Limiting Error of Electrical Instruments

In most indicating instruments, the accuracy is guaranteed to a certain percentage of full-scale reading. Circuit components (such as capacitors, resistors, etc.) are guaranteed within a certain percentage of their rated value. The limits of these deviations from the specified values are known as limiting errors or guarantee errors. For example, if the resistance of a resistor is given as 500 Ω ± 10 per cent, the manufacturer guarantees that the resistance falls between the limits 450 Ω and 550 Ω. The maker is not specifying a standard deviation or a probable error, but promises that the error is no greater than the limits set.

Example-

A 0-150-V voltmeter has a guaranteed accuracy of 1 per cent full-scale reading. The voltage measured by this instrument is 83 V. Calculate the limiting error in per cent

Solution-
The magnitude of the limiting error is
                                                                    0.01 × 150 V = 1.5 V
The percentage error at a meter indication of 83 V is
                                                                1.5 83 × 100 per cent = 1.81%
 It is important to note in Example  that a meter is guaranteed to have an accuracy of better than 1 per cent of the full-scale reading, but when the meter reads 83 V, the limiting error increases to 1.81 per
cent. Correspondingly, when a smaller voltage is measured, the limiting error will increase further. If the meter reads 60 V, the per cent limiting error is 1.5 60 × 100 = 2.5 per cent; if the meter reads 30 V, the limiting error is 1.5 30 × 100 = 5 per cent. The increase in per cent limiting error, as smaller voltages are measured, occurs because the magnitude of the limiting error is a fixed quantity based on the full-scale reading of the meter. Example  shows the importance of taking measurements as close to full scale as possible.

Measurements or computations, combining guarantee errors, are often made. Example  illustrates such a computation.

Example-
Three decade boxes, each guaranteed to ±0.1 per cent, are used in a Wheatstone bridge to measure the resistance of an unknown resistor Rx. Calculate the limits on Rx imposed by the decade boxes.

Solution-
The equation for bridge balance shows that Rx can be determined in terms of the resistance of the three decade boxes and Rx = R1R2/R3, where R1, R2, and R3 are the resistances of the decade boxes, guaranteed to ±0.1 per cent. One must recognize that the two terms in the numerator may both be positive to the full limit of 0.1 per cent and the denominator may be negative to the full 0.1 per cent, giving a resultant error of 0.3 per cent. The guarantee error is thus obtained by taking the direct sum of all the possible errors, adopting the algebraic signs which give the worst possible combination.

As a further example, using the relationship P =  I²R, as shown in Example, consider computing the power dissipation in a resistor.


Example -
The current passing through a resistor of 100 ± 0.2 Ω is 2.00 ± 0.01 A. Using the relationship            P = I 2R, calculate the limiting error in the computed value of power dissipation.

Solution-
Expressing the guaranteed limits of both current and resistance in percentages instead of units, we obtain
                                                          I = 2.00 ± 0.01 A = 2.00 A ± 0.5%
                                                         R = 100 ± 0.2 Ω = 100 Ω ± 0.2%

If the worst possible combination of errors is used, the limiting error in the power dissipation is        (P = I²R):
                                                          (2 × 0.5%) + 0.2% = 1.2%

Power dissipation should then be written as follows:

                                        P = I² R = (2.00)²× 100 = 400 W ± 1.2% = 400 ± 4.8 W