Problem 16-2 Source:
Problem elaborated by the author of the site.
Consider the circuit shown in Figure 16-02.1. Calculate read error when se
insert an ammeter in series with each resistor of the circuit, knowing that in the scale
of 3 A the instrument has an internal resistance of equal value
the 0.2 Ω.
Solution of the Problem 16-2
To measure with the instrument the current circulating through the resistor R1,
we must open the circuit and insert the ammeter in series. In this way, the total resistance
circuit supply to the voltage source is Rtotal1 = 2 + 0.2 = 2.2 Ω. Therefore, the current that the instrument will measure will be:
Note that the insertion of the ammeter in series with the circuit, caused a
measurement error of the order of 1.5-1.36 = 0.14 A . So, the
measured value is less than the actual value.
To measure the error in R2, the procedure is the same, remembering that the
total resistance will be: Rtotal2 = 1.5 + 0.2 = 1.7 Ω. Logo:
In this case, the error was of 0.24 A.
