In electricity and electronics there are the most varied types of measuring instruments. The most common for day to day use
are the ammeter and the voltmeter. Today, with the advent of digital technology, these instruments were allocated
in the same device that became known as the multimeter. There are the most varied
types of multimeters. The most common incorporates a voltage meter (continuous and alternating), electric current and an ohmmeter.
These models are very low cost. The most sophisticated ones, at a higher cost, can incorporate AC current meters, capacitance meters,
inductance, frequency, as well as
incorporate a diode and transistor tester, besides as other features.
On this site we will limit ourselves to studying the ammeter and the voltmeter
from the analogical point of view , since the idea is to transmit a knowledge of the principles of their operation.
The basic instrument used to manufacture ammeters is called d'Arsonval's galvanometer. It is a device composed of a coil with iron core mounted between the poles of a permanent magnet. Helical springs are used to cushion the movement of the coil, as well as to carry the current
to the coil. The coil, when there is a flow of the electric current, generates a magnetic field that interacts with the magnetic field of the permanent magnet. This interaction generates a torque, which causes the coil to rotate around its own axis. Attached to the mobile coil, we have a pointer that will mark the deflection on a scale. As the deflection
is directly proportional to the current flowing through the coil, then we have a current meter based on electromagnetic principles.
Figure 16-01
Usually the coil is constructed with very fine copper wire (finer than a human hair) and as its length is very long, obviously this will result in a resistance electrical to the passage of electric current.
This resistance is called internal resistance of a galvanometer represented by Ri in the Figure 16-01. And this resistance is very important for the design of ammeters. What's more, every ammeter has what we call the full scale, which is nothing more than the maximum current value that the ammeter can measure.
Normally galvanometers have a full scale, that is, the maximum current that can be measured, of the order of microamper or milliamper. However, if we want to measure higher valued currents, we have to make some modifications to reach our goal. Suppose we have a galvanometer with Ig = 1 mA of full scale and resistance
internal of Ri =50 ohms. We want this instrument to measure a current of 500 mA. One of the ways used is to employ the so called SHUNT circuit, which is simply to add a suitable value resistor in parallel with the instrument.
Notice that if our instrument measures 1 mA, this means that by the resistor we put in parallel it should pass a current of 499 mA, totalizing
the 500 mA that is our goal. Then, to calculate the value of this resistor, we must first calculate which voltage causes the maximum deflection in our milliammeter without the resistor in parallel, that is:
Vg = Ri Ig = 50 0.001 = 0.05 volts = 50 mV
Now we know that we must have a resistor in parallel with the milliamperimeter, which we will call Rsh
(shunt resistor), which has a value of such form that when we have on it a potential difference of 50 mV , a current of
499 mA will circulate.
Naturally, it is very easy to develop an equation that allows us to calculate the value of this resistor for any instrument.
For this, we must know the full scale of the instrument, which will be represented by Ig.
The value of the internal resistance of the instrument, we will represent by Ri. The current flowing through the resistor in parallel will call Ish.
Notice that Ish = I - Ig, where I is the current that will be measured by the instrument. So:
eq. 16-01
Notice that the product Ri Ig is the value of
Vg. So for our problem let's calculate the value of Rsh.
Rsh = Vg / Ish = 0.05 / 0.499 = 0.1 ohm
See how easy it is to design an ammeter with several scales suitable for day-to-day use.
Another useful instrument is the voltmeter. It basically consists of a milliammeter or microammeter associated in series with a resistor of adequate value, for
determine the scale that we are interested.
Let's design a voltmeter using a milliamperimeter with the same characteristics as the previous item, that is,
1 mA of full scale and internal resistance of 50 ohms. We already know that when we apply a voltage of
50 mV, our milliammeter has the maximum deflection, indicating that the current that is
circulating through it is 1 mA. Therefore, if we want to measure a voltage of 10 volts, we must put a resistor in series in such a way that there is a potential difference of 10 - 0.05 = 9.95 volts.
Since the circulating current is 1 mA, then the value of the series resistor, here called Rs, and again using Ohm's law, it will be:
Rs = Vg/ Ig = 9.95 / 0.001 = 9 950 ohms
In instruments of electrical measurements it is common to define a property called sensitivity, S, as the inverse
of the current that the instrument is able to measure at its maximum deflection (ie full scale). But the inverse of the current is nothing more than ohms per volt.
In this way, the instrument of the above example has a sensitivity of:
S = 1/ Ig = 1 / 0.001 = 1 000 ohms/volt
In practice, it is possible to use a voltmeter with sensitivity value
equal to or greater than 20 000 Ω/V. This means that our galvanometer must be a microammeter with a scale of
50 µA.
See more practical measurement instruments by clicking in the tab problems, or
Here!
We must be aware of the fact that when we want to measure a
current in a circuit with an ammeter, we must open the circuit at the point where we want to carry out the measurement, and insert the ammeter IN SERIES with the circuit. NEVER put a ammeter in paralell with the circuit because, as a rule, it has a low resistance, it will put short circuit the point to be measured. This could cause serious damage to the circuit and the measuring instrument itself.
By inserting the ammeter in series with the circuit, this causes an imprecision in the reading of the electric current, since with the series ammeter there was a increase of the electrical resistance in the circuit and consequently the reading will be INFERIOR to real.
Figura 16-02
The real current I that flows through the circuit without the ammeter is given by I = V / R. However, as we can see in the Figure 16-02 interrupting the circuit and introducing the ammeter in series with the circuit (between points A and P), of course we must add the value of Ri to R and the measured current will be given by:
Imeasure = V/ (Ri + R)
Therefore, as there was an increase in the resistance of the circuit, the current measured by the ammeter will be less than the actual current I, that is, Imeasure < I.
As was evident in item 3, when we use a voltmeter on a given scale,
this instrument has a certain resistance value between its terminals.
To measure the voltage between two points of a circuit, we know
that the instrument should be in parallel with the components that are between
the two points mentioned. Therefore, this resistance will interfere with the final
measurement. Let's understand how this happens.
Figura 16-03
By inspection in the circuit of the Figure 16-03, we know that between points a-b, considering
that we do not have the voltmeter inserted in points a-b, there is a
potential difference of 10 volts (just apply a resistive voltage divider).
When we insert the voltmeter at the points a-b, of course we put the resistance
of the instrument in parallel with the resistor of 4 kΩ. Suposing that
the voltmeter has a sensitivity of 1 000 Ω / V and we have chosen the
10 volts scale, this means that the internal resistance of the instrument is
10,000 Ω. Now, what we have between points a-b is a value of
resistance resulting from the parallel of 4 kΩ and 10 kΩ. This results
at a resistance of value equal to 2 857 Ω.
Now we will calculate what the new voltage value will be between points a-b.
Applying a resistive voltage divider, we have:
Vab = 20 (2 857 / 4 000 + 2 857) = 8.33 volts
Notice that we have measured a voltage below the actual value, causing a very reasonable error. Using the same idea, let's assume that the instrument used had a sensitivity of 20,000 Ω / V. In this case, for a scale of
10 volts we have an internal resistance of the instrument of 200,000 Ω.
Calculating the parallel of 4 kΩ and 200 kΩ yields a value equal to
3 922 Ω. Applying the resistive voltage divider, we have:
Vab = 20 (3 922 / 4 000 + 3 922) = 9.90 volts
Now you can understand why we should use an instrument with great sensitivity,
because this allows us to measure values very close to the real value.
