Question

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 =

Answer #1

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 I_{o} is given as,

=

= 0.0744 A

% change =

=

= 0.8%

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