A galvanometer having a resistance of 30.0 ? has a 1.00-? shunt resistance insta

Question

A galvanometer having a resistance of 30.0 ? has a 1.00-? shunt resistance installed to convert it to an ammeter. It is then used to measure the current in a circuit consisting of a 16.0-? resistor connected across the terminals of a 45.0-V battery having no appreciable internal resistance.
(a) What current does the ammeter measure?
A

(b) What should be the true current in the circuit (that is, the current without the ammeter present)?
A

(c) By what percentage is the ammeter reading in error from the true current?
%

Explanation / Answer

PART-A

Rm=(30*1)/(30+1)= 30/31ohm

Ic= 45/(16+30/31)= 2.652[Amp]

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PART-B

I=E/R

I= 45/16

I= 2.8125[A]

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PART-C

2.652/2.8125= 0.9429

94.29%

100-94.29=5.706

About 5.706% smaller

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