ELECTRICAL MEASURING INSTRUMENTS

By | December 5, 2020

It is understood that moving electrons will produce a magnetic field which of course can be attracted or repelled by other magnetic sources. This condition is used as the basis for making electric motors and simple electric meters to measure currents and voltages. The basic construction of an electric meter is shown in Figure 1.

Figure 1. Basic construction of electric meter

This basic meter consists of a permanent magnet in the form of a horseshoe with round poles. A coil with a core of malleable iron is positioned between the U and S poles so that it rotates freely. A pointer is attached to the coil and will move when the coil rotates.

The electric current to be measured is passed to the coil so that the coil will produce a magnetic field (electro-magnetic). The electro-magnetic poles of will interact with the permanent magnetic poles so that the coil rotates according to the amount of current in it.

Basic Meter 

The most important use is as a current measuring tool and a voltage measuring device. In use as an ampere meter (ammeter), striving for all current at a branch point that is measured can be through the ammeter. The goal is that at the branch point it appears as if a short circuit occurs, that is, it has a low resistance and a low voltage drop. For use as a voltmeter (installed between two points), try to keep the current passing through the meter (voltmeter) as small as possible. The goal is that the two connection points appear as if they were open circuits, that is, they have a very large resistance or very little current passes. Figure 2 shows how the two electricity meters are installed in the circuit. A basic meter typically requires a current of 1 mA (and about 0.1 V) to create afull-scale deflection.

Figure 2.  Installation of a voltmeter and ammeter in the circuit.

Base Meters as Ammeters 

We can make a meter with a full-scale current indication (measuring limit) greater than its basic capability (but with the same full-scale voltage indication capability), by installing aresistance shunt in parallel with the meter.

Figure 3.3 Full-scale representation of the base meter:
a) ammeter and
b) voltmeter.

Figure 3 (a) shows a meter with a full scale designation (measuring limit) of 1 mA which will be converted to 1 A. Using the current dividing principle, the shunt resistance value is obtained:

$R_{p}=\frac{{R_{m}}}{(n-1)}$                                (1)

where n shows the magnification of the meter limit. For the above case, n is about 1,000 times and thus  Rp = 25 $\Omega$/ 999 = 0,025 $\Omega$

A multimeter usually have several scales measuring limit by connecting with corresponding terminals. In this case the shunt resistance is installed in the meter circuit. Figure 3.4 shows meters with measuring limits of 2 and 10 A made using the above principle.

Figure 4 Installation of a shunt to change the meter measurement limit.

Basic Meters as Voltmeters 

We can also increase the measuring limit of a voltmeter by n times its basic measurement limit (with the same full scale current), that is, by installing an external resistance in series. For the circuit in Figure 3.3-b shows a basic meter with a maximum current measuring limit of 1 mA will be used to measure a voltage of 2 V. The total resistance (outer resistor + meter resistor) is

2 V / 1 mA = 2000 Ω

thus resistance The outside that must be installed is

RS = (2000 – 25) Ω = 1975 Ω

On a voltmeter with several measuring limits it is usually equipped with a switch to select the appropriate series resistor.

Figure 5 Setting the meter limit by installing a resistor.

Example 

Suppose that a meter base 50$\mu$A has a resistance of 3000 Ω. Try to design a multimeter that can be used for measurements up to the measuring limits of 100 $\mu$A, 1 mA, 1 V and 10 V. A suitable circuit is shown in Figure 5.

Answer: 

At the limit measure 100A, a current of 50$\mu$A must flow through the meter and the resistance  (R1 + R2). Thus (R1 + R2)= 3000 Ω

At the 1 mA measurement limit, a current of 50 $\mu$A flows through ( R2 + 3000 Ω) and the balance of 950 $\mu$A through R1. Thus,

At the limit measure 1 V, drain current of 100 $\mu$A through meter and 50 $\mu$A via (R1 + R2). On the meter there is a voltage of

50$\mu$A x  3000 Ω =  0.15 V

thus the voltage at R3 is, or

R3 = 0.85 V / 100 $\mu$A = 8500 Ω

In the same way,  R3 = 9.85 / 100$\mu$A = 98.5  kΩ

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