A nonideal ammeter that has an internal resistance of 0.303 ? is connected in series with a 3.00-V battery and a 40.0-? resistor.
By what percentage does the presence of the ammeter change the current measurement?
Iammeter?I0/I0 =
Current flow through battery is given as,
=
= 0.075 A
When Ammeter is connected in series, total resistance = R + r
where 'r' is resistance of the ammeter,
Hence Current in Ammeter Io is given as,
=
= 0.0744 A
% change =
=
= 0.8%
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