Learning Event 3:
DETERMINE METER SENSITIVITY
1.
you determine the resistance of the meter to decrease errors in readings.
This
points to the conclusion that meters are delicate devices which can respond to
small forces.
2.
The resistance of a meter movement and the maximum current permitted to flow
through it are so small that the use of an unshunted meter movement as a measuring
device is very limited.
A typical meter movement has 50 ohms of resistance and
gives full-scale deflection with 1 milliampere of current through the meter coil.
Such a meter movement has a 50 millivolt voltage across it at full-scale deflection
as shown by the formula:
E = 1 x R = 0.001 x 50 = 50 millivolts
The above meter movement is limited to measuring
current
values
from
0
to
1
milliampere and voltage from 0 to 50 millivolts.
3.
much larger values of current and voltage?
Let's discuss sensitivity in more
detail and see how it is possible to measure large values of current, voltage, and
resistance with the applicable meters.
The sensitivity of a meter movement is inversely proportional
to the amount of current that causes the indicator to deflect full scale.
The
smaller the current required for full-scale deflection, the more sensitive the
meter movement.
For measuring current in electronic equipment, ammeters with a
sensitivity of 0.1 ampere or even 1 milliamperes are used.
Meters with a
sensitivity of 100 microamperes are common.
(1) To understand how to determine applicable shunt resistors for an
ammeter, let's study the circuit in Figure 1-13.
Since current through the two
parallel branches divides in a ratio inversely proportional to the branch
resistances, it is possible to calculate the current through the coil as well as
the total current in the circuit in which current is being measured.
(2) In the circuit shown in Figure 1-12 you can find the current in the
shunt (Is) and the total current (It) in the circuit.
For example, if the shunt
resistance (Rs) is equal to 1/5th the value of the resistance of the coil (Rc), and
current through the coil (Ic) is 0.5 milliampere, there is 5 times as much current
through the shunt (Is) as through the coil (Ic), because the current divides in
2.5(5x0.5) milliamperes.
The total current in the circuit is 3 (0.5-2.5)
milliamperes. The total current in the circuit is 3 (0.5+2.5) milliamperes.
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