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precision class greater than 4 are used as demos. The accuracy class of
            the instrument is indicated on its scale.

                    Example of calculation of developments measured electricity

                                 for classes of accuracy of equipment.
                  In  many  cases,  the  values  of  the  desired  electrical  value  are
            obtained by the results of one measurement. Therefore, a special role

            is  played  by  way  of  estimating  the  errors  of  measurements.  Let's
            consider individual cases.


                  Direct measurement of electric size.
                  For  example,  an  ammeter,  whose  scale  is  calculated  at  1A,  we
            will conduct a one-time measurement of current strength of 0.8A. The
            accuracy class of the device is 1. Given that the accuracy class of the

            instrument is its error, expressed in percentages according to formula
            (1.3), we will have

                                                                 I 
                                                      0,01        ,
                                                               1A

            where  the  absolute  error  of  measuring  the  current  strength  of  this
            instrument  is  I = 0.011A = 0,01А  and  therefore  the  value  of  the

            current strength obtained should be considered equal to

                                                  І = (0,8 0,01) А

                  Indirect measurement of electric size.
                  Indirect  measurements  of  some  physical  value  are  such
            measurements  when  the  desired  value  is  obtained  as  a  result  of

            calculations  on  the  results  of  measurements  of  a  number  of  other
            quantities.  For  example,  according  to  the  Ohm  law  for  a  circle,  the
            resistance of a conductor can be calculated by the formula

                                                            U
                                                       R                                        (1.4)
                                                             I

                  But  for  this  purpose,  it  is  necessary  to  conduct  direct
            measurements of the current and voltage. Let the current is 0.8A, and

            the voltage is 160V. Then, by the formula (1.4), we find the resistance,
            which is equal to





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