible", but rather that the total resistance of the meter must be negligible compared to the resistance of whatever you're measuring.

How an ammeter works

Suppose you would like to measure the current flowing through a resistor of resistance $R_l$ (the $l$ stands for "load"). In series with $R_l$ I attach my ammeter, which has the circuit as in the diagram. The series resistance of the entire ammeter circuit is

\begin{align} R_{\text{ammeter}} &= R_{sh} || R_c \\ &= \frac{R_{sh}R_c}{R_{sh}+R_c} \, . \end{align}

If we assume $R_{\text{ammeter}} \ll R_l$ then adding the meter to the original circuit doesn't change the total resistance appreciably Therefore, we can be sure that the total current with the meter hooked up is very close the total current without the meter hooked up, which is what we wanted to measure in the first place. Denote this current $I$. Since $I$ flows through the resistance $R_{\text{ammeter}}$ we know that the voltage across the ammeter is

$$V_{\text{ammeter}} = I R_{\text{ammeter}} \, .$$

This voltage is dropped across both $R_c$ and $R_{sh}$. In particular, the current through $R_c$ is

$$ I_c = V_{\text{ammeter}} / R_c = I \frac{R_{\text{ammeter}}}{R_c} \approx I \frac{R_{sh}}{R_c + R_{sh}} \, . \qquad (*) $$

We get to pick $R_c$ and $R_{sh}$ because we design the meter. Therefore we know the value of $R_{sh}/(R_c + R_{sh})$, so by looking at the deflection of the needle, which tells us $I_c$, we can extract $I$.

What is $R_{sh}$ for?

If $R_{sh}$ were not there (i.e. $R_{sh}\rightarrow \infty$), then we would have $I_c = I$. In other words, all the current would flow through the coil. But what if the current we're measuring maxes out the deflection of the needle? That's the point of $R_{sh}$. By adding $R_{sh}$ the fraction of the total current $I$ which goes through the coil is reduced (see equation $(*)$). In particular, if $R_{sh} \ll R_c$ we find $I_c = I (R_{sh}/R_c)$.

Your actual question

So, which resistance is negligible in the ammeter? Neither! The thing which has to be true is that $R_{\text{ammeter}} \ll R_l$. You can't say that $R_{sh}$ or $R_c$ is small in an absolute sense. It always has to be compared against the resistance of the load.$^{[a]}$

For the voltmeter case the situation is similar but sort of flipped. If you do the calculation yourself you'll find that the requirement for the voltmeter to work well is that its resistance must be much higher than the load resistance: $R_{\text{voltmeter}} \gg R_l$.

In real life, voltmeters and ammeters use active circuits to effectively ensure that their resistance is smaller than (for the ammeter) or larger than (for the voltmeter) whatever it's connected to.

$[a]$: This is, in a sense, the thing you have to understand to be good at electronics. You always have to remember that when you connect circuits together, whether they're meters or various stages of an amplifier, what happens to each circuit depends on the properties of the ones to which you're connecting it. There are ways to make this fact less of an issue, often by constructing circuits where the output impedances are always lower than the input impedances.

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  • $\begingroup$ Thank you for a great answer. Can you elaborate a little more on how you can read $I_c$? You say by looking at the deflection of the needle, so how is the needle deflection giving us $I_c$? I know it is proportional to, but how do you get the exact value? $\endgroup$ – Steeven Jan 6 '15 at 10:57
  • $\begingroup$ @Steeven: The current through the coil $I_c$ creates a magnetic field. Presumably your meter is constructed by attaching a needle to a magnet and a spring. As the magnetic field increases it pushes against the spring and deflects the needle. The stiffness of the string and the shape of the coil have to be calibrated if you want to be able read the magnetic field in absolute units. However, with an uncalibrated meter you can still determine that e.g. one current is twice as big as another, etc. $\endgroup$ – DanielSank Jan 6 '15 at 16:47

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