In the previous section
, we learned that power supplies are not the ideal
instruments we may have thought they were. The applied load can have an
effect on the terminal voltage. Fortunately, since today's supplies have such
small load regulation factors, the change in terminal voltage with load can
usually be ignored for most applications. If we now turn our attention to the
various meters we use in the lab, we again find that they are not totally ideal:
Whenever you apply a meter to a circuit, you change the circuit and
the response of the system. Fortunately, however, for most
applications, considering the meters to be ideal is a valid
approximation as long as certain factors are considered.
Any ammeter connected in a series circuit will introduce resistance to
the series combination that will affect the current and voltages of the
Fig. 1: Including the effects of the internal resistance of an ammeter: (a) 2 mA scale; (b) 2 A scale.
The resistance between the terminals of an ammeter is determined by
the chosen scale of the ammeter. In general,
for ammeters, the higher the maximum value of the current for a particular scale, the smaller will the internal resistance be.
In Fig. 1(a)
, the ammeter scale is $2mA$, which means the maximum current flow through ammeter should not exceed $2mA.$ while Fig. 1(b)
, shows the ammeter with scale $2A$, which is far higher than $2mA$ scale. The $2mA$ scale ammeter series resistance must be high e.g $250Ω$ enough to resist the current flow in the meter circuit to limit the value upto $2mA$ current . While the $2A$ scale ammeter series resistor must be small enough e.g $1.5Ω$ to prevent current flow in the meter circuit up to $2A$ current. In the ideal situation of ammeter connection to the circuit, we ignore the internal resistance of the circuit inside ammeter instrument due to low internal series resistance.
How does ammeter effect circuit analysis?
In the Fig. 2(a)
, the ideal consideration of ammeter does not have any effect on the current flow in the series circuit. while in Fig. 2(b)
, the internal resistance $250Ω$ is summed with the load resistors and the result is shown as:
Fig. 2: Applying ammeter to a circuit with resistors in KΩ: (a) Ideal. (b) Practical.
For example, in Fig. 2(a)
, for an ideal
, the current displayed is $0.6 mA$ as determined from
$$I_s = E / R_T = 12 V/20 kΩ = 0.6 mA$$
If we now insert a meter with an
internal resistance of $250 Ω$ as shown in Fig. 2(b)
, the additional
resistance in the circuit will drop the current to $0.593 mA$ as determined
$$I_s = E/R_T = 12 V/20.25 kΩ = 0.593 mA$$
Now, certainly the current has dropped from the ideal level, but the difference in results is
only about 1% - nothing major, and the measurement can be used for most purposes. If the series resistors were in the same range as the $250 Ω$ resistors, we would have a different problem, and we would have to look
at the results very carefully.